chapter 1 systems of linear equations and...

Chapter 1

Systems of Linear Equations andMatrices

System of linear algebraic equations and their solution constitute one of the major topicsstudied in the course known as “linear algebra.” In the first section we shall introducesome basic terminology and discuss a method for solving such system.

1.1 Systems of Linear Equations

Linear Equations

Any straight line in the xy-plane can be represented algebraically by an equation of theform

a1x+ a2y = b

where a1, a2, and b are real constants and a1 and a2 are not both zero. An equation ofthis form is called a linear equation in the variables x and y. More generally, we definea linear equation in the n veriables x1, x2, . . . , xn to be one that can be expressed inthe form

a1x1 + a2x2 + · · ·+ anxn = b

where a1, a2, . . . , an, and b are real constants. The variables in a linear equation aresometimes called unknowns. For example, the equations

3x+ y = 7, y = 15x+ 2z + 4, and x1 + 3x2 − 2x3 + 5x4 = 7

are linear. The equations

√x+ 3y = 2, 3x− 2y − 5z + yz = 4, and y = cosx

are not linear.

Observe that a linear equation does not involve any products or roots of variables. Allvariables occur only to the first power and do not appear as arguments for trigonometric,logarithmic, or exponential functions.

1

Prepared by Dr.Archara Pacheenburawana (MA131 Sec 750001) 2

A solution of a linear equation a1x1 + a2x2 + · · · + anxn = b is a sequence of nnumbers s1, s2, . . ., sn such that the equation satisfied when we substitute x1 = s1,x2 = s2, . . ., xn = sn. The set of all solutions of the equation is called its solution set

or sometimes the general solution of the equation.

Example 1.1 Find the solution set of

(a) 5x− 2y = 3 and

(b) x1 − 3x2 + 4x3 = 2.

Solution . . . . . . . . .

Linear Systems

A finite set of linear equations in the variables x1, x2, . . ., xn is called system of linear

equations or a linear system. A sequence of numbers s1, s2, . . ., sn is called asolution of the system if x1 = s1, x2 = s2, . . ., xn = sn is a solution of every equationin the system. For example, the system

4x1 − x2 + 3x3 = −1

3x1 + x2 + 9x3 = −4

the the solution x1 = 1, x2 = 2, and x3 = −1 since these values satisfy both equations.However, x1 = 1, x2 = 8, x3 = 1 is not a solution since these values satisfy only the firstequation in the system. Thus, not all system of linear equations have solutions.

A system of equations that has no solutions is said to be inconsistent ; if there isat least one solution of the system, it is called consistent. To illustrate the possibilitiesthat can occur in solving systems of linear equations, consider a general system of twolinear equations in the unknowns x and y:

a1x+ b1y = c1 (a1, b1 not both zero)

a2x+ b2y = c2 (a2, b2 not both zero)

The graphs of these equations are lines; call them `1 and `2. Since a point (x, y) lies ona line if and only if the number x and y satisfy the equation of the line, the solutions ofthe system of equations correspond to points of the intersection of `1 and `2. There arethree possibilities, illustrated in Figure 1.1

• The lines `1 and `2 may be parallel, in which case there is no intersection andconsequently no solution to the system.

• The lines `1 and `2 may intersect at only one point, in which case the system hasexactly one solution.

• The lines `1 and `2 may coincide, in which case there are infinitely many points ofintersection and consequently infinitely many solutions to the system.


x

y

`1 `2

(a) No solution

x

y

`1 `2

(b) One solution

x

y

`1 and `2

(c) Infinitely many solutions

Figure 1.1:

Although we have considered only two equations with two unknowns here, we willshow later that the same three possibilities hold for arbitrary linear systems:

Every system of linear equations has no solutions, or has exactly one solution,or has infinitely many solutions.

An arbitrary system of m linear equations in n unknowns can be written as

a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2 (1.1)...

......

