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The Montjuïc Communications Tower, or Torre Telefónica,was built in the center of the Olympic Park in Barcelona,Spain, for the 1992 Olympic Games. The tower, built bySpanish architect Santiago Calatrava, was designed tocarry coverage of the Olympic Games to broadcast sta-tions around the world. The structure was designed torepresent an athlete holding up an Olympic torch.

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© Jones & Bartlett Learning, LLC. NOT FOR SALE OR DISTRIBUTION.

© Jones & Bartlett Learning, LLCNOT FOR SALE OR DISTRIBUTION




















1Linear Equations,Vectors, and Matrices

P A R T

3

1 Linear Equations and Vectors

2 Matrices and Linear Transformations

3 Determinants and Eigenvectors

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Informally referred to as “the Gherkin,” 30 St. Mary Axe inLondon, England, is located in London’s financial district.The building employs energy-saving methods, such asmaximizing the use of natural light and ventilation, whichallow it to use half the power a similar structure wouldtypically consume.

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5

Mathematics is, of course, a discipline in its own right. It is, however, more

than that—it is a tool used in many other fields. Linear algebra is a branch

of mathematics that plays a central role in modern mathematics, and also

is of importance to engineers and physical, social, and behavioral scientists. In this course

the reader will learn mathematics, will learn to think mathematically, and will be instructed

in the art of applying mathematics. The course is a blend of theory, numerical techniques,

and interesting applications.

When mathematics is used to solve a problem it often becomes necessary to find

a solution to a so-called system of linear equations. Historically, linear algebra developed

from studying methods for solving such equations. This chapter introduces methods for

solving systems of linear equations and looks at some of the properties of the solutions.

It is important to know not only what the solutions to a given system of equations are

but why they are the solutions. If the system describes some real-life situation, then an

understanding of the behavior of the solutions can lead to a better understanding of the

circumstances. The solutions form subsets of spaces called vector spaces. We develop

the basic algebraic structure of vector spaces. We shall discuss two applications of sys-

tems of linear equations. We shall determine currents through electrical networks and

analyze traffic flows through road networks.

Matrices and Systems of Linear EquationsAn equation in the variables x and y that can be written in the form wherea, b, and c are real constants (a and b not both zero), is called a linear equation. The graphof such an equation is a straight line in the xy-plane. Consider the system of two linearequations,

2x 2 y 5 4

2x 1 y 5 5

ax 1 by 5 c,

1.1

1Linear Equations and Vectors

C H A P T E R

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6 CHAPTER 1 Linear Equations and Vectors

Our aim in this chapter is to analyze larger systems of linear equations. A linear equa-tion in n variables is one that can be written in the form

,

where the coefficients and b are constants. The following is an example ofa system of three linear equations.

It can be seen on substitution that is a solution to this system.(We arrive at this solution in Example 1 of this section.)

A linear equation in three variables corresponds to a plane in three-dimensional space.Solutions to a system of three such equations will be points that lie on all three planes. Asfor systems of two equations, there can be a unique solution, no solution, or many solu-tions. We illustrate some of the various possibilities in Figure 1.4.

As the number of variables increases, a geometrical interpretation of such a system ofequations becomes increasingly complex. Each equation will represent a space embeddedin a larger space. Solutions will be points that lie on all the embedded spaces. While a gen-eral geometrical way of thinking about a problem is often useful, we rely on algebraic meth-ods for arriving at and interpreting the solution. We introduce a method for solving systems

a1x1 1 a2x2 1 a3x3 1 c1 anxn 5 b

x1 5 21, x2 5 1, x3 5 2

1x1 2 1x2 2 2x3 5 26

2x1 1 3x2 1 2x3 5 23

1x1 1 1x2 1 1x3 5 22

a1, a2, c, an

x1, x2, x3, c, xn

A pair of values of x and y that satisfies both equations is called a solution. It can be seenby substitution that is a solution to this system. A solution to such a systemwill be a point at which the graphs of the two equations intersect. The following examples,Figures 1.1, 1.2, and 1.3, illustrate that three possibilities can arise for such systems ofequations. There can be a unique solution, no solution, or many solutions. We use thepoint/slope form where m is the slope and b is the y-intercept, to graph theselines.

y 5 mx 1 b,

x 5 3, y 5 2

x + y = 5

x + y = 5

2x – y = 4

2x – y = 4

(3, 2)

Unique solution

Write as y = –x + 5 and y = 2x – 4.

The lines have slopes –1 and 2, and

y-intercepts 5 and –4. They intersect

at a point, the solution. There is a unique

solution, x = 3, y = 2.

x

y

–4x + 2y = 2–2x + y = 3

Write as y = 2x + 3 and y = 2x + 1.

The lines have slope 2, and y-intercepts

3 and 1. They are parallel. There is no

point of intersection. No solution.

–2x + y = 3–4x + 2y = 2

No solution

y

x

4x – 2y = 6

4x – 2y = 6

6x – 3y = 9

6x – 3y = 9

Each equation can be written as

y = 2x – 3. The graph of each

equation is a line with slope 2

and y-intercept –3. Any point

on the line is a solution.

Many solutions.

Many solutions

y

x

Figure 1.1 Figure 1.2 Figure 1.3

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1.1 Matrices and Systems of Linear Equations 7

of linear equations called Gauss-Jordan elimination.1 This method involves systemati-cally eliminating variables from equations. In this section, we shall see how this methodapplies to systems of equations that have a unique solution. In the following section, weshall extend the method to more general systems of linear equations.

We shall use rectangular arrays of numbers called matrices to describe systems of lin-ear equations. At this time we introduce the necessary terminology.

1Carl Friedrich Gauss (1777–1855) was one of the greatest mathematical scientists ever. Among his discoveries was a wayto calculate the orbits of asteroids. He taught for forty-seven years at the University of Göttingen, Germany. He made con-tributions to many areas of mathematics, including number theory, probability, and statistics. Gauss has been described as“not really a physicist in the sense of searching for new phenomena, but rather a mathematician who attempted to formulatein exact mathematical terms the experimental results of others.” Gauss had a turbulent personal life, suffering financial andpolitical problems because of revolutions in Germany.

Wilhelm Jordan (1842–1899) taught geodesy at the Technical College of Karlsruhe, Germany. His most important workwas a handbook on geodesy that contained his research on systems of equations. Jordan was recognized as being a masterteacher and an excellent writer.

Figure 1.4

A

C

B

P

Unique solution

No solution

AP

C

B

A

A

B

B

C C

Three planes A, B, and C intersect at a single point P.P corresponds to a unique solution.

Planes A, B, and C have no points in common.There is no solution.

P

Q

Many solutions

A

A B CB

C

Three planes A, B, and C intersect in a line PQ. Any point on the lineis a solution.

Three equations represent the sameplane. Any point on the plane isa solution.

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Matrices are usually denoted by capital letters. Examples of matrices in standard nota-tion are

, ,

Rows and Columns Matrices consist of rows and columns. Rows are labeled from the topof the matrix, columns from the left. The following matrix has two rows and three columns.

The rows are:

row 1 row 2

The columns are:

, ,

column 1 column 2 column 3

Submatrix A submatrix of a given matrix is an array obtained by deleting certain rowsand columns of the matrix. For example, consider the following matrix A. The matrices P,Q, and R are submatrices of A.

matrix A submatrices of A

Size and Type The size of a matrix is described by specifying the number of rows andcolumns in the matrix. For example, a matrix having two rows and three columns is saidto be a matrix; the first number indicates the number of rows, and the second indicatesthe number of columns. When the number of rows is equal to the number of columns, thematrix is said to be a square matrix. A matrix consisting of one row is called a row matrix.A matrix consisting of one column is a column matrix. The following matrices are of thestated sizes and types.

matrix matrix matrix matrixa square matrix a row matrix a column matrix

Location The location of an element in a matrix is described by giving the row and col-umn in which the element lies. For example, consider the following matrix.

The element 7 is in row 2, column 1. We say that it is in location (2, 1).

3 3 1

c2 3 24

7 5 21d

2 3 3 3 3 3 1 3 4

c 1 0 3

22 4 5d £

2 5 7

29 0 1

23 5 8

§ 34 23 8 5 4 £8

3

2

§

2 3 3

R 5 c1 4

5 22dQ 5 £

7

3

1

§P 5 £1 7

2 3

5 1

§A 5 £1 7 4

2 3 0

5 1 22

§

c24

21dc3

5dc2

7d

37 5 21 432 3 24 4,

c2 3 24

7 5 21d

B 5 £7 1

0 5

28 3

§ C 5 £3 5 6

0 22 5

8 9 12

§A 5 c2 3 24

7 5 21d

DEFINITION A matrix is a rectangular array of numbers. The numbers in the array are called the elements of thematrix.

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The element in location (1, 3) is Note that the convention is to give the row inwhich the element lies, followed by the column.

Identity Matrices An identity matrix is a square matrix with 1s in the diagonal locations(1, 1), (2, 2), (3, 3), etc., and zeros elsewhere. We write for the identity matrix.The following matrices are identity matrices.

,

We are now ready to continue the discussion of systems of linear equations. We use matri-ces to describe systems of linear equations. There are two important matrices associatedwith every system of linear equations. The coefficients of the variables form a matrix calledthe matrix of coefficients of the system. The coefficients, together with the constant terms,form a matrix called the augmented matrix of the system. For example, the matrix of coef-ficients and the augmented matrix of the following system of linear equations are as shown:

matrix of coefficients augmented matrix

Observe that the matrix of coefficients is a submatrix of the augmented matrix. The aug-mented matrix completely describes the system.

Transformations called elementary transformations can be used to change a systemof linear equations into another system of linear equations that has the same solution. Thesetransformations are used to solve systems of linear equations by eliminating variables. Inpractice it is simpler to work in terms of matrices using analogous transformations calledelementary row operations. It is not necessary to write down the variables ateach stage. Systems of linear equations are in fact described and manipulated on comput-ers in terms of such matrices. These transformations are as follows.

Elementary Transformations Elementary Row Operations1. Interchange two equations. 1. Interchange two rows of a matrix.2. Multiply both sides of an equation 2. Multiply the elements of a row by

by a nonzero constant. a nonzero constant.3. Add a multiple of one equation to 3. Add a multiple of the elements of

another equation. one row to the corresponding elements of another row.

Systems of equations that are related through elementary transformations are called equiv-alent systems. Matrices that are related through elementary row operations are called rowequivalent matrices. The symbol is used to indicate equivalence in both cases.

Elementary transformations preserve solutions since the order of the equations doesnot affect the solution, multiplying an equation throughout by a nonzero constant does notchange the truth of the equality, and adding equal quantities to both sides of an equalityresults in an equality.

The method of Gauss-Jordan elimination uses elementary transformations to eliminatevariables in a systematic manner, until we arrive at a system that gives the solution. Weillustrate Gauss-Jordan elimination using equations and the analogous matrix implemen-tation of the method side by side in the following example. The reader should note the wayin which the variables are eliminated in the equations in the left column. At the same time,

24.

<

n 3 nIn

x1, x2, x3,

2x1 2 2x2 2 2x3 5 26

2x1 1 3x2 1 2x3 5 23

2x1 1 2x2 1 2x3 5 22

I3 5 £1 0 0

0 1 0

0 0 1

§I2 5 c1 0

0 1d

£1 1 1 2

2 3 1 3

1 21 22 26

§£1 1 1

2 3 1

1 21 22

§

9781284120097_CH01.qxd 9/15/17 9:31 AM Page 9






















Equation MethodInitial System

Eliminate from 2nd and 3rd equations.

Eliminate from 1st and 3rd equations.

Make coefficient of in 3rd equation 1 (i.e., solve for ).

Eliminate from 1st and 2nd equations.

The solution is

2x1 1 2x2 1 2x3 5 22

2x1 1 3x2 1 2x3 5 23

x1 5 21, x2 5 1, x3 5 2.

x3 5 x3 5 x3 5 22

x2 5 x2 5 x2 5 21

x1 5 x1 5 x1 5 21

x1

2x1 2 2x2 2 2x3 5 26

x3

x3 5 x3 5 2x3 5 22

x2 2 x2 2 2x3 5 21

x1 1 x1 1 2x3 5 23

x3 x3

x3 5 x3 25x3 5 210

x2 2 x2 2 3x3 5 021

x1 1 x1 1 2x3 5 023

x2

x122x2 2 3x3 5 28

x2 2 2x3 5 21

x1 1 x2 1 2x3 5 22


observe how this is accomplished in terms of matrices in the right column by creating zerosin certain locations. We shall henceforth be using the matrix approach.

Solve the system of linear equations

SOLUTION

1x1 1 1x2 1 1x3 5 22

EXAMPLE 1

2x1 1 3x2 1 2x3 5 23

1x1 2 1x2 2 2x3 5 26

Eq3 1 121 2Eq1

Eq2 1 122 2Eq1

<

Eq3 1 12 2Eq2

Eq1 1 121 2Eq2

<

Eq2 1 Eq3

Eq1 1 122 2Eq3

<

121/5 2Eq3

<

Analogous Matrix MethodAugmented Matrix

Create zeros in column 1.

Create appropriate zeros in column 2.

Make the (3, 3) element 1 (called “normalizing” the element).

Create zeros in column 3.

Matrix corresponds to the system.

The solution is x1 5 21, x2 5 1, x3 5 2.

x3 5 x3 5 x3 5 22

x2 5 x2 5 x2 5 21

£1 1 1 2

2 3 1 3

1 21 22 26

§

x1 5 x1 5 x1 5 21

£1 0 0 21

0 1 0 1

0 0 1 2

§

£1 0 2 3

0 1 21 21

0 0 1 2

§

£1 0 2 3

0 1 21 21

0 0 25 210

§

£1 1 1 2

0 1 21 21

0 22 23 28

§R3 1 121 2R1

R2 1 122 2R1<

R3 1 12 2R2

R1 1 121 2R2

<

R2 1 R3

R1 1 122 2R3

<

121/5 2R3

<

9781284120097_CH01.qxd 9/15/17 9:32 AM Page 10























Geometrically, each of the three original equations in this example represents a plane inthree-dimensional space. The fact that there is a unique solution means that these threeplanes intersect at a single point. The solution gives the coordinates of this pointwhere the three planes intersect. We now give another example to reinforce the method.

Solve the following system of linear equations.

SOLUTION

Start with the augmented matrix and use the first row to create zeros in the first column.(This corresponds to using the first equation to eliminate from the second and thirdequations.)

Next multiply row 2 by to make the (2, 2) element 1. (This corresponds to making thecoefficient of in the second equation 1.)

Create zeros in the second column as follows. (This corresponds to using the secondequation to eliminate from the first and third equations.)

Multiply row 3 by (This corresponds to making the coefficient of in the third equa-tion 1.)

Finally, create zeros in the third column. (This corresponds to using the third equationto eliminate from the first and second equations.)

121, 1, 2 2

<R1 1 122 2R3

R2 1 R3

£1 0 0 2

0 1 0 1

0 0 1 3

§

x3

£1 0 2 8

0 1 21 22

0 0 1 3

§<1 1

2 2R3

x312.

<R1 1 12 2R2

R3 1 121 2R2

£1 0 2 8

0 1 21 22

0 0 2 6

§

x2

<1 1

3 2R2£

1 22 4 12

0 1 21 22

0 1 1 4

§

x2

13

<R2 1 122 2R1

R3 1 R1

£1 22 4 12

0 3 23 26

0 1 1 4

§£1 22 4 12

2 21 5 18

21 3 23 28

§

x1

28

18

12

22x1 1 3x2 2 3x3 5

22x1 2 2x2 1 5x3 5

22x1 2 2x2 1 4x3 5

EXAMPLE 2

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This matrix corresponds to the system

The solution is

This Gauss-Jordan method of solving a system of linear equations using matrices involvescreating 1s and 0s in certain locations of matrices. These numbers are created in a system-atic manner, column by column. The following example illustrates that it may be necessaryto interchange two rows at some stage in order to proceed in the preceding manner.

Solve the system

SOLUTION

We start with the augmented matrix and proceed as follows. (Note the use of zeroin the augmented matrix as the coefficient of the missing variable in the thirdequation.)

At this stage we need a nonzero element in the location (2, 2) in order to continue. Toachieve this we interchange the second row with the third row (a later row) and thenproceed.

The solution is x1 5 2, x2 5 3, x3 5 21.

<R1 1 122 2R2

£1 0 1 1

0 1 22 5

0 0 1 21

§<

R1 1 121 2R3

R2 1 12 2R3

£1 0 0 2

0 1 0 3

0 0 1 21

§

<1 1

3 2R2£

1 2 23 11

0 1 22 5

0 0 1 21

§<R2 4 R3

£1 2 23 11

0 3 26 15

0 0 1 21

§

<R2 1 123 2R1

R3 1 12 2R1

£1 2 23 11

0 0 1 21

0 3 26 15

§

£1 2 23 11

3 6 28 32

22 21 0 27

§<1 1

4 2R1£

4 8 212 44

3 6 28 32

22 21 0 27

§

x3

44

32

2722x1 2 8x2 2 13x3 5

23x1 1 6x2 2 18x3 5

24x1 1 8x2 2 12x3 5

x1 5 2, x2 5 1, x3 5 3.

x3 5 x3 5 x3 5 3

x2 5 x2 5 x2 5 1

x1 5 x1 5 x1 5 2

EXAMPLE 3

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SummaryWe now summarize the method of Gauss-Jordan elimination for solving a system of n lin-ear equations in n variables that has a unique solution. The augmented matrix is made upof a matrix of coefficients A and a column matrix of constant terms B. Let us writefor this matrix. Use row operations to gradually transform this matrix, column by column,into a matrix where is the identity matrix.

This final matrix is called the reduced echelon form of the original augmentedmatrix. The matrix of coefficients of the final system of equations is and X is the columnmatrix of constant terms. This implies that the elements of X are the unique solution. Observethat as is being transformed to A is being changed to Thus:

If A is the matrix of coefficients of a system of n equations in n variables that has aunique solution, then it is row equivalent to

If cannot be transformed in this manner into a matrix of the form the sys-tem of equations does not have a unique solution. More will be said about such systems inthe next section.

Many SystemsCertain applications involve solving a number of systems of linear equations, all havingthe same square matrix of coefficients A. Let the systems be

The constant terms might for example be test data, and one wants to knowthe solutions that would lead to these results. The situation often dictates that the solutionsbe unique. One could of course go through the method of Gauss-Jordan elimination foreach system, solving each system independently. This procedure would lead to the reducedechelon forms

and the solutions would be However, the same reduction of A to wouldbe repeated for each system; this involves a great deal of unnecessary duplication. The sys-tems can be represented by one large augmented matrix and the Gauss-Jordan method can be applied to this one matrix. We would get

leading to the solutions

Solve the following three systems of linear equations, all of which havethe same matrix of coefficients.

for in turn.

x1 2 2x2 1 3x3 5 b1

2x1 2 2x2 1 4x3 5 b2

2x1 1 2x2 2 4x3 5 b3

£b1

b2

b3

§ 5 £8

11

211

§ , £0

1

2

§ , £3

3

24

§

In

EXAMPLE 4

X1, X2, c, Xk.