...

am1x1 + am2x2 + · · ·+ amnxn = bm

where x1, x2, . . ., xn are the unknowns and the subscripted a’s and b’s denote constants.For example, a general system of three linear equations in four unknowns can be writtenas

a11x1 + a12x2 + a13x3 + a14x4 = b1

a21x1 + a22x2 + a23x3 + a24x4 = b2

a31x1 + a32x2 + a33x3 + a34x4 = b3

The double subscripting on the coefficients of the unknowns is a useful device thatis used to specify the location of the coefficient in the system. The first subscript onthe coefficient aij indicates the equation in which the coefficient occurs, and the secondsubscript indicates which unknown it multiplies. Thus, a12 is in the first equation andmultiplies unknown x2.


Augmented Matrices

A system of m linear equations in n unknowns can be abbreviated by writing only therectangular array of numbers

a11 a12 · · · a1n b1a21 a22 · · · a2n b2...

......

...am1 am2 · · · amn bm

.

This is called the augmented matrix for the system. For example, the augmentedmatrix for the system of equations

x1 + x2 + 2x3 = 9

2x1 + 4x2 − 3x3 = 1

3x1 + 6x2 − 5x3 = 0

is

1 1 2 92 4 −3 13 6 −5 0

.

Remark When constructing an augmented matrix, we must write the unknowns in thesame order in each equation, and the constants must be on the right.

Exercise 1.1

1. Which of the following are linear equations in x1, x2 and x3?

(a) x1 + 5x2 −√2x3 = 1 (b) x1 + 3x2 + x1x3 = 2

(c) x1 = −7x2 + 3x3 (d) x−21 + x2 + 8x3 = 5

(e) x3/51 − 2x2 + x3 = 4 (f) πx1 −

√2x2 +

13x3 = 71/3

2. Find the solution set of each of the following linear systems.

(a) 7x− 5y = 3 (b) 3x1 − 5x2 + 4x3 = 7

(c) −8x1 + 2x2 − 5x3 + 6x4 = 1 (d) 3v − 8w + 2x− y + 4z = 0

3. Which of the following are linear systems?

(a) x1 − 3x2 = x3 − 4x4 = 1− x1

x1 + x4 + x3 − 2 = 0

(b) 2x−√y + 3z = −1

x + 2y − z = 24x − y = −1

(c) 3x− xy = 1x+ 2xy − y = 0

(d) y = 2x− 1y = −x

(e) 2x1 − sin x2 = 3x2 = x1 + x3

−x1 + x2 − 3x3 = 0

(f) x1 + x2 + x3 = 1−2x2

1 − 2x3 = −13x2 − x3 = 2


4. Find the augmented matrix for each of the following systems of linear equations.

(a) 3x1 − 2x2 = −14x1 + 5x2 = 37x1 + 3x2 = 2

(b) 2x1 + 2x3 = 13x1 − x2 + 4x3 = 76x1 + x2 − x3 = 0

(c) x1 = 1x2 = 2

x3 = 3

(d) x1 + 2x2 − x4 + x5 = 13x2 + x3 − x5 = 2

x3 + 7x4 = 1

5. Find a system of linear equations corresponding to the augmented matrix.

(a)

2 0 03 −4 00 1 1

(b)

3 0 −2 57 1 4 −30 −2 1 7

(c)

[7 2 1 −3 51 2 4 0 1

]

(d)

1 0 0 0 70 1 0 0 −20 0 1 0 30 0 0 1 4

6. (a) Find a linear equation in the variables x and y that has the general solution

x = 5 + 2t, y = t.

(b) Show that

x = t, y = 12t− 5

2

is also the general solution of the equation in part (a).

7. Consider the system of equations

ax+ by = k

cx+ dy = `

ex+ fy = m

Indicate what we can say about the relative positions of the line ax + by = k,cx+ dy = `, and ex+ fy = m when

(a) the system has no solutions.

(b) the system has exactly one solution.

(c) the system has infinitely many solutions.

8. Show that the linear system

ax+ by = e

cx+ dy = f

where a, b, c, d, e, f are constants, has a unique solution if ad− bc 6= 0. Express thissolution in terms of a, b, c, d, e, and f .


Answer to Exercise 1.1

1. (a), (c), (f)

2. (a) x = 37+ 5

7t, y = t (b) x1 =

53s− 4

3t− 1

8, x2 = s, x3 = t

(c) x1 =14r − 5

8s+ 3

4t− 1

8, x2 = r, x3 = s, x4 = t

(d) v = 83q − 2

3r + 1

3s− 4

3t, w = q, x = r, y = s, z = t

3. (a), (d)

4. (a)

3 −2 −14 5 37 3 2

(b)

2 0 2 13 −1 4 76 1 −1 0

(c)

1 0 0 10 1 0 20 0 1 3

(d)

1 2 0 −1 1 10 3 1 0 −1 20 0 1 7 0 1

5. (a) 2x1 = 03x1 − 4x2 = 0

x2 = 1

(b) 3x1 − 2x3 = 57x1 + x2 + 4x3 = −3

−2x2 + x3 = 7

(c) 7x1 + 2x2 + x3 − 3x4 = 5x1 + 2x2 + 4x3 = 1

(d) x1 = 7x2 = −2

x3 = 3x4 = 4

6. (a) x− 2y = 5 (b) Let x = t; then t− 2y = 5. Solving for y yields y = 12t− 5

2.

7. (a) The line have no common point of intersection.

(b) The line intersect in exactly one point.

(c) The three lines coincide.

1.2 Elementary Operations

Elementary Row Operations

The basic method for solving a system of linear equations is to replace the given systemby a new system that has the same solution set but is easier to solve. This new system isgenerally obtained in a series of steps by applying the following three types of operationsto eliminate unknowns systematically:


1. Multiply any equation, say the i th, by a nonzero constant α. This is denoted byα ri.

2. Interchange of two equations, say the i th and j th. This is denoted by ri ↔ rj.

3. Add α times the i th equation to the j th equation. This is denoted by rj + α ri.(This keeps the i th equation unchanged.)

Since the rows of an augmented matrix correspond to the equations in the associatedsystem, these three operations correspond to the following operations on the rows of theaugmented matrix:

1. Multiply any row by a nonzero constant: α ri.

2. Interchange of two rows: ri ↔ rj.

3. Add a multiple of one row to another: rj + α ri.

These are called elementary row operations. The following illustrates how theseoperations can be used to solve system of linear equations.

Example 1.2 Solve the following system by using elementary row operations.

x+ y + 2z = 9

2x+ 4y − 3z = 1

3x+ 6y − 5z = 0

Solution In the left column we solve a system of linear equations by operating on theequations in the system, and in the right column we solve the same system by operatingon the rows of the augmented matrix.

x+ y + 2z = 92x+ 4y − 3z = 13x+ 6y − 5z = 0

1 1 2 92 4 −3 13 6 −5 0

Add −2 times the first equationto the second to obtain

Add −2 times the first rowto the second to obtain

x+ y + 2z = 92y − 7z = −17

3x+ 6y − 5z = 0

1 1 2 90 2 −7 −173 6 −5 0

Add −3 times the first equationto the third to obtain

Add −3 times the first rowto the third to obtain

x+ y + 2z = 92y − 7z = −173y − 11z = −27

1 1 2 90 2 −7 −170 3 −11 −27

Multiply the second equationby 1

2to obtain

Multiply the second rowby 1

2to obtain


x+ y + 2z = 9

y − 72z = −17

2

3y − 11z = −27

1 1 2 9

0 1 −72

−172

0 3 −11 −27

Add −3 times the second equationto the third to obtain

Add −3 times the second rowto the third to obtain

x+ y + 2z = 9

y − 72z = −17

2

−12z = −3

2

1 1 2 9

0 1 −72

−172

0 0 −12

−32

Multiply the third equationby −2 to obtain

Multiply the third rowby −2 to obtain

x+ y + 2z = 9

y − 72z = −17

2

z = 3

1 1 2 9

0 1 −72

−172

0 0 1 3

Add −1 times the second equationto the first to obtain

Add −1 times the second rowto the first to obtain

x + 112z = 35

2

y − 72z = −17

2

z = 3

1 0 112

352

0 1 −72

−172

0 0 1 3

Add −112times the third equation

to the first and 72times the third

equation to the second to obtain

Add −112times the third row

to the first and 72times the third

row to the second to obtain

x = 1y = 2

z = 3

1 0 0 10 1 0 20 0 1 3

The solution x = 1, y = 2, z = 3 is now evident. ♣

Echelon Forms

In previous example, we solved a linear system in the unknowns x, y, and z by reducingthe augmented matrix to the form

1 0 0 10 1 0 20 0 1 3

.