3A : B1 B2cBk 4 < c< 3In : X1 X2

cXk 4

X1, X2, c, Xk.

3A : B1 B2cBk 4,

3In : X1 4, 3In : X2 4, c, 3In : Xk 4

B1, B2, c, Bk,

3A : B1 4, 3A : B2 4, c, 3A : Bk 4

3In : X 4,3A : B 4In.

In.3In : X 4,3A : B 4In

3In : X 43A : B 4 < c< 3In : X 4

n 3 nIn3In : X 4,3A : B 4

9781284120097_CH01.qxd 9/15/17 9:32 AM Page 13























SOLUTION

Construct the large augmented matrix that describes all three systems and determine thereduced echelon form as follows.

The solutions to the three systems of equations are given by the last three columns of thereduced echelon form. They are

In this section we have limited our discussion to systems of n linear equations in n vari-ables that have a unique solution. In the following section, we shall extend the method ofGauss-Jordan elimination to accommodate other systems that have a unique solution, andalso to include systems that have many solutions or no solutions.

x1 5 22, x2 5 21, x3 5 2

x1 5 20, x2 5 23, x3 5 1

x1 5 21, x2 5 21, x3 5 2

£1 0 0 1 0 22

0 1 0 21 3 1

0 0 1 2 1 2

§<

R1 1 121 2R3

R2 1 2R3

£1 0 1 3 1 0

0 1 22 25 1 23

0 0 1 2 1 2

§<

R1 1 R2

R3 1 121 2R2

£1 21 3 8 0 3

0 1 22 25 1 23

0 1 21 23 2 21

§<

R2 1 122 2R1

R3 1 R1

£1 21 3 8 0 3

2 21 4 11 1 3

21 2 24 211 2 24

§

EXERCISE SET 1.1*

Matrices1. Give the sizes of the following matrices.

(a) (b)

(c) (d)

(e)

(f)

2. Give the (1, 1), (2, 2), (3, 3), (1, 5), (2, 4), (3, 2) elementsof the following matrix.

3. Give the (2, 3), (3, 2), (4, 1), (1, 3), (4, 4), (3, 1) elementsof the following matrix.

4. Write down the identity matrix I4.

≥1 2 7 0

21 2 4 5

3 5 0 21

6 9 0 2

¥

£1 2 3 0 21

22 4 25 3 6

5 8 9 2 3

§

32 23 4 7 4

£1 2 9 28 7

4 2 5 7 2

4 26 4 0 0

§

£0 9

26 4

23 2

§£1 2 3

0 1 2

4 5 3

§

£27

4

3

§c1 2 3 0

1 2 4 5d

* Answers to exercises marked in red are provided in the back of the book.

9781284120097_CH01.qxd 9/15/17 9:32 AM Page 14























Matrices and Systems of Equations5. Determine the matrix of coefficients and augmented matrix

of each of the following systems of equations.

(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

6. Interpret the following matrices as augmented matrices ofsystems of equations. Write down each system of equations.

(a) (b)

(c) (d)

(e) (f)

(g) (h)

Elementary Row Operations7. In the following exercises you are given a matrix followed

by an elementary row operation. Determine each resultingmatrix.

(a)

(b)

(c)

(d)

(e)

(f)

8. Interpret each of the following row operations as a stage inarriving at the reduced echelon form of a matrix. Why havethe indicated operations been selected? What particular aimsdo they accomplish in terms of the systems of linear equa-tions that are described by the matrices?

(a)

(b)

(c)

(d)

£1 0 1 6

0 1 2 23

0 0 7 211

§£1 2 5 0

0 1 2 23

0 23 1 22

§<

R1 1 122 2R2

R3 1 13 2R2

£1 3 24 5

0 1 3 28

0 0 22 6

§£1 3 24 5

0 0 22 6

0 1 3 28

§ <R2 4 R3

£1 2 24 7

0 3 9 26

0 4 7 28

§ <1 1

3 2R2£

1 2 24 7

0 1 3 22

0 4 7 28

§

£1 24 3 5

0 27 13 15

0 16 215 214

§<

R2 1 12 2R1

R3 1 124 2R1

£1 24 3 5

22 1 7 5

4 0 23 6

§

<121

2 2R3£

1 0 2 7

0 1 5 23

0 0 22 8

§

<R1 1 124 2R3

R2 1 13 2R3

£1 0 4 23

0 1 23 2

0 0 1 5

§

<R1 1 122 2R2

R3 1 14 2R2

£1 2 3 24

0 1 2 1

0 24 3 25

§

<R2 1 R1

R3 1 122 2R1

£1 2 3 21

21 1 7 1

2 24 5 23

§

<R1 4 R2

£0 28 4 3

2 7 5 1

3 25 8 9

§

<1 1

2 2R1£

2 6 24 0

1 2 23 6

8 3 2 5

§

£8 7 5 21

4 6 2 4

9 3 7 6

§

£1 2 21 6

0 1 4 5

0 0 1 22

§£1 0 0 3

0 1 0 8

0 0 1 4

§

£0 22 4

5 7 23

6 0 8

§£2 23 6 4

7 25 22 3

0 2 4 0

§

c1 9 23

5 0 2d

c7 9 8

6 4 23dc1 2 3

4 5 6d

4x1 1 6x2 2 9x3 5 7

22x1 1 3x2 2 5x3 5 23

2x2 1 3x3 2 5x4 5 20

21x1 1 6x2 2 8x3 2 7x4 5 115

24x1 1 2x2 2 9x3 1 4x4 5 21

x3 2 x3 2 x3 5 28

x2 2 x2 2 x2 5 112

x1 2 x1 2 x1 5 23

2x1 1 8x2 1 8x2 5 21

2x1 2 1x1 2 4x3 5 111

2x1 1 3x2 2 9x3 5 24

5x1 2 5x1 2 5x3 5 7

4x2 1 4x2 1 3x3 5 0

5x1 1 2x2 2 4x3 5 8

2x1 1 2x2 5 23

2x1 2 8x2 5 24

5x1 1 4x2 5 29

22x1 1 3x2 1 3x2 5 26

22x1 2 2x2 1 4x3 5 28

5x1 1 3x2 1 6x3 5 4

5x1 1 2x2 2 4x3 5 8

2x1 2 5x2 5 23

2x1 1 3x2 5 27

9781284120097_CH01.qxd 9/15/17 9:33 AM Page 15























9. Interpret each of the following row operations as a stage inarriving at the reduced echelon form of a matrix. Why havethese operations been selected?

(a)

(b)

(c)

(d)

Solving Systems of Linear Equations10. The following systems of equations all have unique solu-

tions. Solve these systems using the method of Gauss-Jordanelimination with matrices.

(a)

(b)

(c)

(d)

(e)

(f)

11. The following systems of equations all have unique solu-tions. Solve these systems using the method of Gauss-Jordanelimination with matrices.

(a)

(b)

(c)

(d)

(e)

12. The following systems of equations all have unique solu-tions. Solve these systems using the method of Gauss-Jordanelimination with matrices.

(a)

(b)

(c)

(d)

(e)

1x1 2 1x2 2 1x3 2 16x4 5 212

2x2 1 3x2 1 4x3 1 15x4 5 228

2x1 1 3x2 1 6x3 1 19x4 5 236

1x1 1 1x2 1 2x3 1 16x4 5 211

2x1 2 2x2 1 2x3 1 2x3 5 12

2x1 1 3x2 1 4x3 1 8x4 5 192x1 1 4x2 1 2x3 1 8x4 5 14

2x1 1 2x2 1 2x3 1 5x4 5 11

£1 0 0 22

0 1 0 5

0 0 1 23

§

£4 3 2 28

0 2 4 21

5 27 1 2

§

£1 0 0 2

0 1 0 5

0 0 1 2

§

2x1 1 3x2 1 x2 1 3x2 5 128

2x1 2 2x2 1 x3 1 2x4 5 2181x1 1 3x2 2 x3 2 4x4 5 212

3x1 1 6x2 2 x4 2 3x4 5 213

24x1 2 7x2 2 17x3 5 24

2x1 1 3x2 1 19x3 5 21

23x1 2 6x2 2 15x3 5 23

2x1 1 5x3 1 5x3 5 222

2x1 1 7x2 2 9x3 5 245

32x1 1 3x3 1 3x3 5 215

3x1 1 x2 2 2x3 5 20

2x1 1 x1 1 2x3 5 28

x1 2 x1 2 2x3 5 23

2x1 1 2x2 1 2x3 5 5

2x1 1 2x2 1 2x2 5 6

2x2 1 2x2 1 4x3 5 8

2x1 2 2x2 1 2x3 5 21

3x1 1 2x2 1 2x3 5 128

2x1 1 2x2 2 4x3 5 114

2x1 1 2x2 1 26x3 5 29

22x1 1 6x2 1 10x3 5 114

x1 2 2x2 2 22x3 5 21

2x1 2 2x2 2 2x2 5 24

2x1 1 5x2 1 8x3 5 136

2x1 1 2x2 1 3x3 5 114

4x1 1 2x2 1 3x3 5 114

3x1 1 2x2 1 2x3 5 110

2x1 1 2x2 2 2x3 5 22

3x1 1 x2 2 2x3 5 3

2x1 2 x2 1 2x3 5 2

2x1 2 x2 1 3x3 5 3

2x1 1 2x2 1 6x3 5 11

2x1 1 2x2 1 4x3 5 29

2x1 1 2x2 1 3x3 5 26

x3 2 2x2 2 2x3 5 25

x2 2 2x2 2 2x3 5 24

x1 1 2x1 1 3x3 5 23

3x1 1 2x2 5 3

2x1 1 2x2 5 4

2x1 2 3x2 5 211

1x1 2 2x2 5 128

<R1 1 12 2R3

R2 1 123 2R3

<121

2 2R3

<R1 4 R2

<R1 1 122 2R3

R2 1 R3

£0 2 4 21

4 3 2 28

5 27 1 2

§

£1 0 3 7

0 1 4 2

0 0 22 6

§

£1 0 22 4

0 1 3 24

0 0 1 23

§

£1 0 3 7

0 1 4 2

0 0 1 23

§

£1 0 2 6

0 1 21 3

0 0 1 2

§

9781284120097_CH01.qxd 9/15/17 9:34 AM Page 16






















DEFINITION

1.2 Gauss-Jordan Elimination 17

13. The following exercises involve many systems of linearequations with unique solutions that have the same matrixof coefficients. Solve the systems by applying the methodof Gauss-Jordan elimination to a large augmented matrixthat describes many systems.

(a)

for in turn.

(b)

for , in turn.

(c)

for in turn.

(d)

for in turn.£b1

b2

b3

§ 5 £21

1

21

§ , £6

24

18

§ , £0

22

24

§

3x1 1 5x2 5 b2

1x1 1 2x2 5 b1

3x1 1 7x2 2 x3 5 b3

2x1 2 2x2 1 x3 5 b2

x1 1 2x2 2 x3 5 b1

£4

3

9

§£b1

b2

b3

§ 5 £6

5

14

§ , £25

23

28

§ ,

2x1 2 3x2 1 6x3 5 b3

2x1 2 2x2 1 2x3 5 b2

2x1 2 2x2 1 3x3 5 b1

c12dcb1

b2d 5 c0

1d , c 5

13d

2x1 1 3x2 5 b2

2x1 1 2x2 5 b1

c37dcb1

b2d 5 c3

8d , c4

9d ,

Gauss-Jordan EliminationIn the previous section we used the method of Gauss-Jordan elimination to solve systemsof n equations in n variables that had a unique solution. We shall now discuss the methodin its more general setting, where the number of equations can differ from the number ofvariables and where there can be a unique solution, many solutions, or no solutions. Ourapproach again will be to start from the augmented matrix of the given system and to per-form a sequence of elementary row operations that will result in a simpler matrix (thereduced echelon form), which leads directly to the solution.

We now give the general definition of reduced echelon form. The reader will observe thatthe reduced echelon forms discussed in the previous section all conform to this definition.

1.2

A matrix is in reduced echelon form if:1. Any rows consisting entirely of zeros are grouped at the bottom of the matrix.2. The first nonzero element of each other row is 1. This element is called a leading 1.3. The leading 1 of each row after the first is positioned to the right of the leading 1 of the

previous row.4. All other elements in a column that contains a leading 1 are zero.

The following matrices are all in reduced echelon form.

≥1 0 5 0 0 8

0 1 7 0 0 9

0 0 0 1 0 5

0 0 0 0 1 4

¥ ≥1 2 0 3 0 4

0 0 1 2 0 7

0 0 0 0 1 6

0 0 0 0 0 0

¥

£1 0 8

0 1 2

0 0 0

§ £1 0 0 7

0 1 0 3

0 0 1 9

§ £1 4 0 0

0 0 1 0

0 0 0 1

§ £1 2 3 0

0 0 0 1

0 0 0 0

§

9781284120097_CH01.qxd 9/15/17 9:34 AM Page 17























The following matrices are not in reduced echelon form for the reasons stated.

Row of zeros First nonzero Leading 1 in Nonzero not at bottom element in row row 3 not to the element above

of matrix 2 is not 1 right of leading leading 1 in 1 in row 2 row 2

There are usually many sequences of row operations that can be used to transform agiven matrix to reduced echelon form—they all, however, lead to the same reduced eche-lon form. We say that the reduced echelon form of a matrix is unique. The method of Gauss-Jordan elimination is an important systematic way (called an algorithm) for arriving at thereduced echelon form. It can be programmed on a computer. We now summarize the method,then give examples of its implementation.

Gauss-Jordan Elimination1. Write down the augmented matrix of the system of linear equations.2. Derive the reduced echelon form of the augmented matrix using elementary row

operations. This is done by creating leading 1s, then zeros above and below eachleading 1, column by column, starting with the first column.

3. Write down the system of equations corresponding to the reduced echelon form.This system gives the solution.

We stress the importance of mastering this algorithm. Not only is getting the correctsolution important, the method of arriving at the solution is important. We shall, for exam-ple, be interested in the efficiency of this algorithm (the number of additions and multipli-cations used) and the comparison of it with other algorithms that can be used to solve systemsof linear equations.

Use the method of Gauss-Jordan elimination to find the reduced echelonform of the following matrix.

SOLUTION

Step 1 Interchange rows, if necessary, to bring a nonzero element to the top of the firstnonzero column. This nonzero element is called a pivot.

pivot

Step 2 Create a 1 in the pivot location by multiplying the pivot row by

<1 1

3 2R1£

1 1 21 3 4

0 0 2 22 2

4 4 22 11 12

§

1pivot.

<R1 4 R2

£3 3 23 9 12

0 0 2 22 2

4 4 22 11 12

§

£0 0 2 22 2

3 3 23 9 12

4 4 22 11 12

§

EXAMPLE 1

£1 2 0 4

0 0 0 0

0 0 1 3

§ £1 2 0 3 0

0 0 3 4 0

0 0 0 0 1

§ £1 0 0 2

0 0 1 4

0 1 0 3

§ ≥1 7 0 8

0 1 0 3

0 0 1 2

0 0 0 0

¥

9781284120097_CH01.qxd 9/15/17 9:35 AM Page 18























Step 3 Create zeros elsewhere in the pivot column by adding suitable multiples of thepivot row to all other rows of the matrix.

Step 4 Cover the pivot row and all rows above it. Repeat Steps 1 and 2 for the remain-ing submatrix. Repeat Step 3 for the whole matrix. Continue thus until the reduced ech-elon form is reached.

first nonzero column pivotof the submatrix

pivot

This matrix is the reduced echelon form of the given matrix.

We now illustrate how this method is used to solve various systems of equations. Thefollowing example illustrates how to solve a system of linear equations that has many solu-tions. The reduced echelon form is derived. It then becomes necessary to interpret thereduced echelon form, expressing the many solutions in a clear manner.

Solve, if possible, the system of equations

SOLUTION

Start with the augmented matrix and follow the Gauss-Jordan algorithm. Pivots and lead-ing ones are circled.

£3 23 3 9

2 21 4 7

3 25 21 7

§ <1 1

3 2R1£

1 21 1 3

2 21 4 7

3 25 21 7

§

3x1 2 5x2 2 3x3 5 7

2x1 2 3x2 1 4x3 5 7

3x1 2 3x2 1 3x3 5 9

EXAMPLE 2

<R1 1 122 2R3

R2 1 R3

£1 1 0 0 17

0 0 1 0 25

0 0 0 1 26

§

<1 1

2 2R2£

1 1 2 1 3 4

0 0 1 2 1 1

0 0 2 2 1 2 4

§<

R1 1 R2

R3 1 122 2R2

£1 1 0 2 5

0 0 1 21 1

0 0 0 1 26

§

£1 1 21 3 4

0 0 2 22 2

0 0 2 21 24

§5£1 1 21 3 4

0 0 2 22 2

0 0 2 21 24

§

£1 1 21 3 4

0 0 2 22 2

0 0 2 21 24

§<R3 1 124 2R1

9781284120097_CH01.qxd 9/15/17 9:35 AM Page 19























We have arrived at the reduced echelon form. The corresponding system of equations is

There are many values of and that satisfy these equations. This is a system ofequations that has many solutions. is called the leading variable of the first equa-tion and is the leading variable of the second equation. To express these many solu-tions, we write the leading variables in each equation in terms of the remaining variables,called free variables. We get

Let us assign the arbitrary value r to The general solution to the system is

As r ranges over the set of real numbers, we get many solutions. r is called a param-eter. We can get specific solutions by giving r different values. For example,

gives

gives

This example illustrates that the general solution can involve a number ofparameters. Solve the system of equations

SOLUTION

On applying the Gauss-Jordan algorithm, we get

<R1 1 123 2R2

R3 1 R2

£1 2 21 0 22

0 0 0 1 2

0 0 0 0 0

§

£1 2 21 3 4

2 4 22 7 10

21 22 1 24 26

§<

R2 1 122 2R1

R3 1 R1

£1 2 21 3 4

0 0 0 1 2

0 0 0 21 22

§

<R2 1 122 2R1

R3 1 123 2R1

£1 21 1 3

0 1 2 1

0 22 24 22

§<

R1 1 R2

R3 1 12 2R2

£1 0 3 4

0 1 2 1

0 0 0 0

§

2x1 2 2x2 1 4x3 2 4x4 5 26

2x1 1 4x2 2 2x3 1 7x4 5 10

x1 1 2x2 2 3x3 1 3x4 5 24

EXAMPLE 3

r 5 22 x1 5 10, x2 5 25, x3 5 22

r 5 21 x1 5 11, x2 5 21, x3 5 21

x1 5 23r 1 4, x2 5 22r 1 1, x3 5 r

x3.

x2 5 22x3 1 1

x1 5 23x3 1 4

x2 1 x2 1 2x3 5 1

x1 1 x1 1 3x3 5 4

x2

x1

x3x1, x2,

9781284120097_CH01.qxd 9/15/17 9:36 AM Page 20























We have arrived at the reduced echelon form. The corresponding system of equations is

Expressing the leading variables in terms of the remaining variables, we get

Let us assign the arbitrary values r to and s to The general solution is

Specific solutions can be obtained by giving r and s various values.

This example illustrates a system that has no solution. Let us try to solvethe system

SOLUTION

Starting with the augmented matrix, we get

The last row of this reduced echelon form gives the equation

This equation cannot be satisfied for any values of and Thus the system hasno solution. (This information was in fact available from the next-to-last matrix.)

Homogeneous Systems of Linear EquationsA system of linear equations is said to be homogeneous if all the constants are zero. As weproceed in the course, we shall find that homogeneous systems of linear equations havemany interesting properties and play a key role in our discussions.

The following system is a homogeneous system of linear equations.

Observe that x1 5 0, x2 5 0, x3 5 0, x4 5 0, is a solution to this system. It is apparent, byletting all the variables be zero, that this result can be extended as follows to any homoge-neous system of equations.

x1 1 x2 2 x4 5 0

2x1 2 2x2 1 2x3 2 3x4 5 0

x1 1 2x2 2 x3 1 2x4 5 0

x1 1 2x2 2 x3 2 x3 5 22

x1 5 22x2 1 x3 2 2, x4 5 2

x4 5 22

x1, x2, x3.

0x1 1 0x2 1 0x3 5 1

<R1 1 121 2R2

R3 1 121 2R2

£1 0 2 4

0 1 3 21

0 0 0 1

§<

R1 1 124 2R3

R2 1 R3

£1 0 2 0

0 1 3 0

0 0 0 1

§

£1 1 5 3

0 1 3 21

1 2 8 3

§ <R3 1 121 2R1

£1 1 5 3

0 1 3 21

0 1 3 0

§

x1 1 2x2 1 8x3 5 23

x2 1 3x2 1 3x3 5 21

x1 1 2x2 1 5x3 5 23

x4 5 2x3 5 s,x1 5 22r 1 s 2 2, x2 5 r,

x3.x2

EXAMPLE 4

9781284120097_CH01.qxd 9/15/17 9:36 AM Page 21























THEOREM 1.1

A homogeneous system of linear equations in n variables always has the solutionThis solution is called the trivial solution.

Let us see if the preceding homogeneous system has any other solutions. We solve thesystem using Gauss-Jordan elimination.

This reduced echelon form gives the system

Expressing the leading variables in terms of the remaining free variable, we get

x1 5 3x4, x2 5 22x4, x3 5 x4

Letting x4 � r, we see that the system has many solutions,

x1 5 3r, x2 5 22r, x3 5 r, x4 5 r

Observe that the solution x1 5 0, x2 5 0, x3 5 0, x4 5 0, is obtained by letting r 5 0.Note that in this example the homogeneous system had more variables (4) than equa-

tions (3). This led to free variables in the general solution, implying many solutions. Guidedby this thinking, we now consider a general homogeneous system of m linear equations inn variables with n > m—the number of variables is greater than the number of equations.The reduced echelon form will have at most m nonzero rows (Gauss-Jordan eliminationmay have created some zero rows). The corresponding system of equations has fewer equa-tions than variables. There will thus be free variables, leading to many solutions.

THEOREM 1.2

A homogeneous system of linear equations that has more variables than equations hasmany solutions. One of these solutions is the trivial solution.

More will be said about the existence and uniqueness of solutions to other classes oflinear equations as we continue in the course.

In the first two sections of this chapter, we introduced the method of Gauss-Jordan elim-ination for solving systems of linear equations. As we proceed in the course, we shall intro-duce other methods and compare the merits of the methods. There is another popular eliminationmethod for solving systems of linear equations, for example, called Gaussian elimination.

<R1 1 121 2R3

R2 1 R3

x3 2 x4 5 0

x2 1 2x4 5 0

x1 2 3x4 5 0

£1 0 0 23 0

0 1 0 2 0

0 0 1 21 0

§£1 0 1 24 0

0 1 21 3 0

0 0 1 21 0

§<

R1 1 121 2R2

R3 1 R2

£1 1 0 21 0

0 1 21 3 0

0 21 2 24 0

§<

R2 1 121 2R1

R3 1 R1

£1 1 0 21 0

1 2 21 2 0

21 22 2 23 0

§

x1 5 0, x2 5 0, c, xn 5 0.

9781284120097_CH01.qxd 9/15/17 9:36 AM Page 22























We introduce that method in Section 1 of the “Numerical Methods” chapter.2 The followingdiscussion reveals some of the numerical concerns when solving systems of equations.

Numerical Considerations In practice, systems of linear equations are solved on computers. Numbers are representedon computers in the form where are integers between 0and 9 and r is an integer (positive or negative). Such a number is called a floating-pointnumber. The quantity is called the mantissa, and r is the exponent. For exam-ple, the number 125.6 is written in floating-point form as An arithmetic operation of multiplication, division, addition, or subtraction on floating-point numbers iscalled a floating-point operation, or flop.

Computers can handle only a limited number of integers in the mantissa of a number.The mantissa is rounded to a certain number of places during each operation, and conse-quently errors called round-off errors occur in methods such as Gauss-Jordan elimina-tion. These errors are propagated and magnified during computation. The fewer flopsthat are performed during computation, the faster and more accurate the result will be.(Ways of minimizing these errors are discussed in the “Numerical Methods” chapter.) Tocompute the reduced echelon form of a system of n equations in n variables, the methodof Gauss-Jordan elimination requires multiplications and additions(Section 1 of the “Numerical Methods” chapter). The number of multiplications requiredto solve a system of, say, ten equations in ten variables is 550, and the numberof additions is 495. The total number of flops is the sum of these, namely 1045. Algorithmsare usually measured and compared using such data.

1n 5 10 212n

3 1 12n

2 12n

3 2 12n

0.1256 3 103.

60. a1 can 3 10r, a1, c, an

a1, c, an

EXERCISE SET 1.2

Reduced Echelon Form of a Matrix1. Determine whether the following matrices are in reduced

echelon form. If a matrix is not in reduced echelon form,give a reason.

(a) (b)

(c) (d)

(e) (f)

(g) (h)

(i)

2. Determine whether the following matrices are in reducedechelon form. If a matrix is not in reduced echelon form,give a reason.

(a) (b)

(c) (d)

(e) (f)

(g) (h)

(i)

3. Each of the following matrices is the reduced echelon formof the augmented matrix of a system of linear equations.Give the solution (if it exists) to each system of equations.

£1 5 23 0 7

0 0 0 1 4

0 0 0 0 0

§

£1 0 0 5 3

0 0 1 0 3

0 1 2 3 7

§ £0 0 1 0 4

0 0 0 1 5

0 1 0 0 3

§

c1 2 5 6

0 1 3 27d c1 4 0 5

0 0 2 9d

c1 0 2

0 1 3d c1 2 0 4

0 0 1 7d

≥1 0 2 0 3

0 0 0 0 0

0 1 2 0 7

0 0 0 1 3

¥ ≥1 0 4 0 0

0 1 2 0 0

0 0 0 1 0

0 0 0 0 1

¥

≥1 5 0 2 0

0 0 1 9 0

0 0 0 0 1

0 0 0 0 0

¥ ≥1 0 4 2 6

0 1 2 3 4

0 0 0 1 2

0 0 0 0 1

¥

£1 0 3 22

0 0 1 8

0 1 4 9

§ £1 2 0 0 4

0 0 1 0 6

0 0 0 1 5

§

£1 0 3 0

0 1 6 0

0 0 0 1

§

£1 0 0 4

0 1 0 5

0 0 1 9

§ £1 0 0 3 2

0 2 0 6 1

0 0 1 2 3

§

£1 5 0

0 0 1

0 0 0

§£1 0 0

0 1 0

0 0 1

§

2Gaussian elimination can in fact be used in place of Gauss-Jordan elimination as the standard method for this course if sodesired.

9781284120097_CH01.qxd 9/15/17 9:37 AM Page 23























(a) (b)

(c) (d)

(e)

(f)

4. Each of the following matrices is the reduced echelon formof the augmented matrix of a system of linear equations.Give the solution (if it exists) to each system of equations.

(a)

(b)

(c)

(d)

Solving Systems of Linear Equations5. Solve (if possible) each of the following systems of three

equations in three variables using the method of Gauss-Jordan elimination.

(a)

(b)

(c)

(d)

(e)

(f)

6. Solve (if possible) each of the following systems of threeequations in three variables using the method of Gauss-Jordan elimination.

(a)

(b)

(c)

(d)

(e)

(f)

7. Solve (if possible) each of the following systems of equa-tions using the method of Gauss-Jordan elimination.

(a)

(b)

(c)

(d)

(A homogeneous system)

22x1 2 4x2 1 3x3 2 2x4 5 0

22x1 1 2x2 1 4x4 1 4x4 5 0

2x1 2 2x2 1 2x3 1 4x4 5 5

2x1 1 2x2 1 2x4 1 2x4 5 4

2x1 1 2x2 2 2x3 2 2x4 5 0

£1 0 0 2

0 1 0 4

0 0 1 23

§ £1 0 23 4

0 1 2 8

0 0 0 0

§

£1 0 0 5 3

0 1 0 6 22

0 0 1 2 24

§

£1 3 0 6

0 0 1 22

0 0 0 0

§ £1 0 5 0

0 1 27 0

0 0 0 1

§

£1 3 0 0 2

0 0 1 0 4

0 0 0 1 5

§

23x1 1 7x2 1 15x3 5 237

22x1 2 6x2 2 14x3 5 238

23x1 2 2x2 1 4x3 5 224

21x1 1 1x2 2 3x3 5 210

1x2 1 1x2 1 13x3 5 14

2x1 1 4x2 1 16x3 5 14

1x1 1 2x2 1 18x3 5 17

x1 1 2x3 1 2x3 5 24

x1 1 2x2 1 5x3 5 13

x2 1 1x2 1 2x3 5 15

2x1 1 2x1 1 6x3 5 11

3x1 1 7x2 1 9x3 5 26

1x1 1 2x2 1 3x3 5 18

3x1 1 6x2 1 2x3 5 21

2x1 1 4x2 2 2x3 5 26

1x1 1 2x2 2 1x3 5 23

2x1 1 4x2 1 3x3 5 18

1x1 1 2x2 1 2x3 5 11

1x1 1 2x2 1 1x3 5 17

23x1 1 6x2 2 2x3 5 110

22x1 2 4x2 2 3x3 5 21

23x1 1 6x2 2 3x3 5 26

2x1 1 2x2 2 4x3 5 211

12x1 2 2x2 1 7x3 5 217

13x1 2 3x2 1 9x3 5 224

3x1 2 5x2 2 1x3 5 7

2x1 2 1x2 1 4x3 5 7

1x1 2 1x2 1 1x3 5 3

2x1 1 6x2 1 x2 5 3

1x1 1 2x2 2 x3 5 0

1x1 1 4x2 1 x3 5 2

2x1 1 1x2 2 3x3 5 11

12x1 1 3x2 1 1x3 5 18

11x1 1 1x2 1 1x3 5 17

2x1 1 3x2 1 5x3 5 20

2x1 1 4x2 1 9x3 5 33

2x1 1 2x2 1 4x3 5 15

2x1 1 6x2 1 27x3 5 3

2x1 1 8x2 1 11x3 5 7

2x1 1 4x2 1 23x3 5 1

£1 0 2 0 3 6

0 1 5 0 4 7

0 0 0 1 9 23

§

£1 22 0 3 0 4

0 0 1 2 0 9

0 0 0 0 1 8

§

£1 23 2 0 4

0 0 0 1 27

0 0 0 0 0

§

£1 0 2 4 1

0 1 23 5 26

0 0 0 0 0

§

9781284120097_CH01.qxd 9/15/17 9:39 AM Page 24























(e)


8. Solve (if possible) each the following systems of equationsusing the method of Gauss-Jordan elimination.

(a)

(b)

(c)

(d)

(e)


(f)

(g)

Understanding Systems of Linear Equations9. Construct examples of the following:

(a) A system of linear equations with more variablesthan equations, having no solution.

(b) A system of linear equations with more equationsthan variables, having a unique solution.

10. The reduced echelon forms of the matrices of systems oftwo equations in two variables, and the types of solutionsthey represent can be classified as follows. (• correspondsto possible nonzero elements.)

unique solution no solutions many solutions

Classify in a similar manner the reduced echelon forms ofthe matrices, and the types of solutions they represent, of

(a) systems of three equations in two variables,

(b) systems of three equations in three variables.

11. Consider the homogeneous system of linear equations

(a) Show that if is a solution, thenis also a solution, for any value of

the constant k.

(b) Show that if and areany two solutions, then isalso a solution.

12. Show that is a solution to the homogeneoussystem of linear equations

Prove that this is the only solution if and only if

13. Consider two systems of linear equations having augmentedmatrices and where the matrix of coeffi-cients of both systems is the same matrix A.

(a) Is it possible for to have a unique solutionand to have many solutions?

(b) Is it possible for to have a unique solutionand to have no solutions?

(c) Is it possible for to have many solutions andto have no solutions?3A : B2 4

3A : B1 43A : B2 4

3A : B1 43A : B2 4

3A : B1 43 3 3

3A : B1 4 3A : B2 4,

22x1 2 2x2 1 2x3 2 8x4 5 0

21x1 1 1x2 2 1x3 1 4x4 5 0

21x2 2 1x2 2 3x3 1 1x4 5 0

c1 • •

0 0 0dc1 • 0

0 0 1dc1 0 •

0 1 •d

12x1 1 3x2 1 x3 2 5x4 5 29

y 5 0x 5 0,

11x1 1 3x2 2 x3 2 6x4 5 27

11x1 1 1x2 1 x3 2 1x4 5 23

ad 2 bc 2 0.

cx 1 dy 5 0

ax 1 by 5 0

x 5 x0 1 x1, y 5 y0 1 y1,x 5 x1, y 5 y1,x 5 x0, y 5 y0,

x 5 kx0, y 5 ky0,x 5 x0, y 5 y0

cx 1 dy 5 0

ax 1 by 5 0

1x1 1 2x2 5 1

1x1 1 3x2 5 0

2x1 1 3x2 5 3

1x1 1 1x2 5 2

23x1 2 6x2 1 19x3 5 221

22x1 1 4x2 2 18x3 5 216

12x1 2 2x2 1 13x3 5 127

24x1 1 8x2 2 12x3 5 228

23x1 1 18x2 2 1x3 2 16x4 5 0

22x1 2 12x2 1 5x3 1 17x4 5 0

21x1 1 16x2 2 1x3 2 14x4 5 0

23x1 1 1x2 2 8x3 2 10x4 5 229

12x1 1 1x2 2 1x3 1 12x4 5 124

22x1 2 2x2 1 2x3 2 14x4 5 212

21x1 2 2x2 1 2x3 1 22x3 5 217

13x1 2 4x2 1 18x3 2 13x4 5 117

2x1 2 3x2 1 11x3 2 11x4 5 24

2x1 1 5x2 2 17x3 1 19x4 5 22

12x1 2 4x2 1 16x3 2 14x4 5 110

12x1 2 5x2 2 15x3 5 246

2x1 2 1x2 2 12x3 5 125

2x1 2 2x2 2 16x3 5 218

2x2 1 1x2 1 12x3 5 217

2x1 2 2x2 2 x3 2 2x2 5 21

9781284120097_CH01.qxd 9/15/17 9:40 AM Page 25























14. Solve the following systems of linear equations by apply-ing the method of Gauss-Jordan elimination to a large aug-mented matrix that represents two systems with the samematrix of coefficients.

(a)

for , , in turn.

(b)

for , , in turn.

15. Write down a matrix at random. Find its reduced ech-elon form. The reduced echelon form is probably the iden-tity matrix ! Explain this. [Hint: Think about the geometry.]

16. If a matrix is written down at random, what type ofreduced echelon form is it likely to have and why?

17. Computers can only carry a finite number of digits. Thiscauses errors called round-off errors to occur when num-bers are truncated. Because of this phenomenon, comput-ers can give incorrect results. Much research goes intodeveloping algorithms that minimize such round-off errors.(Readers who are interested in these algorithms should readSection 3 of the “Numerical Methods” chapter.) A computeris used to determine the reduced echelon form of an aug-mented matrix of a system of linear equations. Which of thefollowing is most likely to happen, and why?

(a) The computer gives a solution to the system, when infact a solution does not exist.

(b) The computer gives that a solution does not exist,when in fact a solution does exist.

3 3 4

I3

3 3 3

2x1 1 4x2 1 16x3 5 b3

1x1 1 2x2 1 18x3 5 b2

1x1 1 1x2 1 15x3 5 b1

£b1

b2

b3

§ 5 £8

5

13

§ £5

3

11

§

2x1 1 3x2 1 6x3 5 b3

1x1 1 1x2 1 2x3 5 b2

1x1 1 2x2 1 4x3 5 b1

£b1

b2

b3

§ 5 £2

5

10

§ £3

2

4

§

The Vector Space In the previous sections, we found that solutions to systems of linear equations can be pointsin a plane if the equations have two variables, points in three-space if they are equations inthree variables, points in four-space if they have four variables, and so on. The solutionsmake up subsets of the larger spaces. We now set out to investigate these spaces and theirsubsets and to develop mathematical structures on them. At the moment, we know how tosolve systems of linear equations, but we do not know anything about the properties of thesets of solutions. The structures that we develop (using operations of addition and a multi-plication called scalar multiplication) lead to information about solutions to systems of lin-ear equations. The spaces that we construct are called vector spaces. These spaces arise inmany areas of mathematics.

The locations of points in a plane are usually discussed in terms of a coordinate sys-tem. For example, in Figure 1.5, the location of each point in the plane can be describedusing a rectangular coordinate system. The point A is the point (5, 3).

1.3 Rn

Furthermore, point A is a certain distance in a certain direction from the origin (0, 0).The distance and direction are characterized by the length and direction of the line segmentfrom the origin, O, to A. We call such a directed line segment a position vector and denoteit O is called the initial point of and A is called the terminal point. There are thusOA

>

. OA>

OA

y

xO

3

5

A(5, 3)→

Figure 1.5

9781284120097_CH01.qxd 9/15/17 9:42 AM Page 26






















1.3 The Vector Space 27Rn

two ways of interpreting (5, 3); it defines the location of a point in a plane, and it also defines the position vector

Sketch the position vectors and

See Figure 1.6.OC>

5 123, 4 2 .EXAMPLE 1 OA

>

5 14, 1 2 , OB>

5 125, 22 2 ,

OA>

.

Denote the set of all ordered pairs of real numbers by (R stands for real number and2 stands for the number of entries; it is pronounced “r-two.”) Note the significance of“ordered” here; for example, the point (5, 3) is not the same point as (3, 5). The order isimportant.

These concepts can be extended to the set of ordered triples, denoted by Elementsof this set such as (2, 4, 3) can be interpreted in two ways: as the location of a point in three-space relative to an xyz coordinate system, or as a position vector. These interpretations areshown in Figure 1.7.