This is an example of a matrix that is in reduced row-echelon form. To be of thisform, a matrix must have the following properties:

1. If the row does not consist entirely of zeros, then the first nonzero number in therow is a 1. We call this a leading 1.


2. If there are any rows that consist entirely of zeros, then they are grouped togetherat the bottom of the matrix.

3. In any two successive rows that do not consist entirely of zeros, the leading 1 inthe lower row occurs farther to the right than the leading 1 in the higher row.

4. Each column that contains a leading 1 has zeros everywhere else in that column.

A matrix that has the first three properties is said to be in row-echelon matrix.(Thus, a matrix in reduced row-echelon form is of necessity in row-echelon form, but notconversely.)

The following matrices are in reduced row-echelon form.

1 0 0 40 1 0 70 0 1 −1

,

1 0 00 1 00 0 1

,

0 1 −2 0 00 0 0 1 30 0 0 0 00 0 0 0 0

,

[0 00 0

]

The following matrices are in row-echelon form.

1 4 −3 70 1 6 20 0 1 −1

,

1 0 00 1 00 0 0

,

0 1 2 6 00 0 1 −1 00 0 0 0 1

Example 1.3 Suppose that the augmented matrix for a linear equations has been reducedby row operations to the given reduced row-echelon form. Solve the system.

(a)

1 0 0 20 1 0 40 0 1 −3

(b)

1 0 0 3 −50 1 0 4 −10 0 1 2 6

(c)

1 0 2 −30 1 1 40 0 0 0

(d)

1 0 4 00 1 −5 00 0 0 1

Solution . . . . . . . . .

Elimination Methods

We have just seen how easy it is to solve a system of linear equations once its augmentedmatrix is in reduced row-echelon form.

• The process of using row operations to transform a linear system into one whoseaugmented matrix is in row-echelon form is called Gaussian elimination.

• The process of using row operations to transform a linear system into one whoseaugmented matrix is in reduced row-echelon form is called Gauss-Jordan elimi-nation.


Remark It can be show that every matrix has a unique reduced row-echelon form.

Example 1.4 Solve by Gauss-Jordan elimination.

x1 + 3x2 − 2x3 + 2x5 = 0

2x1 + 6x2 − 5x3 − 2x4 + 4x5 − 3x6 = −1

5x3 + 10x4 + 15x6 = 5

2x1 + 6x2 + 8x4 + 4x5 + 18x6 = 6

Solution . . . . . . . . .

Back-Substitution

It is sometimes preferable to solve a system of linear equations by using Gaussian elimi-nation to bring the augmented matrix into row-echelon form without continuing all theway to the reduced row-echelon form. When this is done, the corresponding system ofequations can be solved by a technique called back-substitution. The next exampleillustrates the idea.

Example 1.5 From Example 1.4, solve by Gaussian elimination and back-substitution.

Solution . . . . . . . . .

Example 1.6 Solve

x+ y + 2z = 9

2x+ 4y − 3z = 1

3x+ 6y − 5z = 0

by Gaussian elimination and back-substitution.

Solution . . . . . . . . .

Homogeneous Linear Systems

A system of linear equations is said to be homogeneous if the constant terms are allzero; that is the system has the form

a11x1 + a12x2 + · · ·+ a1nxn = 0

a21x1 + a22x2 + · · ·+ a2nxn = 0...

......

...

am1x1 + am2x2 + · · ·+ amnxn = 0

Every homogeneous system of linear equations is consistent, since all such systemshave x1 = 0, x2 = 0, . . ., xn = 0 as a solution. This solution is called the trivial

solution ; if there are other solutions, they are called nontrivial solution.Because a homogeneous linear system always has the trivial solution, there are only

two possibilities for its solutions:


• The system has only the trivial solution.