R3.

R2.

We now generalize these concepts. Let be a sequence of n real num-bers. The set of all such sequences is called n-space and is denoted is the first com-ponent of is the second component, and so on.

For example, is the set of sequences of four real numbers; (1, 2, 3, 4) andare in is the set of sequences of five real numbers;

is in this set.Many of the results and techniques that we develop for with will be useful

mathematical tools, without direct geometrical significance. The elements of can, how-ever, be interpreted as points in n-space or as position vectors in n-space. It is difficult to

RnRn n . 3

121, 3, 5.2, 0 2 R4. R5 121, 2, 0, 3, 9 2R41u1, u2, c, un 2 , u2

Rn. u1

1u1, u2, c, un 2

OA

y

xO

4

1

–2

4–3–5

OC→

→

OB→

Figure 1.6

z

yx

O2

Point (2, 4, 3)

4

3

z

yx

O2

Position vector(2, 4, 3)

4

3

Figure 1.7

9781284120097_CH01.qxd 9/15/17 9:42 AM Page 27























visualize an n-space for but the reader is encouraged to try to form an intuitive pic-ture. A geometrical “feel” for what is taking place often makes an algebraic discussion eas-ier to follow. The mathematics that we shall develop on will be motivated by the geometrythat we are familiar with on and

Addition and Scalar MultiplicationWe begin the development of an algebraic theory of vectors by introducing equality ofvectors.

Let and be two elements of We say that u andv are equal if Thus two elements of are equal if their corresponding components are equal.

When working with elements of , it is customary to refer to numbers as scalars. Wenow define addition and scalar multiplication.

Rn

u1 5 v1, c, un 5 vn. Rnu 5 1u1, c, un 2 v 5 1v1, c, vn 2 Rn.

R2 R3.Rn

n . 3,

DEFINITION Let and be elements of and let c be a scalar. Addition andscalar multiplication are performed as follows.

Addition:Scalar multiplication: cu 5 1cu1, c, cun 2

u 1 v 5 1u1 1 v1, c, un 1 vn 2u 5 1u1, c, un 2 v 5 1v1, c, vn 2 Rn

To add two elements of , we add corresponding components. To multiply an element ofby a scalar, we multiply every component by that scalar. Observe that the resulting ele-

ments are in We say that is closed under addition and under scalar multiplication. with operations of componentwise addition and scalar multiplication is an example

of a vector space, and its elements are called vectors. (We will use boldface for vectorsand plain text for scalars.)

We shall henceforth in this course interpret to be a vector space.

Let and be vectors in Findand 3u.

SOLUTION

We get

Note that the resulting vector under each operation is in the original vector space

We now give examples to illustrate geometrical interpretations of these vectors and theiroperations.

This example gives us a geometrical interpretation of vector addition.Consider the sum of the vectors (4, 1) and (2, 3). We get

In Figure 1.8 we interpret these vectors as position vectors. Construct the parallelogramhaving the vectors (4, 1) and (2, 3) as adjacent sides. The vector (6, 4), the sum, will bethe diagonal of the parallelogram.

3u 5 3 121, 4, 3, 7 2 5 123, 12, 9, 21 2u 1 v 5 121, 4, 3, 7 2 1 122, 23, 1, 0 2 5 123, 1, 4, 7 2

Rn

14, 1 2 1 12, 3 2 5 16, 4 2EXAMPLE 3

R4.

u 1 vR4.v 5 122, 23, 1, 0 2u 5 121, 4, 3, 7 2

RnRnRn.

RnRn

EXAMPLE 2

9781284120097_CH01.qxd 9/15/17 9:43 AM Page 28























Such vectors are used in the physical sciences to describe forces. In this example,(4, 1) and (2, 3) might be forces, acting on a body at the origin, O. The vectors wouldgive the directions of the forces and their lengths (using the Pythagorean Theorem) would be the magnitudes of the forces, and The vectorsum (6, 4) is the resultant force, a single force that would be equivalent to the two forces.

The magnitude of the resultant force would be "62 1 42 5 7.21.

"22 1 32 5 3.61."42 1 12 5 4.12

In general, if u and v are vectors in the same vector space, then is the diagonalof the parallelogram defined by u and v. See Figure 1.9. This way of visualizing vectoraddition is useful in all vector spaces.

This example gives us a geometrical interpretation of scalar multiplica-tion. Consider the scalar multiple of the vector (3, 2) by 2. We get

Observe in Figure 1.10 that (6, 4) is a vector in the same direction as (3, 2), and 2 timesit in length.

2 13, 2 2 5 16, 4 2

EXAMPLE 4

u 1 v

In general, if u is a vector in any vector space, and c a nonzero scalar, the direction ofcu will be the same as the direction of u if and the opposite direction to u ifThe length of cu is times the length of u. See Figure 1.11.

Special Vectors The vector having n zero components, is called the zero vector of and isdenoted 0. For example, (0, 0, 0) is the zero vector of We shall find that zero vectorsplay a central role in the development of vector spaces.

R3.10, 0, c, 0 2 , Rn

0 c 0 c . 0, c , 0.

O

(2, 3)

(4, 1)

(6, 4)

y

x

v

u

(u + v)

O

Figure 1.8 Figure 1.9

O

(3, 2)

(6, 4)

y

x

Figure 1.10 Figure 1.11

O

cu

cu

cu

cu

u u u u

OO O

c > 1 c < –10 < c < 1 –1 < c < 0

9781284120097_CH01.qxd 9/15/17 9:43 AM Page 29























The vector is written and is called the negative of u. It is a vector havingthe same magnitude as u, but lies in the opposite direction to u. For example,is the negative of

Subtraction Subtraction is performed on elements of by subtracting correspondingcomponents. For example, in

Observe that this is equivalent to

Thus subtraction is not a new operation on —it is the sum of the first vector and the neg-ative of the second. There are only two independent operations on the vector spacenamely addition and scalar multiplication.

We now summarize some of the properties of vector addition and scalar multiplication.

THEOREM 1.3

Properties of Vector Addition and Scalar MultiplicationLet u, v, and w be vectors in and let c and d be scalars.

(a) Commutative property(b) Associative property(c) Property of the zero vector (d) Property of the negative vector(e)(f) Distributive properties(g)(h) Scalar multiplication by 1

Proof These results are proved by writing the vectors in terms of components and usingthe definitions of vector addition and scalar multiplication, and the properties of realnumbers. We give the proofs of (a) and (e). We ask you to give further proofs in the exer-cises that follow.

Let and Then

Rn,

5 cu 1 cv5 c 1u1, c, un 2 1 c 1v1, c, vn 25 1cu1, c, cun 2 1 1cv1, c, cvn 25 1cu1 1 cv1, c, cun 1 cvn 25 1c 1u1 1 v1 2 , c, c 1un 1 vn 2 25 c 1u1 1 v1, c, un 1 vn 2

c 1u 1 v 2 5 c 1 1u1, c, un 2 1 1v1, c, vn 2 2c 1u 1 v 2 5 cu 1 cv:

5 v 1 u5 1v1, c, vn 2 1 1u1, c, un 25 1v1 1 u1, c, vn 1 un 25 1u1 1 v1, c, un 1 vn 2

u 1 v 5 1u1, c, un 2 1 1v1, c, vn 2u 5 1u1, c, un 2 v 5 1v1, c, vn 2 .

u 1 v 5 v 1 u:

1u 5 uc 1du 2 5 1cd 2u1c 1 d 2u 5 cu 1 duc 1u 1 v 2 5 cu 1 cvu 1 12u 2 5 0u 1 0 5 0 1 u 5 uu 1 1v 1 w 2 5 1u 1 v 2 1 wu 1 v 5 v 1 u

Rn

Rn

15, 3, 26 2 1 121 2 12, 1, 3 2 5 13, 2, 29 2

15, 3, 26 2 2 12, 1, 3 2 5 13, 2, 29 2R3,

Rn

12, 23, 1 2 . 122, 3, 21 22u121 2u

¶

9781284120097_CH01.qxd 9/15/17 9:44 AM Page 30























Some of the preceding properties can be illustrated geometrically. The commutative prop-erty of vector addition is illustrated in Figure 1.12. Note that we get the same diagonal tothe parallelogram regardless of whether we add the vectors in the order or in theorder .v 1 u

u 1 v

Linear Combinations of VectorsOne implication of the preceding properties is that we can write certain algebraic expres-sions, such as , without parentheses. We call a linear com-bination of the vectors u, v, and w.

Let Determine the lin-ear combination

SOLUTION

We get

We say that the vector is a linear combination

of the three vectors and (4, 0, 2).

Let v1, v2, . . . , vm be vectors in . The vector v in is a linear combination of v1, v2,. . . , vm if there exist scalars c1, . . . , cm such that v can be written

v = c1v1 + + cmvm

In general, determining whether a given vector is a linear combination of other vectors ornot involves examining a system of linear equations.

c

Rn Rn

au 1 bv 1 cw

12, 5, 23 2 , 124, 1 , 9 2 ,120, 7, 231 2 5 2 12, 5, 23 2 2 3 124, 1 , 9 2 1 14, 0, 2 2

120, 7, 231 25 120, 7, 231 25 14 1 12 1 4, 10 2 3 1 0, 26 2 27 1 2 25 14, 10, 26 2 2 1212, 3, 27 2 1 14, 0, 2 2

2u 2 3v 1 w 5 2 12, 5, 23 2 2 3 124, 1, 9 2 1 14, 0, 2 2

2u 2 3v 1 w.w 5 14, 0, 2 2 .v 5 124, 1 , 9 2 ,u 5 12, 5, 23 2 ,

au 1 bv 1 cw

EXAMPLE 5

O

u

u + vv + u

v

Commutativity of vector additionu + v = v + u

Figure 1.12

9781284120097_CH01.qxd 9/15/17 3:29 PM Page 31























Determine whether the vector 18, 0, 52 is a linear combination of the vec-tors 11, 2, 32, 10, 1, 42, and 12, �1, 12.SOLUTION

We examine the following identity for values of c1, c2, and c3.

c111, 2, 32 1 c210, 1, 42 1 c312, 21, 12 5 18, 0, 52Multiplying the scalars and then adding the vectors, we get

1c1 � 2c3, 2c1 1 c2 2 c3, 3c1 1 4c2 1 c32 5 18, 0, 52Equating the components of the vectors, we get the following system of linear equations:

It can be shown that this system of equations has the unique solution,

c1 5 2, c2 5 21, c3 5 3

Thus, the vector 18, 0, 52 can be written in one way as a linear combination,

18, 0, 52 5 211, 2, 32 2 110, 1, 42 1 312, 21, 12If the above system of linear equations had many solutions 1c1, c2, and c3 many values2,there would have been many ways of writing 18, 0, 52 as a linear combination of the othervectors. If the system had no solutions, then 18, 0, 52 would not have been a linear com-bination of the three vectors. We shall see the geometrical significance of these conceptsin the following sections.

Column Vectors To this time we have defined only row vectors; that is, the components of a vector werewritten in row form. We shall find that it is more suitable at times to use column vectors.Matrix notation is often used for column vectors—many of the properties of vectors andmatrices are the same and we shall often find it useful to interpret column vectors as matrices.We define addition and scalar multiplication of column vectors in in a componentwisemanner:

and

For example, in

and

3c1 1 4c2 1 c3 5 5

2c1 1 c2 2 c3 5 0

c1 1 2c3 5 8

EXAMPLE 6

Rn

£u1

:

un

§ 1 £v1

:

vn

§ 5 £u1 1 v1

:

un 1 vn

§ c £u1

:

un

§ 5 £cu1

:

cun

§

4 c21d 5 c8

4dc1

2d 1 c4

7d 5 c5

9d

R2,

9781284120097_CH01.qxd 9/15/17 3:29 PM Page 32






















1.4 Subspaces of 33Rn

EXERCISE SET 1.3

Position Vectors1. Sketch the position vectors (1, 0) and (0, 1) in The

notation i and j is often used in science for these vectors.

2. Sketch the position vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) in The notation i, j, and k is often used in science for these vectors.

3. Sketch the following position vectors in

(a)

(b)(c)


(a)(b)

Addition and Scalar Multiplication5. Multiply the following vectors by the given scalars.

(a) (1, 4) by 3 (b) by

(c) (2, 6) by (d) (2, 4, 2) by

(e) by 3 (f) by 4

(g) by

(h) by 3

6. Compute the following linear combinations forand

(a) (b)(c) (d)(e)

7. Compute the following linear combinations for and

(a) (b)(c) (d)(e)

8. If u, v, and w are the following column vectors in deter-mine the linear combinations.

(a) (b)

(c) (d)

9. If u, v, and w are the following column vectors in deter-mine the linear combinations.

(a) (b)(c) (d)

, ,

10. Determine whether the first vector is a linear combinationof the other vectors.

(a) (1, 7); (1, 2), (�1, 3)

(b) (1, 2); (1, 1), (3, 2)

(c) (3, 5); (1, �3), (�2, 6)

(d) (6, 2); (2, 4), (�4, �8)

11. Determine whether the first vector is a linear combinationof the other vectors.

(a) (7, 9, 15); (1, 1, 2), (1, 2, 1), (2, 3, 4)

(b) (6, 13, 9); (1, 1, 1), (1, 2, 4), (0, 1, �3)

(c) (1, 2, �1); (1, 2, 0), (�1, �1, 2), (1, 3, 2)

(d) (�1, �1, 2); (1, 2, 3), (2, 5, 7), (0, 0, 1)

(e) (5, 8, 1); (1, 1, 1), (1, 2, �1), (5, 7, 1)

(f) (5, �1, 7); (1, 1, 2), (2, 2, 4), (1, �1, 1)

12. Prove the following properties of vector addition and scalarmultiplication that were introduced in this section.

(a)(b)(c)(d) 1u 5 u

1c 1 d 2u 5 cu 1 duu 1 12u 2 5 0u 1 1v 1 w 2 5 1u 1 v 2 1 w

u 5 £1

2

21

§ v 5 £3

0

1

§ w 5 £21

0

5

§

2u 1 3v 2 8w3u 2 2v 1 4w24v 1 3wu 1 2v

R3,

u 5 c23d , v 5 c21

24d , w 5 c 4

26d

23u 2 2v 1 4w2u 1 4v 2 w

u 1 v 2v 2 3w

R2,

2u 2 3v 2 4w5u 2 2v 1 6wu 1 3w

u 1 w 2u 1 v

w 5 12, 4, 22 2 .u 5 12, 1, 3 2 , v 5 121, 3, 2 2 ,23u 1 4v 2 2w

2u 1 3v 2 wv 1 wu 1 w u 1 3v

w 5 123, 5 2 . u 5 11, 2 2 ,v 5 14, 21 2 ,

13, 0, 4, 2, 21 211, 24, 3, 22, 5 2 25

121, 2, 3, 22 2121, 2, 3 221

212

22121, 3 2

w 5 10, 0, 23 2u 5 11, 1, 1 2 , v 5 121, 22, 24 2 ,OC

>

5 121, 3, 4 2OA>

5 12, 3, 1 2 , OB>

5 10, 5, 21 2 ,R3.

w 5 15, 3 2u 5 11, 1 2 , v 5 121, 24 2 ,OR

>

5 13, 23 2OP>

5 12, 4 2 , OQ>

5 124, 5 2 ,OC

>

5 11, 23 2OA>

5 15, 6 2 , OB>

5 123, 2 2 ,R2.

R3.

R2.

Subspaces of In this section we introduce subsets of that have all the algebraic properties of . Thesesubsets are vector spaces in their own right. They are called subspaces. There are two prop-erties that have to be checked to see if a subset is a subspace or not. The sum of two arbi-trary vectors of the subset must lie in the subset, and the scalar multiple of an arbitraryvector must also lie in the subset. In other words, the subset must be closed under addition

1.4 Rn

RnRn

9781284120097_CH01.qxd 9/15/17 9:47 AM Page 33























and under scalar multiplication. All the other vector space properties such as the commu-tative property u 1 v 5 v 1 u, and the associative property u 1 1v 1 w2 5 1u 1 v2 1 ware inherited from the larger space .

A nonempty subset of the vector space that has all the algebraic properties of iscalled a subspace. A subset of is a subspace if it is closed under addition and underscalar multiplication.

Consider the vector space and a plane W through the origin (Figure 1.13). The planeis a subset of . We can see as follows that W is a subspace of . Let u, v, and w bearbitrary vectors in W and k be an arbitrary scalar. The sum u 1 v lies in the plane W, and the scalar multiple kw also lies in W. W is closed under addition and scalar multipli-cation. W is a subspace of . In general, the vectors that make up a plane in form asubspace of .Rn

R3 Rn

R3R3R3

RnRnRn

Rn

Figure 1.13 A set of vectors that forms a plane is a subspace. W is closed under addition and scalarmultiplication. W is a subspace.

z

y

x

W

kwu + v

O

u

w

v

Let us next look at a specific such plane in .

A subset of a vector space is often defined in terms of an arbitrary vector.Consider the subset W of consisting of vectors of the form where thethird component is the sum of the first two. The vector 12, 5, 72 is in W, for example,whereas 12, 5, 92 is not. Let us show that W is a subspace of . Let u 5 1a, b, a 1 b2 andv 5 1c, d, c 1 d 2 be vectors in W and k be a scalar. Then

u 1 v is in W since the third component is the sum of the first two components. W isclosed under addition. Furthermore,

ku is in W. W is also closed under scalar multiplication. Thus W is a subspace of . Ithas all the algebraic properties of a vector space.

R3

R3

5 1ka, kb, ka 1 kb 25 1ka, kb, k 1a 1 b 2 2

ku 5 k 1a, b, a 1 b 2

5 1a 1 c, b 1 d, 1a 1 c 2 1 1b 1 d 2 25 1a 1 c, b 1 d, a 1 b 1 c 1 d 2

u 1 v 5 1a, b, a 1 b 2 1 1c, d, c 1 d 2

R3

R3 1a, b, a 1 b 2 ,EXAMPLE 1

9781284120097_CH01.qxd 9/15/17 9:48 AM Page 34























Let us look at the geometrical interpretation of this subspace W. Consider the arbi-trary vector u 5 1a, b, a 1 b2. Separate the variables a and b as follows:

W consists of all the vectors in that are linear combinations of 11, 0, 12 and 10, 1, 12.W is in fact the plane through the origin defined by the vectors 11, 0, 12 and 10, 1, 12. SeeFigure 1.14.

When every vector of a vector space can be written as a linear combination of a setof vectors, as here, we say that the vectors span the space. Thus the vectors 11, 0, 12 and10, 1, 12 span W.

5 1a, 0, a 2 1 10, b, b 2u 5 1a, b, a 1 b 2

R3

5 a 11, 0, 1 2 1 b 10, 1, 1 2

z

y

x

Wu

(1, 0, 1) (0, 1, 1)

O

= a(1, 0, 1) + b(0, 1, 1)

Figure 1.14 W is the plane defined by the vectors (1, 0, 1) and (0, 1, 1).

Let us now consider another class of subspaces of . Let u and v be vectors that lie ona line V in . The sum u 1 v lies on the line, and so does the scalar multiple ku. SeeFigure 1.15. V is closed under addition and under scalar multiplication. Thus V is a sub-space. In general, the vectors that make up a line in form a subspace of .

R3

Rn Rn

R3

Figure 1.15 A set of vectors that forms a line is a subspace. V is a subspace.

z

x

V

yO

u + vuv

V is closed under addition.

z

x

V

yO

ku

u

V is closed under scalar multiplication.

The following example illustrates a subspace that is a line.

9781284120097_CH01.qxd 10/13/17 8:22 AM Page 35























Consider the subset V of of vectors of the form 1a, 2a, 3a2, where thesecond component is twice the first, and the third is three times the first. Let us showthat V is a subspace of . Let 1a, 2a, 3a2 and 1b, 2b, 3b2 be two vectors in V, and let k bea scalar. Then

This is a vector in V since the second component is twice the first, and the third is threetimes the first. V is closed under addition. Further,

This vector is in V. V is closed under scalar multiplication. V is a subspace of . We now proceed to look at the geometry. Consider the arbitrary vector u 5 1a, 2a, 3a2. We get

The vectors of V are scalar multiples of the single vector 11, 2, 32. These vectors form the line through the origin defined by the vector 11, 2, 32. See Figure 1.16. The vector 11, 2, 32 spans V.

EXAMPLE 2 R3

R3

5 a 11, 2, 3 2u 5 1a, 2a, 3a 2

5 1a 1 b, 2 1a 1 b 2 , 3 1a 1 b 2 21a, 2a, 3a 2 1 1b, 2b, 3b 2 5 1a 1 b, 2a 1 2b, 3a 1 3b 2

R3

k 1a, 2a, 3a 2 5 1ka, 2ka, 3ka 2

The following example illustrates that not every subset of a vector space is a subspace.

Let W be the set of vectors of the form (a, a2, b). Show that W is not a sub-

space of .

SOLUTION

W consists of all elements of for which the second component is the square of thefirst. For example, the vector 12, 4, 32 is in W, whereas the vector 12, 5, 32 is not.

Let us check for closure under addition. Let 1a, a2, b) and 1c, c2, d) be elements of W. We get

2 1a 1 c, 1a 1 c 2 2, b 1 d 21a, a2, b 2 1 1c, c2, d 2 5 1a 1 c, a2 1 c2, b 1 d 2

R3

EXAMPLE 3

R3

z

x

V

(1, 2, 3)

yO

Figure 1.16

9781284120097_CH01.qxd 9/15/17 9:49 AM Page 36























Thus 1a, a2, b2 1 1c, c2, d 2 is not an element of W. The set W is not closed under addition.Thus, W is not a subspace.

This completes the proof that W is not a subspace. Let us, however, illustrate thecheck for closure under scalar multiplication. Let k be a scalar. We get

Thus k1a, a2, b2 is not an element of W. W is not closed under scalar multiplication either.

The following example illustrates how a set of solutions to a homogeneous systemof linear equations can form a subspace.

Consider the following homogeneous system of linear equations:

x1 � x2 � 3x3 5 0

x2 � 5x3 5 0

2x1 � x2 � x3 5 0

It can be shown that there are many solutions x1 5 2r, x2 5 5r, x3 5 r. We can write thesesolutions as vectors in R3 as follows.

12r, 5r, r2The set of solutions W thus consists of vectors for which the first component is twice thethird, and the second component is five times the third component. The vector 14, 10, 22for example is in W while 14, 9, 22 is not. Let us show that W is closed under addition andunder scalar multiplication. Let u 5 12r, 5r, r2 and v 5 12s, 5s, s2 be arbitrary vectors inW. We get

and

In u 1 v and ku, the first component is twice the third, and the second component is fivetimes the third. They are both in W. W is closed under addition and under scalar multi-plication. It is a subspace of R3. Writing the vector 12r, 5r, r2 in the form r 12, 5, 12, wesee that the set of solutions is the line defined by the vector 12, 5, 12. See Figure 1.17. Thevector 12, 5, 12 spans the set of solutions.

5 12kr, 5kr, kr 2ku 5 k 12r, 5r, r 2

2 1ka, 1ka 22, kb 2k 1a, a2, b 2 5 1ka, ka2, kb 2

EXAMPLE 4

5 12 1r 1 s 2 , 5 1r 1 s 2 , r 1 s 2u 1 v 5 12r 1 2s, 5r 1 5s, r 1 s 2

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The following example illustrates a set of solutions to a homogeneous system of lin-ear equations that forms a plane in R4. It is not possible to sketch a plane in R4, but we canvisualize it in the mind’s eye.

Consider the following system of homogeneous linear equations.

x1 � x2 � x3 � 2x4 � 0

x1 � 3x3 � 2x4 � 0

2x1 � x2 � 2x3 � 4x4 � 0

It can be shown that there are many solutions x1 � 3r � 2s, x2 � 4r, x3 � r, x4 � s. Writethese solutions as vectors in R4,

13r � 2s, 4r, r, s2It can be shown that this set W of vectors is closed under addition and scalar multipli-cation, and is thus a subspace of R4. Separate the variables in the general solution.

13r � 2s, 4r, r, s2 � r 13, 4, 1, 02 � s 1�2, 0, 0, 12This implies that every vector in W can be expressed as a linear combination of 13, 4, 1, 02 and 1�2, 0, 0, 12. The vectors 13, 4, 1, 02 and 1�2, 0, 0, 12 span W. W will be aplane through the origin in a four-space.

We shall see in Section 2 of the “Matrices and Linear Transformations” chapter thatthe set of solutions to every homogeneous system of linear equations is a subspace. Animplication of this result is that the sum of any two solutions is a solution. The scalarmultiple of a solution is also a solution. However, we shall find that the set of solutionsto a nonhomogeneous system is not a subspace. We shall discuss ways of relating solu-tions of nonhomogeneous systems to solutions of “corresponding” systems of homoge-neous linear equations.

EXAMPLE 5

Figure 1.17

z

x

W

yO

(2, 5, 1)

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1.5 Basis and Dimension in 39Rn

EXERCISE SET 1.4

Subspaces1. Show that the sets consisting of vectors of the following

form are subspaces of R2 by showing that they are closedunder addition and under scalar multiplication.

(a) (a, 3a) (b) (a, �a) (c) (a, 0) (d) (2a, 3a)

2. Show that the sets consisting of vectors of the followingform are subspaces of R3 or R4.

(a) (a, b, b) (b) (a, �a, b) (c) (a, 2a, �a)

(d) (a, a, b, b)

3. Determine whether the sets defined by the following vec-tors are subspaces of R3.

(a) (a, b, 2a � 3b) (b) (a, b, 3)

(c) (a, a � 2, b) (d) (a, �a, 0)

Systems of Linear EquationsIn Exercises 4–7, solve the homogeneous systems of linear equa-tions. Show that the sets of solutions form subspaces of R3. Givethe geometrical interpretation of the subspaces.

4. x1 � 3x2 � x3 � 0

x2 � x3 � 0

2x1 � 7x2 � 3x3 � 0

5. x1 � x2 � 7x3 � 0

x2 � 4x3 � 0

x1 � 3x3 � 0

6. x1 � 2x2 � 3x3 � 0

�x1 � 2x2 � x3 � 0

x1 � 2x2 � 4x3 � 0

7. x1 � 2x2 � x3 � 0

x1 � 3x2 � x3 � 0

3x1 � 7x2 � x3 � 0

8. Consider the following homogeneous system of linear equa-tions in four variables. For convenience, the general solu-tion is given. Show that the set of solutions forms a subspaceof R4.

x1 � x2 � 3x3 � 5x4 � 0

x2 � x3 � 3x4 � 0

x1 � 2x2 � 4x3 � 8x4 � 0

General solution is (2r � 2s, r � 3s, r, s).

In Exercises 9–11, consider the homogeneous systems of linearequations. For convenience, the general solutions are given. Itcan be shown, as in Exercise 8, that the sets of solutions formsubspaces of R4.

(a) Use the general solution to construct two specific solu-tions.

(b) Use the operations of addition and scalar multiplicationto generate four vectors from these two solutions.

(c) Use the general solution to check that these vectors areindeed solutions, giving the values of r and s for whichthey are solutions.

9. x1 � x2 � 2x3 � 3x4 � 0

2x1 � x2 � 5x3 � 2x4 � 0

3x1 � x2 � 8x3 � x4 � 0

General solution is (3r � s, �r � 4s, r, s).

10. x1 � x2 � 5x3 � 0

2x1 � 3x2 � 13x3 � 0

x1 � 2x3 � 0

General solution is (2r, 3r, r)

11. x1 � x2 � x3 � 3x4 � 0

x1 � 2x2 � 4x3 � 5x4 � 0

x1 � 2x3 � x4 � 0

General solution is (2r � s, �3r � 2s, r, s).

Basis and DimensionIn the previous sections, we discussed the vector space and certain subsets calledsubspaces that are vector spaces in their own right. Vector spaces usually contain aninfinite number of vectors. These spaces can often best be discussed in terms of afinite subset of vectors called a basis. The basis is a set of vectors that represents thewhole space.

Rn

1.5

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Let us develop these ideas. We shall find that they can be applied to all vector spacesand subspaces. We first generalize the idea of a spanning concept that was introduced inthe previous section.

Span: The vectors v1, v2, . . . , vn span a vector space V if every vector v in the spacecan be expressed as a linear combination of them, v � c1v1 � c2v2 � . . . � cnvn. A set ofvectors that spans a space represents the whole space.

We next introduce a suitable concept of vector independence.

Linear independence: The vectors v1, v2, . . . , vn are linearly independent if the iden-tity c1v1 � c2v2 � . . . � cnvn � 0 is only true for c1 � 0, c2 � 0, . . . , cn � 0. If there arenonzero values of c1, c2, . . . , cn for which this identity holds, the set is said to be linearlydependent.

Spanning and independence come together to define a suitable set of vectors for describ-ing a vector space in a precise way—a spanning set that has no redundant vectors (seeExercise 8). This set is called a basis.

Basis and dimension: A set of vectors is a basis for a vector space if the vectors (1) spanthe space and (2) are linearly independent. There are usually many bases for a given vec-tor space. However the bases all have the same number of vectors. This number is calledthe dimension of the space.

Since subspaces are vector spaces in their own right, the preceding concepts also applyto subspaces.

Dimension of Rn

Let us apply these ideas to the vector space . Let (a, b, c) be an arbitrary vector in .Observe that we can write (a, b, c) � a(1, 0, 0) � b(0, 1, 0) � c(0, 1, 0). The vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) thus span .

Further, p(1, 0, 0) � q(0, 1, 0) � r (0, 0, 1) � (0, 0, 0) implies that p � 0, q � 0, r � 0.* The vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) are linearly independent. Thus the set of vec-tors {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is a basis for .

There are many other bases for , sets that span and are linearly independent. Weshall see in Section 4 of the “General Vector Spaces” chapter, for example, that {(1, 3, �1),(2, 1, 0), (4, 2, 1)} is a basis for . The set {(1, 0, 0), (0, 1, 0), (0, 0, 1)}, however, is themost important basis for . It is called the standard basis of . We shall see that all thesedifferent bases for have the same number of vectors, namely 3. The dimension of is 3. Similarly, {(1, 0), (0, 1)} is the standard basis for R2, and the dimension of R2 is 2. The set{(1, 0, . . . , 0), (0, 1, . . . , 0), . . . , (0, 0, . . . , 1)} of n vectors is the standard basis of . Thedimension of is n.

The set {(1, 0, . . . , 0), (0, 1, . . . , 0), . . . , (0, 0, . . . , 1)} of n vectors is the standardbasis of Rn. The dimension of Rn is n.

RnRn

R3 R3R3 R3

R3

R3 R3R3

R3

R3 R3

*p(1, 0, 0) � q(0, 1, 0) � r(0, 0, 1) � (0, 0, 0) �> (p, 0, 0) � (0, q, 0) � (0, 0, r) � (0, 0, 0) �> (p, q, r) � (0, 0, 0) �> p � 0, q � 0, r � 0.

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Consider the subspace W of R3 of vectors of the form (a, b, a � b) thatwas discussed in the previous section. Let u � (a, b, a � b) be an arbitrary vector in W.We know that we can write u in the form

u � a(1, 0, 1) � b(0, 1, 1)

The vectors (1, 0, 1) and (0, 1, 1) thus span W. Further, the identity p(1, 0, 1) � q(0, 1, 1) � (0, 0, 0) leads to p � 0, q � 0. These vectors are linearly independent. The set {(1, 0, 1), (0, 1, 1)} is a basis for W. The dimension of W is 2. We know that W is a planethrough the origin, fitting in with our intuitive thinking of a plane as being a two-dimensional space. See Figure 1.18.

EXAMPLE 1

Figure 1.18 Vector space of dimension 2 with basis {(1, 0, 1), (0, 1, 1)}

z

y

x

W

(1, 0, 1) (0, 1, 1)

Subset of vectors ofthe form (a, b, a + b)

O

The following example illustrates how to recognize subspaces of dimension 1 (lines).

Consider the subset V of R2 of vectors of the form (a, 2a), where the second component is twice the first component. Let us show that V is a subspace. Let u � (a, 2a) and v � (b, 2b) be vectors in V and k be a scalar. Then

and

5 1ka, 2ka 2ku 5 k 1a, 2a 2

5 1a 1 b, 2 1a 1 b 2 25 1a 1 b, 2a 1 2b 2

u 1 v 5 1a, 2a 2 1 1b, 2b 2

EXAMPLE 2

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Figure 1.19

y

x

V

(1, 2)

O

Consider the following homogeneous system of linear equations:

x1 � x2 � 7x3 � 2x4 � x5 � 0

x1 � 2x2 � 10x3 � 2x4 � 2x5 � 0

2x1 � 3x2 � 17x3 � 4x4 � 3x5 � 0

It can be shown that the general solution to this system in vector form is (4r � 2s, 3r � t, r, s, t) and that this set is closed under addition and under scalar multiplication.It is a subspace W of R5. Let us find a basis and the dimension of this subspace. Separatethe variables r, s, and t as follows:

(4r � 2s, 3r � t, r, s, t) � r(4, 3, 1, 0, 0) � s(�2, 0, 0, 1, 0) � t(0, �1, 0, 0, 1)

The vectors (4, 3, 1, 0, 0), (�2, 0, 0, 1, 0), (0, �1, 0, 0, 1) span W. Let us check for lin-ear independence. Consider the following identity for arbitrary variables p, q, h.

p(4, 3, 1, 0, 0) � q(�2, 0, 0, 1, 0) � h(0, �1, 0, 0, 1) � (0, 0, 0, 0, 0)

We get

(4p �2q, 3p � h, p, q, h) � (0, 0, 0, 0, 0)

EXAMPLE 3

Both u � v and ku are in V since their second components are twice the first. V is closedunder addition and scalar multiplication. It is a subspace of R2; it is a vector space in itsown right.

We now proceed to find a basis for V. Write the arbitrary vector as follows:

(a, 2a) � a(1, 2)

The elements of V are scalar multiples of the single vector (1, 2). The implication is thatthe set {(1, 2)} is a basis for V. The dimension of V is 1. The vectors form the line throughthe origin defined by (1, 2). See Figure 1.19.

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Determine whether the following sets of vectors are linearly dependentor independent in R3. (a) {(1, 2, 0), (0, 1, �1), (1, 1, 2)}. (b) {(1, 2, 3), (�2, 0, 1), (4, �4, �9)}.

SOLUTION

(a) We examine the identity

c1(1, 2, 0) � c2(0, 1, �1) � c3(1, 1, 2) � (0, 0, 0)

This identity leads to the following system of linear equations:

c1 � c3 � 0

2c1 � c2 � c3 � 0

� c2 � 2c3 � 0

This system has the unique solution c1 � 0, c2 � 0, c3 � 0. Thus, the vectors are linearlyindependent.

(b) We examine c1(1, 2, 3) � c2(�2, 0, 1) � c3(4, �4, �9) � (0, 0, 0).This identity leads to the following system of linear equations:

c1 � 2c2 � 4c3 � 0

2c1 � 4c3 � 0

3c1 � c2 � 9c3 � 0

This system has many solutions, c1 � 2r, c2 � 3r, c3 � r, where r is any real number.The set is thus linearly dependent. For example, when r � 1, we get 2(1, 2, 3) �3(�2, 0, 1) � (4, �4, �9) � (0, 0, 0).

EXAMPLE 4

*We shall often refer to sets of vectors as being linearly dependent or independent.

This implies that p � 0, q � 0, h � 0. Thus, the vectors are linearly independent.The set {(4, 3, 1, 0, 0), (�2, 0, 0, 1, 0), (0, �1, 0, 0, 1)} is a basis for W. The set

of solutions is a three-dimensional vector space embedded in a five-dimensionalspace.

We have introduced linear independence as an important natural requirement for a setof vectors to be a basis of a vector space. The concept of independence is one that occursin many situations in mathematics. The following example reinforces this important con-cept. Note that determining whether a set of vectors* is linearly independent or not gener-ally involves solving a system of linear equations.

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EXERCISE SET 1.5

Standard Bases1. Give the standard basis for

(a) Show that the vectors in the basis span

(b) Show that the vectors in the basis are linearly inde-pendent.

2. Give the standard basis for

(a) Show that the vectors in the basis span

(b) Show that the vectors in the basis are linearly inde-pendent.

Bases and Subspaces3. Consider the sets of vectors of the following form. Prove

that they are subspaces of Find a basis for each subspaceand give its dimension.

(a) (b)(c) (d)

4. Consider the sets of vectors of the following form. Determinewhether the sets are subspaces of If a set is a subspace,give a basis and its dimension.

(a) (b)(c) (d)

5. Consider the sets of vectors of the following form. Determinewhether the sets are subspaces of an appropriate vector space.If a set is a subspace, give a basis and its dimension.

(a) (b)(c) (d)

6. State with a brief explanation whether the following state-ments are true or false.

(a) The vectors (1, 0) and (0, 1) span (b) The vectors (1, 0), (0, 1), and (1, 1) span (c) The vectors (1, 0) and (0, 1) are linearly independent.(d) The vectors (1, 0), (0, 1), and (0, 2) are linearly inde-

pendent. (e) The set of vectors is a basis for

(f) The set of vectors {(2, 0), (0, 3)} is a basis for

7. State with a brief explanation whether the following state-ments are true or false.

(a) The set {(1, 0, 0), (0, 1, 0)} is the basis for a two-dimensional subspace of

(b) The set {(1, 0, 0)} is the basis for a one-dimensionalsubspace of

(c) The vector is an arbitrary vector in theplane spanned by the vectors (1, 2, 0) and (0, 0, 1).