• The system has infinitely many solutions in addition to the trivial solution.

In the special case of a homogeneous linear system of two equations in two unknowns,say

a1x+ b1y = 0 (a1, b1 not both zero)

a2x+ b2y = 0 (a2, b2 not both zero)

the graphs of the equations are lines through the origin, and the trivial solution corre-sponds to the point of intersection at the origin.

x

y

a1x+ b1y = 0

a2x+ b2y = 0

(a) Only the trivial solution

x

y

a1x+ b1y = 0

anda2x+ b2y = 0

(b) Infinitely many solutions

There is one case in which a homogeneous system is assured of having nontrivialsolution — namely, whenever the system involves more unknowns than equations. Tosee why, consider the following example of four equations in five unknowns.

Example 1.7 Solve the following homogeneous system of linear equations by usingGauss-Jordan elimination.

2x1 + 2x2 − x3 + x5 = 0

−x1 − x2 + 2x3 − 3x4 + x5 = 0

x1 + x2 − 2x3 − x5 = 0

x3 + x4 + x5 = 0

Solution . . . . . . . . .

Theorem 1.1 A homogeneous system of linear equations with more unknowns thanequations has infinitely many solutions.

Remark Note that Theorem 1.1 applied only to homogeneous systems. A nonhomoge-neous system with more unknowns than equations need not be consistent; however, ifthe system is consistent, it will have infinitely many solutions.


Exercise 1.2

1. Which of the following 3× 3 matrices are in reduced row-echelon form?

(a)

1 0 00 1 00 0 1

(b)

0 1 01 0 00 0 0

(c)

0 1 00 0 10 0 0

(d)

1 0 00 0 10 0 0

(e)

1 0 00 0 00 0 1

(f)

1 1 00 1 00 0 0

(g)

1 0 20 1 30 0 0

(h)

0 0 00 0 00 0 0

2. Which of the following 3× 3 matrices are in row-echelon form?

(a)

1 0 00 1 00 0 1

(b)

1 3 00 0 10 0 0

(c)

1 1 10 1 20 0 3

(d)

1 0 00 1 00 2 0

(e)

1 4 60 0 10 1 3

(f)

1 1 00 1 00 0 0

(g)

1 3 40 0 10 0 0

(h)

1 2 30 0 00 0 1

3. In each part determine whether the matrix is in row-echelon form, reduced row-echelon form, both, or neither.

(a)

0 1 3 5 70 0 1 2 30 0 0 0 0

(b)

1 0 0 50 0 1 30 1 0 4

(c)

[1 0 3 10 1 2 4

]

(d)

[1 −7 5 50 1 3 2

]

(e)

1 0 0 1 20 1 0 2 40 0 1 3 6

(f)

0 10 00 0

4. Find the reduced row-echelon form of A when

A =

2 2 −1 0 1 0−1 −1 2 −3 1 01 1 −2 0 −1 00 0 1 1 1 0

.

5. Find the reduced row-echelon form of B when

B =

0 0 −2 0 7 122 4 −10 6 12 282 4 −5 6 −5 −1

.

6. In each part suppose that the augmented matrix for a system of linear equationshas been reduced by row operations to the given reduced row-echelon form. Solvethe system.


(a)

1 0 0 −30 1 0 00 0 1 7

(b)

1 0 0 −7 80 1 0 3 20 0 1 1 −5

(c)

1 −3 0 20 0 1 −20 0 0 0

(d)

[1 2 0 1 50 0 1 3 4

]

(e)

1 −6 0 0 3 −20 0 1 0 4 70 0 0 1 5 80 0 0 0 0 0

(f)