(d) The vector is an arbitrary vector in theplane spanned by the vectors (1, 0, 2) and

(e) The set {(1, 0, 0), (0, 1, 0), (1, 1, 0)} is a basis for asubspace of

(f) is a subspace of

8. (a) Show that the vectors (1, 0), (0, 1) span and are alsolinearly independent.

(b) Show that the vectors (1, 0), (0, 1), (0, 2) spanDemonstrate that it is not an efficient spanning set by showing that an arbitrary vector in can beexpressed in more than one way as a linear combina-tion of these vectors. (We can think of (0, 2) as beinga redundant vector.)

(c) Show that {(1, 0), (0, 1), (0, 2)} is linearly dependentand thus is not a basis for A basis consists of a setof vectors, all of which are needed.

9. (a) Show that the vectors (1, 0), (0, 1) span and are alsolinearly independent.

(b) Show that the vectors (1, 0), (0, 1), (1, 1) span butthat it is not an efficient spanning set—show that anarbitrary vector in can be expressed in more thanone way as a linear combination of these vectors.

(c) Show that {(1, 0), (0, 1), (1, 1)} is linearly dependentand is thus not a basis for

10. (a) Show that the vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) spanR3 and that they are also linearly independent.

(b) Show that the vectors (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 1, 1) span R3. Demonstrate that it is not an effi-cient spanning set by showing that an arbitrary vectorin R3 can be expressed in more than one way as a lin-ear combination of these vectors. We can think of (0, 1, 1) as being a redundant vector.

(c) Show that {(1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 1, 1)} islinearly dependent and is thus not a basis for R2. Abasis consists of a set of vectors, all of which areneeded.

11. (a) Show by means of an example that it is possible to havea set of vectors that span R3 but are not linearly inde-pendent.

(b) Show that it is possible to have a set of vectors thatare linearly independent but do not span R3.

Systems of Linear EquationsIn Exercises 12–15 consider the homogeneous systems of lin-ear equations. (Aspects of these equations were considered inExercises 8–11, Section 1.4.) For convenience, their generalsolutions are given. It can be shown as in Section 1.4 that thesolutions of each system form a subspace W of some Rn. Find abasis for each subspace of solutions. Give the dimension of thesubspace and give a written description of the subspace (a line,plane, or otherwise).

12. x1 � x2 � 3x3 � 5x4 � 0

x2 � x3 � 3x4 � 0

x1 � 2x2 � 4x3 � 8x4 � 0

General solution is (2r � 2s, r � 3s, r, s).

R2.

R2

R2,

R2

R2.

R2.

R2.

R2

R2.

R2

R2 R3.R3.

10, 1, 21 2 .1a, b, 2a 2 b 21a, 2a, b 2R3.

R3.

R2.

5 11, 0 2 , 10, 21 2 6 R2.

R2.R2.

1a, 2a 2 1a, b, c, 1 2a 11, 2, 3 2 1a, 0, 0 2

1a, b, 2 2 1a, a, a 1 3 21a, b, a 2 1a, b, 0 2

R4.

R4.

R3.

R3.

1a, 2a, 4a 2 1a, 2a, 0 21a, a, b 2 1a, 2a, b 2

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DEFINITION

1.6 Dot Product, Norm, Angle, and Distance 45

13. x1 � x2 � 2x3 � 3x4 � 0

2x1 � x2 � 5x3 � 2x4 � 0

3x1 � x2 � 8x3 � x4 � 0

General solution is (3r � s, �r � 4s, r, s).

14. x1 � x2 � 5x3 � 0

2x1 � 3x2 � 13x3 � 0

x1 � 2x3 � 0

General solution is (2r, 3r, r)

15. x1 � x2 � x3 � 3x4 � 0

x1 � 2x2 � 4x3 � 5x4 � 0

x1 � 2x3 � x4 � 0

General solution is (2r � s, �3r � 2s, r, s).

In Exercises 16–18 consider the homogeneous systems of linearequations. For convenience their general solutions are given. It canbe shown as in Section 1.4 that the solutions of each system forma subspace W of some Rn. Find a basis for each subspace of solu-tions. Give the dimension of the subspace and give a written descrip-tion of the subspace (a line, plane, or otherwise).

16. x1 � x2 � x3 � 5x4 � 0

x2 � x3 � 3x4 � 0

x1 � x3 � 2x4 � 0

General solution is (2r, �r, 4r, r).

17. x1 � 3x2 � x3 � 0

2x1 � 6x2 � 2x3 � 0

�x1 � 3x2 � x3� 0

General solution is (3r � s, r, s)

18. x1 � x2 � 5x3 � 2x4 � x5 � 0

2x1 � x2 � 7x3 � 2x4 � 2x5 � 0

x1 � 2x2 � 8x3 � 4x4 � x5 � 0

General solution is (2r � t, 3r � 2s, r, s, t).

19. Determine whether the following sets of vectors are linearlydependent or independent in R3.

(a) {(1, 0, 2), (1, 1, 0), (5, 3, 6)}

(b) {(1, 1, 1), (2, �1, 1), (3, �3, 0)}

(c) {(1, �1, 1), (2, 1, 0), (4, �1, 2)}

(d) {(1, 2, 1), (�2, 1, 3), (�1, 8, 9)}

(e) {(�2, 0, 3), (5, 2, 1), (10, 6, 9)}

(f) {( 3, 4, 1), (2, 1, 0), (9, 7, 1)}

Dot Product, Norm, Angle, and DistanceIn this section we develop a geometry for the vector space As we construct this geometry the reader should pay close attention to the approach we use. Although the resultsare of course important, the way we arrive at the results is also very important. The mag-nitude of a vector, the angle between two vectors, and the distance between two points aredefined in by generalizing these concepts from We extend the Pythagorean Theoremto This process of gradually extending familiar concepts to more general surroundingsis fundamental to mathematics.

We start the discussion with the definition of the dot product of two vectors. The dotproduct is a tool that is used to build the geometry of Rn.

Rn.Rn R2.

Rn.

1.6

Let and be two vectors in The dot product of u and v isdenoted and is defined by

The dot product assigns a real number to each pair of vectors.

u # v 5 u1v1 1 c1 unvn

u # vu 5 1u1, c, un 2 v 5 1v1, c, vn 2 Rn.

For example, if and , their dot product is

u # v 5 11 3 3 2 1 122 3 0 2 1 14 3 2 2 5 3 1 0 1 8 5 11

u 5 11, 22, 4 2 v 5 13, 0, 2 2

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DEFINITION


Properties of the Dot ProductLet u, v, and w be vectors in and let c be a scalar. Then1.2.3.4. and if and only if Proof We prove Parts 1 and 4, leaving the other parts for the reader to prove in theexercises that follow.

1. Let and We get

by the commutative property of real numbers

4.since it is the sum of squares. Thus

Further, if and only if Thus if and only if

We shall use these properties to simplify expressions involving dot products. In thissection, for example, we shall use the properties in arriving at a suitable expression for theangle between two vectors in In later sections we shall use these properties to general-ize the concept of a dot product to spaces of matrices and functions.

Norm of a Vector in Let be a vector in We know by the Pythagorean Theorem that the length

of this vector is See Figure 1.20. We generalize this result to define thelength (the technical name is norm) of a vector in Rn.

"1u1 2 2 1 1u2 2 2.

u 5 1u1, u2 2 R2.

Rn

Rn.

u # u 5 0 u 5 0.1u1 2 2 1 c1 1un 2 2 5 0 u1 5 0, c, un 5 0.

1u1 2 2 1 c1 1un 2 2 $ 0, u # u $ 0.u # u 5 u1u1 1 c1 unun 5 1u1 2 2 1 c1 1un 2 2.

5 v # u5 v1u1 1 c1 vnun,


u 5 1u1, c, un 2 v 5 1v1, c, vn 2 .

u # u $ 0, u # u 5 0 u 5 0cu # v 5 c 1u # v 2 5 u # cv

1u 1 v 2 # w 5 u # w 1 v # wu # v 5 v # u

Rn

Figure 1.20

xO

y

u2

u1

u = (u1, u2)

Norm of vector (u1, u2) √(u1)2 + (u2)2

=√(u1)2 + (u2)2

The norm (length or magnitude) of a vector in is denoted and defined by

Note: The norm of a vector can also be written in terms of the dot product

iui 5 "u # u

iui 5 "1u1 2 2 1 c1 1un 2 2

u 5 1u1, c, un 2 Rn iui

9781284120097_CH01.qxd 9/15/17 9:52 AM Page 46






















DEFINITION


For example, if in and in then

and

ivi 5 "13 2 2 1 10 2 2 1 11 2 2 1 14 2 2 5 "9 1 0 1 1 1 16 5 "26

iui 5 "11 2 2 1 13 2 2 1 15 2 2 5 "1 1 9 1 25 5 "35,

u 5 11, 3, 5 2 R3 v 5 13, 0, 1, 4 2 R4,

A unit vector is a vector whose norm is one. If v is a nonzero vector, then the vector

is a unit vector in the direction of v. (See Exercise 28.) This procedure of constructing a unit vector in the same direction as a given vector is called

normalizing the vector.

u 51

iviv

Find the norm of the vector Normalize this vector.

SOLUTION

The norm of is The normalized vector is

This vector may also be written This vector is a unit vector in the

direction of

Properties of Norm1. (the length of a vector cannot be negative) 2. if and only if u � 0 (the length of a vector is zero if and only if the vector is

the zero vector) 3. (the length of is times the length of u)

You are asked to prove these properties in the exercises that follow.

Angle Between VectorsLet and be position vectors in See Figure 1.21. Let us arrive atan expression for the cosine of the angle formed by these vectors.u

u 5 1a, b 2 v 5 1c, d 2 R2.

icui 5 0 c 0 iui cu 0 c 0iui 5 0iui $ 0

12, 21, 3 2 .a 2

"14,

21

"14,

3

"14b.

1

"1412, 21, 3 2

i 12, 21, 3 2 i 5 "22 1 121 2 2 1 32 5 "14. 12, 21, 3 2 "14.

EXAMPLE 1 12, 21, 3 2 .

xO

y

v

u

u v

B

Acos =

||u|| ||v||u

u

u ·

Figure 1.21

9781284120097_CH01.qxd 9/15/17 9:53 AM Page 47






















DEFINITION


The law of cosines gives

Solve for cos

We get that

Further,

Thus, the angle between two vectors in is given by

We know that the cosine of any angle such as has to satisfy the condition The Cauchy-Schwarz inequality, which we discuss later in this section, assures us that

We can thus extend this concept of angle between two vectors to as follows.Rn

` u # viui ivi

` # 1

u 0 cosu 0 # 1.

cosu 5u # v

iui ivi

u R2

2 1OA 2 1OB 2 5 2 iui ivi

5 2ac 1 2db 5 2u # v

5 a2 1 b2 1 c2 1 d2 2 3 1c 2 a 2 2 1 1d 2 b 2 2 4

u.

AB2 5 OA2 1 OB2 2 2 1OA 2 1OB 2cosu

OA2 1 OB2 2 AB2 5 iui 2 1 ivi 2 2 iv 2 ui 2

cosu 5OA2 1 OB2 2 AB2

2 1OA 2 1OB 2

Let u and v be two nonzero vectors in The cosine of the angle between these vectors is

cosu 5u # v

iui ivi, 0 # u # p

Rn. u

Determine the angle between the vectors and in

SOLUTION

We get

Thus

The angle between u and v is (or ).p/4 45°

cosu 5u # v

iui ivi5

1

"2

ivi 5 "12 1 02 1 12 5 "2.

u # v 5 11, 0, 0 2 # 11, 0, 1 2 5 1, iui 5 "12 1 02 1 02 5 1,

R3.EXAMPLE 2 u 5 11, 0, 0 2 v 5 11, 0, 1 2

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We say that two nonzero vectors are orthogonal if the angle between them is a right angle.The following theorem, which the reader is asked to prove in Exercise 29, gives us a con-dition for orthogonality.

THEOREM 1.4

Two nonzero vectors u and v are orthogonal if and only if

For example, the vectors and are orthogonal since

Properties of Standard Basis of Consider the standard basis of Observe that

The vector is a unit vector. Similarly,and are unit vectors. Further note that The vectorsand are orthogonal. Similarly, the pairs andare orthogonal. We call a set of unit pairwise orthogonal vectors an orthonormal set. Thusthe standard basis of is an orthonormal set. This resultapplies to all the vector spaces

The standard basis for is anorthonormal set.

(a) Let w be a vector in Let W be the set of vectors that are orthogonal to w. Showthat W is a subspace of

(b) Find a basis for the subspace W of vectors in that are orthogonal toGive the dimension and a geometrical description of W.

SOLUTION

(a) Let u and v be vectors in W. Each vector is orthogonal to w. Thus andUsing the properties of dot product,

Thus is orthogonal to w, implying that is in W. W is closed underaddition. Further, if c is a scalar,

This means that the vector cu is orthogonal to w and hence is in W. W is closed underscalar multiplication. W is thus a subspace of

(b) W is the set of vectors in that are orthogonal to Let be avector in W. Thus

c 5 2a 2 3b

a 1 3b 1 c 5 0

1a, b, c 2 # 11, 3, 1 2 5 0

R3 w 5 11, 3, 1 2 . 1a, b, c 2Rn.

cu # w 5 0

c 1u # w 2 5 0

u 1 v u 1 v

1u 1 v 2 # w 5 0

u # w 1 v # w 5 0

v # w 5 0.u # w 5 0

R3 w 5 11, 3, 1 2 .Rn.Rn.

EXAMPLE 3

Rn, 5 11, 0, c, 0 2 , 10, 1, c, 0 2 , c, 10, 0, c, 1 2 6,Rn.

5 11, 0, 0 2 , 10, 1, 0 2 , 10, 0, 1 2 6 R3

10, 1, 0 2 11, 0, 0 2 , 10, 0, 1 2 10, 1, 0 2 , 10, 0, 1 210, 0, 1 2 11, 0, 0 2 # 10, 1, 0 2 5 0. 11, 0, 0 2i 11, 0, 0 2 i 5 "12 1 02 1 02. 11, 0, 0, 2 10, 1, 0 25 11, 0, 0 2 , 10, 1, 0 2 , 10, 0, 1 2 6 R3.

Rn

12, 23, 1 2 # 11, 2, 4 2 5 12 3 1 2 1 123 3 2 2 1 11 3 4 2 5 2 2 6 1 4 5 0

12, 23, 1 2 11, 2, 4 2

u # v 5 0.

9781284120097_CH01.qxd 9/15/17 9:56 AM Page 49























Therefore W is the set of vectors of the form Separate the vari-ables a and b,

Every vector in W can be written as a linear combination of andThus the vectors span W. It can be shown that these vectors arealso linearly independent. The set is therefore a basis for W.

Now for the geometry. Each vector in W is of the formAs a and b vary we get the various vectors that make up W. When andfor example, we get the vector Since there are two vectors in the basis,the dimension of W is 2. A subspace of dimension 2 is a plane. Thus W is the planein defined by and See Figure 1.22.R3 11, 0, 21 2 10, 1, 23 2 .

11, 2, 27 2 . a 5 1 b 5 2,a 11, 0, 21 2 1 b 10, 1, 23 2 .

5 11, 0, 21 2 , 10, 1, 23 2 611, 0, 21 2 , 10, 1, 23 2 11, 0, 21 2 10, 1, 23 2 .1a, b, 2a 2 3b 2 5 a 11, 0, 21 2 1 b 10, 1, 23 2

1a, b, 2a 2 3b 2 .

x

W (1, 3, 1)

(0, 1, –3)(1, 0, –1)

y

z

Plane defined by the vectors(1, 0, –1) and (0, 1, –3)

Figure 1.22

The following example illustrates how these concepts also apply to vectors written in col-umn form.

(a) Show that the following vectors are orthogonal. (b) Compute the normof each vector.

SOLUTION

(a) The dot product of u and v is obtained by multiplying corresponding componentsand adding.

The dot product is zero; thus the vectors are orthogonal.(b) The norm of each vector is obtained by taking the square root of the sum of the

squares of the components.

7v 7 5 "121 2 2 1 32 1 12 5 "11

7u 7 5 "12 1 22 1 125 2 2 5 "30

u # v 5 11 3 21 2 1 12 3 3 2 1 125 3 1 2 5 0

u 5 £1

2

25

§ , v 5 £21

3

1

§

EXAMPLE 4

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DEFINITION


Let and be two points in The distance between x and y isdenoted and is defined by

Note: We can also write this distance formula as

d 1x, y 2 5 ix 2 yi

d 1x, y 2 5 "1x1 2 y1 2 2 1 c1 1xn 2 yn 2 2

d 1x, y 2x 5 1x1, c, xn 2 y 5 1y1, c, yn 2 Rn.

See Figure 1.23 for distance in and R2, R3, Rn.

Rn: d 1x, y 2 5 "1x1 2 y1 2 2 1 c1 1xn 2 yn 2 2

R3: d 1x, y 2 5 "1x1 2 y1 2 2 1 1x2 2 y2 2 2 1 1x3 2 y3 2 2

R2: d 1x, y 2 5 "1x1 2 y1 2 2 1 1x2 2 y2 2 2

O O

R2 R3 Rnx(x1, x2)

y(y1, y2) y(y1, y2, y3)y(y1, y2, ..., yn)

x(x1, x2, ..., xn)

x(x1, x2, x3)

Figure 1.23

Determine the distance between the points andin

SOLUTION

Applying the above formula for distance, we get

The norms of vectors and distances between points have certain mathematical proper-ties, which we list on the next page. The following is an important property of distance,for example.

5 "74

5 "9 1 4 1 36 1 25

d 1x, y 2 5 "11 2 4 2 2 1 122 2 0 2 2 1 13 1 3 2 2 1 10 2 5 2 2

y 5 14, 0, 23, 5 2 R4.EXAMPLE 5 x 5 11, 22, 3, 0 2

Having defined angles between vectors and magnitudes of vectors, we now complete thegeometry of a vector space by defining distances between points.

Distance Between Points

The distance between the points and in isWe define distance in by generalizing this expression as

follows."1x1 2 y1 2 2 1 1x2 2 y2 2 2. Rn

x 5 1x1, x2 2 y 5 1y1, y2 2 R2 d 1x, y 2 5

9781284120097_CH01.qxd 9/15/17 9:58 AM Page 51























Prove that distance in has the following symmetric property:

SOLUTION

Let and We get

This result tells us that the distance from x to y is the same as the distance from y to x.This property is one we would naturally want a distance function to have.

We now summarize these geometrical structures of and their properties. This geometryis called Euclidean geometry.

Euclidean Geometry of

Dot product of vectors u and v:

Norm of a vector u:

Angle between vectors u and v:

Distance between points x and y:

Properties of Norm Properties of Distance

1. 1. (the length of a vector (the distance between two points cannot be negative) cannot be negative)

2. if and only if 2. if and only if (the length of a vector is zero if (the distance between two points is zeroand only if the vector is the if and only if the points are coincident)zero vector)

3. 3. (symmetry property)(the length of cu is times (the distance between x and y is the samethe length of u) as the distance between y and x)

In the exercises that follow, you are asked to derive those properties not already dis-cussed. The norms and distances introduced in this section were all derived from the dotproduct. In numerical work, it is often convenient to define other norms and distancesusing these properties as axioms—basic properties that any norm or distance functionshould have. The reader will meet some of these ideas in later sections.