1 −3 0 00 0 1 00 0 0 1

7. Solve each of the following systems by Gauss-Jordan elimination.

(a) x1 + x2 + 2x3 = 8−x1 − 2x2 + 3x3 = 13x1 − 7x2 + 4x3 = 10

(b) 2x1 + 2x2 + 2x3 = 0−2x1 + 5x2 + 2x3 = 18x1 + x2 + 4x3 = −1

(c) 2x1 − 3x2 = −22x1 + x2 = 13x1 + 2x2 = 1

(d) −2b+ 3c = 13a + 6b− 3c = −26a+ 6b+ 3c = 5

(e) 4x1 − 8x2 = 123x1 − 6x2 = 9

−2x1 + 4x2 = −6

(f) x1 + 3x2 + x3 + x4 = 32x1 − 2x2 + x3 + 2x4 = 83x1 + x2 + 2x3 − x4 = −1

(g) x− y + 2z − w = −12x+ y − 2z − 2w = −2−x+ 2y − 4z + w = 1

3x − 3w = −3

(h) 3x1 + 2x2 − x3 = −155x1 + 3x2 + 2x3 = 03x1 + x2 + 3x3 = 11

−6x1 − 4x2 + 2x3 = 30

(i) x1 + x2 + x3 + x4 = 02x1 + x2 − x3 + 3x4 = 0x1 − 2x2 + x3 + x4 = 0

(j) 10y − 4z + w = 1x+ 4y − z + w = 23x+ 2y + z + 2w = 5

−2x− 8y + 2z − 2w = −4x− 6y + 3z = 1

8. Solve each of the systems in Exercise 2 by Gaussian elimination.

9. Determine which of the following homogeneous systems have nontrivial solutions.

(a) 2x1 − 3x2 + 4x3 − x4 = 07x1 + x2 − 8x3 + 9x4 = 02x1 + 8x2 + x3 − x4 = 0

(b) x1 + 3x2 − x3 = 0x2 − 8x3 = 0

4x3 = 0

(c) a11x1 + a12x2 + a13x3 = 0a21x1 + a22x2 + a23x3 = 0

(d) 3x1 − 2x2 = 06x1 − 4x2 = 0


10. Solve the following homogeneous systems of linear equations by any method.

(a) 2x1 + x2 + 3x3 = 0x1 + 2x2 = 0

x2 + x3 = 0

(b) 3x1 + x2 + x3 + x4 = 05x1 − x2 + x3 − x4 = 0

(c) −x1 + x2 − x3 + 3x4 = 03x1 + x2 − x3 − x4 = 02x1 − x2 − 2x3 − x4 = 0

(d) −12x1 +

13x2 +

12x3 = 0

14x1 − 2

3x2 +

14x3 = 0

14x1 +

13x2 − 3

4x3 = 0

(e) 2x+ 2y + 4z = 0w − y − 3z = 0

2w + 3x+ y + z = 0−2w + x+ 3y − 2z = 0

(f) 2x− y − 3z = 0−x+ 2y − 3z = 0x+ y + 4z = 0

11. Consider a linear system whose augmented matrix is of the form

1 2 1 1−1 4 3 22 −2 α 3

For what values of α will the system have a unique solution?


1 2 1 02 5 3 0−1 1 β 0

(a) Is it possible for the system to be consistent? Explain.

(b) For what values of β will the system have infinitely many solutions?


1 1 3 21 2 4 31 3 α β

(a) For what values of α and β will the system have infinitely many solutions?

(b) For what values of α and β will the system be inconsistent?

14. Solve the system

2x1 − x2 = λ x1

2x1 − x2 + x3 = λ x2

−2x1 + 2x2 + x3 = λ x3

for x1, x2, and x3 in the two cases λ = 1, λ = 2.



1. (a), (b), (d), (g), (h) 2. (a), (b), (c), (f), (g)

3. (a) Row-echelon (b) Neither (c) Both (d) Row-echelon (e) Both (f) Both

4.

1 1 0 0 1 00 0 1 0 1 00 0 0 1 0 00 0 0 0 0 0

5.

1 2 0 3 0 70 0 1 0 0 10 0 0 0 1 2

6. (a) x1 = 3, x2 = 0, x3 = 7 (b) x1 = 7t+ 8, x2 = −3t + 2, x3 = −t− 5, x4 = t

(c) x1 = 2+ 3t, x2 = t, x3 = −2 (d) x1 = 5− 2s− t, x2 = s, x3 = 4− 3t, x4 = t

(e) x1 = 6s− 3t− 2, x2 = s, x3 = −4t + 7, x4 = −5t + 8, x5 = t

(f) Inconsistent

7. (a) x1 = 3, x2 = 1, x3 = 2 (b) x1 = −17− 3

7t, x2 =

17− 4

7t, x3 = t

(c) Inconsistent (d) Inconsistent (e) x1 = 3 + 2t, x2 = t

(f) x1 =34− 5

8t, x2 = −1

4− 1

8t, x3 = t, x4 = 3

(g) x = t− 1, y = 2s, z = s, w = t (h) x1 = −4, x2 = 2, x3 = 7

(i) x1 = −43t, x2 = 0, x3 =

13t, x4 = t

(j) x = 145− 6

5t, y = − 7

10+ 3

10t− 1

10s, z = −2 + t, w = t

9. (a), (c), (d)

10. (a) x1 = 0, x2 = 0, x3 = 0

(b) x1 = −s, x2 = −t− s, x3 = 4s, x4 = t

(c) x1 = t, x2 = −t, x3 = t (d) x1 =53t, x2 = t, x3 = t

(e) w = t, x = −t, y = t, z = 0 (f) x = 0, y = 0, z = 0

11. α 6= −2 12. β = 2 13. (a) α = 5, β = 4, (b) α = 5, β 6= 4

14. If λ = 1, then x1 = x2 = s, x3 = 0. If λ = 2, then x1 = −12s, x2 = 0, x3 = s

1.3 Matrices and Matrix Operations

In this section we begin our study of matrix theory by giving some of the fundamentaldefinitions of the subject. We shall see how matrices can be combined through thearithmetic operations of addition, subtraction, and multiplication.


Matrix Notation and Terminology

Definition 1.1 A matrix is a rectangular array of numbers, each of which is called anentry of the matrix.

For example,

2 1−1 20 5

(1.2)

(1 −1 00 5 0

)

(1.3)

and[1

√2 π

](1.4)

Each horizontal line of number is called a row of the matrix; each vertical one iscalled a column. For example, in matrix (1.2) the third row consist of the entries 0 and5, whereas the entries in the first column are 2, −1, and 0.

If the given matrix has m rows and n columns, it is called m× n matrix (read, mby n matrix), and the numbers m and n are the sizes of the matrix. The matrices (1.2),(1.3), and (1.4), are respectively, 3× 2, 2× 3, and 1× 3 matrices.

A matrix with only one column or m × 1 matrix (with m > 1) is called a column

matrix (or a column vector), and a matrix with only one row or 1× n matrix (withn > 1) is called a row matrix (or a row vector). For example, the matrix (1.4) iscalled a row matrix or a row vector.

Capital (uppercase) letters, such as A, B, and C, will be used to denote matrices,whereas the entries of a matrix will be denoted by the corresponding lowercase letterwith “double subscripts.” The entry that occurs in row i and column j of a matrix Awill be denoted by aij. Thus, a general 2× 4 matrix might be written as

A =

[a11 a12 a13 a14a21 a22 a23 a24

]

and a general m× n matrix as

A =

a11 a12 · · · a1na21 a22 · · · a2n...

......

am1 am2 · · · amn

.

When compactness of notation is desired, the preceding matrix can be written as

[aij]m×n and [aij]

the first notation being used when it is important in the discussion to know the size, andthe second being used when the size need not be emphasized.


The entry in row i and column j of a matrix A is also commonly denoted by thesymbol (A)ij . Thus, for the matrix

A =

3 6 −14 2 01 −3 −5

we have (A)12 = a12 = 6, (A)21 = a21 = 4, and (A)33 = a33 = 5.Row and column matrices are of special importance, and it is common practice to

denote them by boldface lowercase letters rather than capital letters. For such matrices,double subscripting of the entries is unnecessary. Thus a general 1×n row matrix a anda general m× 1 column matrix b would be written as

a =[a1 a2 · · · an

]and b =

b1b2...bn

A matrix A with n rows and n columns is called a square matrix of order n, andthe entries a11, a22, . . ., ann in (1.5) are said to be on the main diagonal of A.

a11 a12 · · · a1na21 a22 · · · a2n...