We complete this section with three useful theoretical results. The first of these is theCauchy-Schwarz inequality. This inequality enabled us earlier to extend the definitionof angle that we have in to Here it enables us to extend the Pythagorean Theoremto Rn.

R2 Rn.

0 c 0icui 5 0 c 0 iui d 1x, y 2 5 d 1y, x 2

iui 5 0 u 5 0 d 1x, y 2 5 0 x 5 y

iui $ 0 d 1x, y 2 $ 0

d 1x, y 2 5 "1x1 2 y1 2 2 1 c1 1xn 2 yn 2 2

cosu 5u # v

iui ivi

iui 5 "1u1 2 2 1 c1 1un 2 2


Rn

Rn

5 "1y1 2 x1 2 2 1 c1 1yn 2 xn 2 2 5 d 1y, x 2d 1x, y 2 5 "1x1 2 y1 2 2 1 c1 1xn 2 yn 2 2

x 5 1x1, c, xn 2 y 5 1y1, c, yn 2 .

d 1x, y 2 5 d 1y, x 2 .EXAMPLE 6 Rn

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THEOREM 1.5

The Cauchy-Schwarz Inequality*

If u and v are vectors in then

Here denotes the absolute value of the number

Proof We first look at the special case when If then andEquality holds.

Let us now consider The proof of the inequality for this case involves a veryinteresting application of the quadratic formula. Consider the vector wherer is any real number. Using the properties of the dot product, we get

The properties of the dot product also tell us that

Combining these results, we get

Let Thus

This implies that the quadratic function is never negative.Therefore whose graph is a parabola, must have either one zero or no zeros. Thezeros of are the roots of the equation

The discriminant gives information about these zeros. There is one zero ifand no zeros if Thus

Taking the square root of both sides gives the Cauchy-Schwarz inequality.

0 u # v 0 # iui ivi

1u # v 2 2 # iui 2 ivi 2

1u # v 2 2 # 1u # u 2 1v # v 212u # v 2 2 # 4 1u # u 2 1v # v 2

b2 # 4ac

b2 2 4ac # 0

b2 2 4ac , 0.b2 2 4ac 5 0

ar2 1 br 1 c 5 0

f 1r 2f 1r 2 , f 1r 2 5 ar2 1 br 1 c

ar2 1 br 1 c $ 0

a 5 u # u, b 5 2u # v, c 5 v # v.

r2 1u # u 2 1 2r 1u # v 2 1 v # v $ 0

1ru 1 v 2 # 1ru 1 v 2 $ 0

1ru 1 v 2 # 1ru 1 v 2 5 r2 1u # u 2 1 2r 1u # v 2 1 v # v

ru 1 v,u 2 0.

iui ivi 5 0.u 5 0. u 5 0, 0 u # v 0 5 0

0 u # v 0 u # v.

0 u # v 0 # iui ivi

Rn,

* Augustin Louis Cauchy (1789–1857) was educated in Paris. He worked as an engineer before returning to teach in Paris.He was one of the greatest mathematicians of all time. He was also very prolific, writing seven books and over eight hun-dred papers. More mathematical concepts and results have been named after him than any other mathematician. Cauchy hada very active life outside mathematics. After the revolution of 1830 he refused to take an oath of allegiance to Louis-Phillipeand went into exile in Turin, but was later able to return to Paris. Cauchy was a devout Catholic and was involved in socialwork. He was, for example, involved in charities for unwed mothers, aid for the starving in Ireland, and the rehabilitation ofcriminals. On giving one year’s salary to the mayor of Sceaux for charity he commented, “Do not worry, it is only my salary;it is not my money, it is the emperor’s.”

Herman Amandus Schwarz (1843–1921) was educated in Berlin and taught in Zurich, Göttingen, and Berlin. His great-est asset was his geometric intuition. The methods he developed have turned out to be more significant than the actual prob-lems he worked on. His teaching and his concern for students did not leave him much time for publishing. Another factorthat may have contributed to his lack of publishing was a weakness he had for devoting as much thoroughness to the trivialas to the important.

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We now extend two familiar geometrical results from to

THEOREM 1.6

Let u and v be vectors in (a) Triangle inequality:

This inequality tells us that the length of one side of a triangle cannot exceed the sumof the lengths of the other two sides. See Figure 1.24(a).

(b) Pythagorean Theorem: If then

The square of the hypotenuse of a right triangle is equal to the sum of the squares ofthe other two sides. See Figure 1.24(b).

Proof(a) By the properties of dot product and norm,

The Cauchy-Schwarz inequality implies that

Taking the square root of each side gives the triangle inequality.

(b) By the properties of dot product, and the fact that we get

proving the Pythagorean Theorem.

5 iui 2 1 ivi 2

5 u # u 1 2u # v 1 v # v

iu 1 vi 2 5 1u 1 v 2 # 1u 1 v 2u # v 5 0,

5 1 iui 1 ivi 2 2

iu 1 vi 2 # iui 2 1 2 iui ivi 1 ivi 2

# iui 2 1 2 0 u # v 0 1 ivi 2

5 iui 2 1 2u # v 1 ivi 2

5 u # u 1 2u # v 1 v # v

5 u # u 1 u # v 1 v # u 1 v # v

iu 1 vi 2 5 1u 1 v 2 # 1u 1 v 2

u # v 5 0, iu 1 vi 2 5 iui 2 1 ivi 2.

iu 1 vi # iui 1 iviRn.

R2 Rn.

u + v u + v

||v|| ||v||

||u||

||u + v||

||u + v|| � ||u|| + ||v||Triangle inequality

(a)

||u + v||2 = ||u||2 + ||v||2

Pythagorean Theorem

(b)

||u + v||

u ||u|| u

v v

Figure 1.24

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EXERCISE SET 1.6

Dot Product1. Determine the dot products of the following pairs of vectors.

(a) (b)(c) (d)

2. Determine the dot products of the following pairs of vectors.

(a) (b)(c)(d)

3. Determine the dot products of the following pairs of vectors.

(a) (b)(c)(d)(e)

4. Determine the dot products of the following pairs of col-umn vectors.

(a) (b)

(c) (d)

Norms of Vectors5. Find the norms of the following vectors.

(a) (b)(c) (d)(e)

6. Find the norms of the following vectors.

(a) (b)(c) (d)(e)


(a) (b)(c) (d)(e) (f)

8. Find the norms of the following column vectors.

(a) (b) (c)

(d) (e)

Normalizing Vectors9. Normalize the following vectors.

(a) (b)(c) (d)(e)

10. Normalize the following vectors.

(a) (b)(c) (d)(e)

11. Normalize the following column vectors.

(a) (b) (c)

(d) (e)

Angles Between Vectors12. Determine the angles between the following pairs of vectors.

(a) (b)(c) (d)

13. Determine the cosines of the angles between the followingpairs of vectors.

(a) (b)(c) (d)(e)

14. Determine the cosines of the angles between the followingpairs of column vectors.

(a) (b)

(c) (d)

Orthogonality15. Show that the following pairs of vectors are orthogonal.

(a) (b)(c) (d)

16. Show that the following pairs of vectors are orthogonal.

(a) (b)(c)

(d)

(e) 11, 21, 2, 25, 9 2 , 14, 7, 4, 1, 0 215, 1, 0, 2 2 , 123, 7, 9, 4 217, 1, 0 2 , 12, 214, 3 213, 25 2 , 15, 3 2 11, 2, 23 2 , 14, 1, 2 2

13, 0 2 , 10, 22 2 17, 21 2 , 11, 7 211, 3 2 , 13, 21 2 122, 4 2 , 14, 2 2

£1

23

0

§ , £2

5

21

§ £22

3

24

§ , £2

5

21

§

c12d , c21

4d c5

1d , c 0

23d

11, 2, 21, 3, 1 2 , 12, 0, 1, 0, 4 212, 21, 0 2 , 15, 3, 1 2 17, 1, 0, 0 2 , 13, 2, 1, 0 214, 21 2 , 12, 3 2 13, 21, 2 2 , 14, 1 ,1 2

12, 3 2 , 13, 22 2 15, 2 2 , 125, 22 2121, 1 2 , 10, 1 2 12, 0 2 , 11, "3 2

£21

2

25

§ ≥3

0

1

8

¥

c43d c 1

23d £

3

4

0

§

10, 0, 0, 7, 0, 0 217, 2, 0, 1 2 13, 21, 1, 2 214, 2 2 14, 1, 1 2

10, 5, 0 211, 2, 3 2 122, 4, 0 211, 3 2 12, 24 2

£22

0

5

§ ≥2

3

5

9

¥

c34d c 2

27d £

1

2

3

§

123, 0, 1, 4, 2 2 10, 0, 0, 7, 0, 0 211, 2, 3, 4 2 14, 22, 1, 3 215, 2 2 124, 2, 3 2

17, 22, 23 215, 1, 1 2 10, 5, 0 211, 3, 21 2 13, 0, 4 2

10, 27 214, 0 2 123, 1 211, 2 2 13, 24 2

£2

0

25

§ , £3

6

24

§ £1

3

27

§ , £22

8

23

§

c13d , c22

5d c5

0d , c 4

26d

11, 2, 3, 0, 0, 0 2 , 10, 0, 0, 22, 24, 9 212, 3, 24, 1, 6 2 , 123, 1, 24, 5, 21 217, 1, 2, 24 2 , 13, 0, 21, 5 215, 1 2 , 12, 23 2 123, 1, 5 2 , 12, 0, 4 2

13, 2 ,0 2 , 15, 22, 8 217, 1, 22 2 , 13, 25, 8 211, 2, 3 2 , 14, 1, 0 2 13, 4, 22 2 , 15, 1, 21 2

12, 0 2 , 10, 21 2 15, 22 2 , 123, 24 212, 1 2 , 13, 4 2 11, 24 2 , 13, 0 2

9781284120097_CH01.qxd 9/15/17 10:07 AM Page 55























17. Show that the following pairs of column vectors areorthogonal.

(a) (b)

(c) (d)

18. Determine nonzero vectors that are orthogonal to the fol-lowing vectors.

(a) (b)(c) (d)

19. Determine nonzero vectors that are orthogonal to the fol-lowing vectors.

(a) (b)(c) (d)(e) (f)

20. Determine a vector that is orthogonal to both and

Subspaces21. Let W be the subspace of vectors in that are orthogonal

to Find a basis for W. What is the dimen-sion of this subspace? Give a geometrical description of thesubspace.

22. Let W be the subspace of vectors in that are orthogonalto Find a basis for W. What is the dimen-sion of this subspace? Give a geometrical description of thesubspace.


24. Let W be the subspace of vectors in that are orthogonalto Find a basis for W. What is the dimen-sion of this subspace?

Distances Between Points25. Find the distances between the following pairs of points.

(a)(b)(c)(d)

26. Find the distances between the following pairs of points.

(a)(b)(c)(d)(e)

Miscellaneous Results27. Prove the following two properties of the dot product.

(a)(b)

28. Prove that if v is a nonzero vector, then the following vec-tor u is a unit vector in the direction of v.

29. Prove that two nonzero vectors u and v are orthogonal ifand only if

30. Show that if v and w are two vectors in a vector space U andfor all vectors u in U, then

31. Let be vectors in a given vector space. Letbe scalars. Prove that

32. Let u, v, and w be vectors in a given Euclidean space, andlet c and d be nonzero scalars. Tell whether each of the fol-lowing expressions is a scalar, is a vector, or is not a validexpression.

(a) (b)(c) (d)(e) (f)(g) (h)

33. Find all the values of c such that

34. Prove that u and v are orthogonal vectors if and only if

35. Let be a vector in Prove that the vectoris orthogonal to

36. Let u and v be vectors in Prove that if andonly if and are orthogonal.

37. Let u be a vector in and c a scalar. Prove that the normof a vector has the following properties.

(a)(b) if and only if

(c)

38. Consider the vector space Let be avector in Prove that both the following satisfy the prop-erties of norm mentioned in Exercise 37. These expressions,even though they do not lead to Euclidean geometry, haveall the algebraic properties we expect a norm to have andare used in numerical mathematics.

(a) sum of magnitudes norm

(b) maximum magnitudenorm

(c) Compute these two norms for the vectors11, 2 2 , 123, 4 2 , 11, 2, 25 2 , 10, 22, 7 2 .

iui 5 maxi51cn

0 un 0iui 5 0 u1 0 1 c1 0 un 0

Rn.Rn. u 5 1u1, c, un 2

icui 5 0 c 0 iuiiui 5 0 u 5 0iui $ 0

Rn

u 1 v u 2 vRn. iui 5 ivi

1a, b 2 .1a, b 2 R2. 12b, a 2

iu 1 vi 2 5 iui 2 1 ivi 2

ic 13, 0, 4 2 i 5 15.

cu # dv 1 iwiv iu # vic 1u # v 2 1 dw iu 1 cvi 1 d

u # v 1 cw u # v 1 c

1u # v 2w 1u # v 2 # w

u # 1a1v1 1 c1 anvn 2 5 a1u # v1 1 c1 anu # vn

a1, c, an

u, v1, c, vn

u # v 5 u # w v 5 w.

u # v 5 0.

u 51

iviv

cu # v 5 c 1u # v 2 5 u # cv

1u 1 v 2 # w 5 u # w 1 v # w

123, 1, 1, 0, 2 2 , 12, 1, 4, 1, 21 215, 1, 0, 0 2 , 12, 0, 1, 3 2123, 1, 2 2 , 14, 21, 1 211, 2, 3 2 , 12, 1, 0 214, 1 2 , 12, 23 2

11, 23 2 , 15, 1 217, 23 2 , 12, 2 213, 1 2 , 124, 0 216, 5 2 , 12, 2 2

w 5 11, 23, 7, 4 2 . R4

w 5 11, 22, 5 2 . R3

w 5 123, 4, 1 2 . R3

w 5 121, 1, 1 2 . R3

13, 1, 0 2 .11, 2, 21 2

16, 21, 2, 3 2 10, 22, 3, 1, 5 215, 1, 21 2 15, 0, 1, 1 215, 21 2 11, 22, 3 2

124, 21 2 123, 0 211, 3 2 17, 21 2

£4

21

0

§ , £2

8

21

§ £22

3

2

§ , £2

6

27

§

c12d , c26

3d c 5

22d , c 4

10d

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1.7 Curve Fitting, Electrical Networks, and Traffic Flow 57

Curve Fitting, Electrical Networks, and Traffic Flow Systems of linear equations are used in such diverse fields as electrical engineering, eco-nomics, and traffic analysis. We now discuss applications in some of these fields.

Curve FittingThe following problem occurs in many different branches of science. A set of data points

is given and it is necessary to find a polynomial whose graph passes through the points.The points are often measurements in an experiment. The x-coordinates are called basepoints. It can be shown that if the base points are all distinct, then a unique polynomial ofdegree (or less)

can be fitted to the points. See Figure 1.25.

y 5 a0 1 a1x 1 c1 an22xn22 1 an21x

n21

n 2 1

1x1, y1 2 , 1x2, y2 2 , c, 1xn, yn 2

*1.7

39. Let x, y, and z be points in Prove that distance has thefollowing properties.

(a)(b) if and only if

(c)

These properties are used as axioms to generalize the con-cept of distance for certain spaces.

d 1x, z 2 # d 1x, y 2 1 d 1y, z 2d 1x, y 2 5 0 x 5 yd 1x, y 2 $ 0

Rn.

y

O x

(x2, y2)(x3, y3)

(x1, y1)

(xn, yn)

Fitting a graph to data points

Figure 1.25

*Sections and chapters marked with an asterisk are optional. The instructor can use these sections to build around the corematerial to give the course the desired flavor.

The coefficients of the appropriate polynomial can be found bysubstituting the points into the polynomial equation and then solving a system of linearequations. (It is usual to write the polynomial in terms of ascending powers of x for the pur-pose of finding these coefficients. The columns of the matrix of coefficients of the systemof equations then often follow a pattern. More will be said about this later.)

We now illustrate the procedure by fitting a polynomial of degree two, a parabola, toa set of three such data points.

a0, a1, c, an22, an21

9781284120097_CH01.qxd 9/15/17 10:11 AM Page 57























Determine the equation of the polynomial of degree two whose graphpasses through the points (1, 6), (2, 3), (3, 2).

SOLUTION

Observe that in this example we are given three points and we want to find a polynomialof degree two (one less than the number of data points). Let the polynomial be

We are given three points and shall use these three sets of information to determine thethree unknowns and Substituting

in turn into the polynomial leads to the following system of three linear equations inand

Solve this system for and using Gauss-Jordan elimination.

We get The parabola that passes through these points isSee Figure 1.26.y 5 11 2 6x 1 x2.

a0 5 11, a1 5 26, a2 5 1.

<R1 1 12 2R3

R2 1 123 2R3

£1 0 0 11

0 1 0 26

0 0 1 1

§

<R1 1 121 2R2

R3 1 122 2R2

£1 0 22 9

0 1 3 23

0 0 2 2

§ <1 1

2 2R3£

1 0 22 9

0 1 3 23

0 0 1 1

§

£1 1 1 6

1 2 4 3

1 3 9 2

§<

R2 1 121 2R1

R3 1 121 2R1

£1 1 1 6

0 1 3 23

0 2 8 24

§

a2, a1, a0

a0 1 3a1 1 9a2 5 2

a0 1 2a1 1 4a2 5 3

a0 1 2a1 1 2a2 5 6

a2.a0, a1,

x 5 1, y 5 6; x 5 2, y 5 3; x 5 3, y 5 2

a0, a1, a2.

y 5 a0 1 a1x 1 a2x2

EXAMPLE 1

y

O x

10

5

1 2 3 4

(1, 6)

(2, 3)(3, 2)

y = 11 – 6x + x2

Figure 1.26

9781284120097_CH01.qxd 9/15/17 10:12 AM Page 58























Electrical Network AnalysisSystems of linear equations are used to determine the currents through various branches ofelectrical networks. The following two laws, which are based on experimental verificationin the laboratory, lead to the equations.

Kirchhoff’s Laws*1. Junctions: All the current flowing into a junction must flow out of it.2. Paths: The sum of the IR terms (I denotes current, R resistance) in any direction

around a closed path is equal to the total voltage in the path in that direction.

Consider the electrical network of Figure 1.27. Let us determine the cur-rents through each branch of this network.EXAMPLE 2

*Gustav Robert Kirchhoff (1824–1887) was educated at the University of Konigsberg and did most of his teaching at theUniversity of Heidelberg. His most important contributions were in the discovery and analysis of the laws of electromag-netic radiation. Kirchhoff was an excellent teacher and writer whose books influenced teaching in German universities.Kirchhoff was described as “not easily drawn out but of a cheerful and obliging disposition.”

8 volts

16 volts

2 ohms

1 ohm

4 ohms

2 ohms

B D

I2I2

I3I3

I1I1

A

C

Figure 1.27

SOLUTION

The batteries (denoted ) are 8 volts and 16 volts. The following convention is used inelectrical engineering to indicate the terminal of the battery out of which the currentflows: . The resistances (denoted ) are one 1-ohm, one 4-ohm, and two 2-ohm.The current entering each battery will be the same as that leaving it.

Let the currents in the various branches of the above circuit be and Kirchhoff’slaws refer to junctions and closed paths. There are two junctions in this circuit, namelythe points B and D. There are three closed paths, namely ABDA, CBDC, and ABCDA.Apply the laws to the junctions and paths.

Junctions:

Junction B,

Junction D,

These two equations result in a single linear equation

Paths:

Path ABDA,

Path CBDC, 4I2 1 1I3 5 16

2I1 1 1I3 1 2I1 5 8

I1 1 I2 2 I3 5 0

I3 5 I1 1 I2

I1 1 I2 5 I3

I1, I2, I3.

9781284120097_CH01.qxd 9/15/17 10:13 AM Page 59























It is not necessary to look further at path ABCDA. We now have a system of three linearequations in three unknowns, and Path ABCDA in fact leads to an equation thatis a combination of the last two equations; there is no new information.

The problem thus reduces to solving the following system of three linear equationsin three variables.

Using the method of Gauss-Jordan elimination, we get

The currents are The units are amps. The solution is unique, asis to be expected in this physical situation.

Determine the currents through the various branches of the electrical net-work in Figure 1.28. This example illustrates how one has to be conscious of directionin applying Law 2 for closed paths.

EXAMPLE 3

I1 5 1, I2 5 3, I3 5 4.

<1 1

6 2R3£

1 0 14 2

0 1 254 22

0 0 1 4

§<

R1 1 1214 2R3

R2 1 1 54 2R3

£1 0 0 1

0 1 0 3

0 0 1 4

§

<121

4 2R2£

1 1 21 0

0 1 254 22

0 4 1 16

§<

R1 1 121 2R2

R3 1 124 2R2

£1 0 1

4 2

0 1 254 22

0 0 6 24

§

£1 1 21 0

4 0 1 8

0 4 1 16

§ <R2 1 124 2R1

£1 1 21 0

0 24 5 8

0 4 1 16

§

4I2 1 4I2 1 I3 5 16

4I1 1 4I1 1 I3 5 18

4I1 1 4I2 2 I3 5 10

I1, I2, I3.

Figure 1.28

16 volts

2 ohms

2 ohmsD

I2I2

I3

A

C

12 volts1 ohm

I1I1

B

I3

9781284120097_CH01.qxd 9/15/17 10:14 AM Page 60























SOLUTION

Junctions:

Junction B,

Junction D,

giving

Paths:

Path ABCDA,

Path ABDA,

Observe that we have selected the direction ABDA around this last path. The currentalong the branch BD in this direction is and the voltage is We now have threeequations in the three variables and

Solving these equations, we get amps.

In practice, electrical networks can involve many resistances and circuits; determiningcurrents through branches involves solving large systems of equations on a computer.

Traffic Flow Network analysis, as we saw in the previous discussion, plays an important role in electri-cal engineering. In recent years, the concepts and tools of network analysis have been foundto be useful in many other fields, such as information theory and the study of transporta-tion systems. The following analysis of traffic flow that was mentioned in the introductionillustrates how systems of linear equations with many solutions can arise in practice.

Consider the typical road network of Figure 1.29. It represents an area of downtownJacksonville, Florida. The streets are all one-way with the arrows indicating the directionof traffic flow. The traffic is measured in vehicles per hour (vph). The figures in and out ofthe network given here are based on midweek peak traffic hours, 7 A.M. to 9 A.M. and 4 P.M.to 6 P.M. Let us construct a mathematical model that can be used to analyze the flowwithin the network.

Assume that the following traffic law applies.

All traffic entering an intersection must leave that intersection.

This conservation of flow constraint (compare it to the first of Kirchhoff’s laws for elec-trical networks) leads to a system of linear equations. These are, by intersection:

A: Traffic in Traffic out Thus B: Traffic in Traffic out Thus C: Traffic in Traffic out Thus D: Traffic in Traffic out Thus x2 1 x3 5 1050.5 x2 1 x3.5 800 1 250.

5 x3 1 x4. 5 600 1 300. x3 1 x4 5 900.5 350 1 125. 5 x1 1 x4. x1 1 x4 5 475.5 x1 1 x2. 5 400 1 225. x1 1 x2 5 625.

x1, c, x4

I1 5 2, I2 5 3, I3 5 5

I1 2 2I2 2 2I2 5 24

I1 1 2I1 1 2I3 5 112

I1 1 2I2 2 2I3 5 20

I1, I2, I3.2I2, 216.

1I1 1 2 12I2 2 5 12 1 1216 21I1 1 2I3 5 12

I1 1 I2 2 I3 5 0.

I3 5 I1 1 I2

I1 1 I2 5 I3

9781284120097_CH01.qxd 9/15/17 3:29 PM Page 61























The constraints on the traffic are described by the following system of linear equations.

The method of Gauss-Jordan elimination is used to solve this system of equations. The aug-mented matrix and reduced echelon form of the preceding system are as follows:

The system of equations that corresponds to this reduced echelon form is

Expressing each leading variable in terms of the remaining variable, we get

As was perhaps to be expected, the system of equations has many solutions—there aremany traffic flows possible. One does have a certain amount of choice at intersections.Let us now use this mathematical model to arrive at information. Suppose it becomes nec-essary to perform road work on the stretch DC of Monroe Street. It is desirable to have assmall a flow as possible along this stretch of road. The flows can be controlled alongvarious branches by means of traffic lights. What is the minimum value of along DCx3

x3

x3 5 2x4 1 900

x2 5 2x4 1 150

x1 5 2x4 1 475

x3 1 x3 1 x3 1 x4 5 900

x2 2 x2 2 x2 2 x4 5 150

x1 1 x1 1 x1 1 x4 5 475

≥1 1 0 0 625

1 0 0 1 475

0 0 1 1 900

0 1 1 0 1050

¥ < c< ≥1 0 0 1 475

0 1 0 21 150

0 0 1 1 900

0 0 0 0 0

¥

x2 1 x2 1 x3 1 x3 5 1050

x3 1 x3 1 x3 1 x4 5 1900

x1 1 x1 1 x1 1 x4 5 1475

x1 1 x2 1 x2 1 x2 5 1625

D

A

C

Duval Street

Monroe Street

Downtown Jacksonville, FL

B

Hogan

Str

eet

Lau

ra S

tree

t

225 vph

250 vph 600 vph

125 vph400 vph

800 vph 300 vph

350 vph

N

x2 x4

x1

x3

Figure 1.29

9781284120097_CH01.qxd 9/15/17 3:32 PM Page 62























that would not lead to traffic congestion? We use the preceding system of equations toanswer this question.

All traffic flows must be nonnegative (a negative flow would be interpreted as trafficmoving in the wrong direction on a one-way street). The third equation tells us that willbe a minimum when is as large as possible, as long as it does not go above 900. The largestvalue can be without causing negative values of or is 475. Thus the smallest valueof is or 425. Any road work on Monroe should allow for at least 425 vph.

In practice, networks are much vaster than the one discussed here, leading to larger sys-tems of linear equations that are handled on computers. Various values of variables can befed in and different scenarios created.

x3 2475 1 900,x4 x1 x2

x4

x3

EXERCISE SET 1.7

Curve FittingIn Exercises 1–5, determine the equations of the polynomials ofdegree two whose graphs pass through the given points.

1.2.3.4. What is the value of y when

5. (0, 1), What is the value of y when

6. Find the equation of the polynomial of degree three whosegraph passes through the points

Electrical NetworksIn Exercises 7–14, determine the currents in the various branchesof the electrical networks. The units of current are amps and theunits of resistance are ohms. (Hint: In Exercise 14 it is difficultto decide the direction of the current along AB. Make a guess.A negative result for the current means that your guess was thewrong one—the current is in the opposite direction. However,the magnitude will be correct. There is no need to rework theproblem.)

7.

8.

9.

10.

14, 33 2 .11, 23 2 , 12, 21 2 , 13, 9 2 ,

x 5 3?121, 21 2 , 11, 23 2 .11, 8 2 , 13, 26 2 , 15, 60 2 . x 5 2?

11, 5 2 , 12, 7 2 , 13, 9 211, 14 2 , 12, 22 2 , 13, 32 211, 2 2 , 12, 2 2 , 13, 4 2

34

28

4

2

1 3

I3

I2

I3

I2

I1I1

3 17

1

2

9

I3 I3

I2

I1I1

9

13

1

4

2

I3

I2

I1I1

4

2

2

1 2

3

I3

I2

I1I1

9781284120097_CH01.qxd 9/15/17 10:17 AM Page 63























11.

12.

13.

14.

15. Determine the currents through the various branches of theelectrical network in Figure 1.30.

(a) when battery C is 9 volts

(b) when battery C is 23 volts

Note how the current through the branch AB is reversed in(b). What would the voltage of C have to be for no currentto pass through AB?

Traffic Flow16. Construct a system of linear equations that describes the

traffic flow in the road network of Figure 1.31. All streetsare one-way streets in the directions indicated. The units arevehicles per hour. Give two distinct possible flows of traf-fic. What is the minimum possible flow that can be expectedalong branch AB?

17. Figure 1.32 represents the traffic entering and leaving a“roundabout” road junction. Such junctions are very com-mon in Europe. Construct a system of equations that describesthe flow of traffic along the various branches. What is theminimum flow possible along the branch BC? What are theother flows at that time? (Units are vehicles per hour.)

31

3

1

7

I3

I2

I1

4

1

1

I3

I2

I1

1

I4

I5

14

1I3

I2

I1

2

22

I4

I5

4

19

1

2

I2

I1

A B

16 volts

1 ohm

2 ohms 3 ohms

5 ohms

A B

C

Figure 1.30

Figure 1.31

Figure 1.32

200

100

100

100

50

50

150

50

x2x4

x1

x3

A B

CD

200

100 150

50

x2

x4

x1

x3

A

B

C

D

9781284120097_CH01.qxd 9/15/17 10:17 AM Page 64






















CHAPTER 1 Review Exercises 65

18. Figure 1.33 represents the traffic entering and leaving anothertype of roundabout road junction in Continental Europe.Such roundabouts ensure the continuous smooth flow oftraffic at road junctions. Construct linear equations thatdescribe the flow of traffic along the various branches. Usethese equations to determine the minimum flow possiblealong What are the other flows at that time? (It is notnecessary to compute the reduced echelon form. Use thefact that traffic flow cannot be negative.)

19. Figure 1.34 describes a flow of traffic, the units being vehi-cles per hour.

(a) Construct a system of linear equations that describesthis flow.

(b) The total time it takes the vehicles to travel anystretch of road is proportional to the traffic alongthat stretch. For example, the total time it takes vehicles to traverse AB is minutes. Assuming thatthe constant is the same for all sections of road,the total time for all these 200 vehicles to be in thisnetwork is Whatis this total time if Give the average time foreach car.

20. There will be many polynomials of degree 2 that pass throughthe points (1, 2) and (3, 4). The situation can be describedby a system of two linear equations in three variables thathas many solutions. Find an equation (involving a parame-ter) that represents this family of polynomials. Determinethe polynomials that open up and those that open down.

21. There will be many polynomials of degree 3 that pass throughthe points (1, 2), (3, 4), and (4, 8). The situation can bedescribed by a system of three linear equations in four vari-ables that has many solutions. Find an equation (involvinga parameter) that represents this family of polynomials.Determine the unique polynomial of degree 3 passing throughthese points for which the coefficient of is 1. x3

k 5 4?kx1 1 2kx2 1 kx3 1 2kx4 1 kx5.

kx1

x1x1.

CHAPTER REVIEW EXERCISES*1

1. Give the sizes of the following matrices.

(a) (b)

(c) (d)

(e)

2. Give the (1, 3), (2, 1), (3, 3), (2, 5), (3, 6) elements of thefollowing matrix.

3. Write down the identity matrix

4. Determine the matrix of coefficients and the augmentedmatrix of each of the following systems of equations.

(a)

(b)

3x1 1 5x2 2 7x3 5 23

1x1 2 2x2 1 8x3 5 20

2x1 1 2x2 2 4x3 5 21

4x1 2 3x2 5 21

4x1 1 2x2 5 26

I5.

£3 2 0 7 8 4

6 7 4 2 1 0

0 2 5 7 8 9

§

≥8 5 3 27 5 9

22 3 5 7 0 2

4 23 5 1 2 3

0 28 21 5 3 8

¥

34 3 2 7 4 £22

3

6

§

c4 3 22

1 5 7d c0 2

4 6d

Figure 1.33

Figure 1.34

90

80 75

100

155130

110 120

x7

x2

x5x6

x4x8

x1x3

200 200

x2

x2

x1

x3

x4

x4

x5

A

F E

B C

D

*Answers to all review exercises are provided in the back of the book.

9781284120097_CH01.qxd 9/15/17 10:18 AM Page 65























(c)

(d)

(e)

5. Interpret the following matrices as augmented matrices ofsystems of equations. Write down each system of equations.

(a) (b)

(c) (d)

(e)

6. Determine whether the following matrices are in reducedechelon form. If a matrix is not in reduced echelon form,give a reason.

(a) (b)

(c) (d)

(e)

7. The following systems of equations have unique solutions.Solve these systems using the method of Gauss-Jordan elim-ination with matrices.

(a)

(b)

(c)

8. Solve (if possible) the following systems of equations usingthe method of Gauss-Jordan elimination.

(a)

(b)

9. Let A be an matrix in reduced echelon form. Showthat if then A has a row consisting entirely of zeros.

10. Let A and B be row equivalent matrices. Show that A and Bhave the same reduced echelon form.


(a) (b)(c)

12. Sketch the position vectors (2, 0, 0), (1, 4, 2)in

13. Compute the following linear combinations forand

(a) (b)(c) (d)(e)

14. Determine whether the vector (�1, 3, �5) is a linear com-bination of the vectors (1, 2, 0), (0, 1, 4), (�3, 2, 7).

15. Determine whether the sets defined by the following vec-tors are subspaces of

(a) (b)

16. Consider the following homogeneous system of linear equa-tions and its general solution. The solutions will form a sub-space W of Find a basis for W and give its dimension.

General solution is

17. Determine whether the following set of vectors is linearlyindependent: {(1, �1, 2), (1, 0, 1), (6, �2, 10)}.

18. Determine the dot products of the following pairs ofvectors.

(a) (1, 2), (b)(c)


(a) (b)(c) 11, 22, 3, 4 2

11, 24 2 122, 1, 3 2

12, 2, 25 2 , 13, 2, 21 213, 24 2 11, 22, 3 2 , 14, 2, 27 2

12r 1 s, 3s, r, s 2 .3x1 1 2x2 2 6x3 2 9x4 5 0

2x1 1 2x2 2 4x3 2 5x4 5 0

2x1 1 2x2 2 2x3 2 4x4 5 0

R4.

1a, b, a 1 2b 2 1a, b, a 1 4 2R3.

2u 2 5v 2 wu 2 2w 4u 2 2v 1 3wu 1 w 3u 1 v

u 5 13, 21, 5 2 , v 5 12, 3, 7 2 , w 5 10, 1, 23 2 .R3.

10, 21, 0 2 ,OC

>

5 14, 21 2OA

>

5 11, 4 2 OB>

5 122, 3 2R2.

A 2 In,n 3 n

21x2 1 1x2 1 12x3 2 3x4 5 210

21x1 1 2x2 1 14x3 1 2x2 5 210

22x1 2 5x2 2 10x3 1 3x4 5 210

21x1 1 3x2 1 16x3 2 2x4 5 127

24x1 2 2x2 1 10x3 5 110

22x1 1 3x2 1 10x3 5 28

21x1 2 1x2 1 11x3 5 23

3x1 2 2x2 2 2x2 2 6x4 5 24

1x1 2 1x2 2 1x3 2 4x4 5 29

1x1 2 1x2 1 1x3 1 5x4 5 112

2x3 1 6x2 1 2x3 1 6x4 5 121

1x1 1 2x2 2 11x3 5 125

2x1 2 6x2 2 16x3 5 246

1x1 2 2x2 2 16x3 5 217

3x1 1 7x2 5 2

2x1 1 4x2 5 2

£1 8 0 0 2

0 0 0 1 27

0 0 1 0 3

§

£1 2 0 6

0 1 0 27

0 0 1 9

§ £1 3 0 0 2

0 0 1 0 5

0 0 0 1 7

§

c1 0 4

0 1 7d £

1 3 0 5

0 0 1 9

0 0 0 0

§

£1 4 21 7

0 1 3 8

0 0 1 25

§

c1 2 3 4

5 0 23 6d £

1 0 0 5

0 1 0 28

0 0 1 2

§

c 4 2 0

23 7 8d c1 9 23

0 3 2d

21x212x2 1 2x3 1 3x4 5 25

21x1 1 5x2 1 5x2 2 6x4 5 20

22x1 1 3x2 2 8x3 1 5x4 5 22

x3 1 x3 1 x3 5 23

x2 1 x2 1 x2 5 25

x1 1 x1 1 x1 5 21

14x1 1 3x2 1 3x2 5 2513x1 2 2x2 1 5x3 5 23

2x1 1 2x2 2 7x3 5 22

9781284120097_CH01.qxd 9/15/17 10:23 AM Page 66






















CHAPTER 1 Review Exercises 67

20. Determine the cosines of the angles between the followingvectors.

(a) (2, 3) (b) (4, 1, 2)

21. Find a vector orthogonal to

22. Determine the distances between the following pairs ofpoints.

(a) (5, 3) (b) (3, 2, 1), (7, 1, 2)

(c) (4, 1, 6, 2)

23. Find all the values of c such that


25. Determine the equation of the polynomial of degree twowhose graph passes through the points (1, 3), (2, 6), (3, 13).

26. Determine the currents through the branches of the networkin Figure 1.35.

27. Consider the traffic entering and leaving a roundabout roadjunction in Britain, shown in Figure 1.36. (Observe that thetraffic goes around this roundabout in the opposite direc-tion to the one on the Continent in Exercise 18, Section 1.7.They drive on the left of the road in Britain.) Construct asystem of linear equations that describe the flow of trafficalong the various branches. Determine the minimum flowpossible along . What are the other flows at that time?x8

w 5 123, 5, 1 2 . R3

ic 11, 2, 3 2 i 5 196.

13, 1, 21, 2 2 ,11, 22 2 ,

121, 1 2 , 11, 2, 23 2 ,122, 1, 5 2 .

Figure 1.35

Figure 1.36

7

1

2

35

I3 I3

I2 I2

I1 I1

80

90 120

160

100100

130 80

x3

x8

x5x4

x6x2

x1x7

9781284120097_CH01.qxd 9/15/17 10:24 AM Page 67






















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