......

an1 an2 · · · ann

. (1.5)

Definition 1.2 A square matrix in which all the entries above the main diagonal arezero is called lower triangular, and a square matrix in which all the entries belowthe main diagonal are zero is called upper triangular. A matrix that is either uppertriangular or lower triangular is called triangular.

Operations on Matrices

Definition 1.3 Given matrices A and B, the entries aij and bkp are corresponding

entries if and only if i = k and j = p. In other words, an entry of A corresponds toan entry of B if and only if they occupy the same position in their respective matrices.

For example, if

A =

[1 2 34 5 6

]

and B =

[7 8 9−1 −2 −3

]

then 1 and 7, 2 and 8, and 4 and −1 are all pairs of corresponding entries. We shallspeak of corresponding entries only when the two matrices in question have the samesize.

Definition 1.4 Equality of matrices. Two matrices A and B are equal, writtenA = B, if they have the same size and their corresponding entries are equal.

In matrix notation, if A = [aij ] and B = [bij ] have the same size, then A = B if andonly if aij = bij for all i and j.


Example 1.8 Which of the following matrices are equal:

A =

[1 2 −14 0 1 + 2

]

, B =

[1 24 0

]

,

C =

[1 4

2−1

4 0 3

]

, D =

[1 2 −14 0 1

]

?

Solution . . . . . . . . .

Definition 1.5 Sum of matrices. Let A and B be matrices with the same size. Thesum of A and B, written A+B, is the matrix obtained by adding corresponding entriesof A and B

In matrix notation, if A = [aij ] and B = [bij ] have the same size, then C = A+B ifand only if cij = aij + bij for all i and j.

Example 1.9 Let A =

[2 −1 53 0 −4

]

, B =

[−3 2 75 −6 2

]

, and C =

[3 5−2 −1

]

. Find

A+B and A+ C, if defined

Solution . . . . . . . . .

Definition 1.6 Multiplication by a scalar. Let A be a matrix and c be a scalar.Then the product cA is the matrix obtained by multiplying each entry of A by scalar c.The matrix cA is said to be a scalar multiple of A.

In matrix notation, if A = [aij ], then B = cA if and only if bij = caij for all i and j.

Definition 1.7 The negative of a matrix B, written −B, is the matrix (−1)B. It isobtained from B by simply reversing the sign of every entry.

Example 1.10 Let

A =

[1 −1 0 32 6 −4 8

]

.

Find 3A and (−1)A.

Solution . . . . . . . . .

Definition 1.8 Difference of matrices. Let A and B be matrices with the same size.The difference of A and B, written A− B, is the matrix C given by C = A+ (−B).

Note The matrix A − B can be obtained by simply subtracting each entry of B fromthe corresponding entry of A.

Example 1.11 Find C = 2A− B where A =

5 4−1 79 −3

and B =

−3 04 25 −7

.


Solution . . . . . . . . .

If A1, A2, . . . , An are matrices of the same size and c1, c2, . . . , cn are scalars, then anexpression of the form

c1A1 + c2A2 + · · ·+ cnAn

is called a linear combination of A1, A2, . . . , An with coefficients c1, c2, . . ., cn. Forexample, If

A =

[2 3 41 3 1

]

, B =

[0 2 7−1 3 −5

]

, and C =

[9 −6 33 0 12

]

,

then

2A− B +1

3C = 2

[2 3 41 3 1

]

+ (−1)

[0 2 7−1 3 −5

]

+1

3

[9 −6 33 0 12

]

=

[4 6 82 6 2

]

+

[0 −2 −71 −3 5

]

+

[3 −2 11 0 4

]

=

[7 2 24 3 11

]

is the linear combination of A, B, and C with scalar coefficients 2, −1, and 13.

Definition 1.9 Matrix multiplication. If A is an m × r matrix and B is an r × nmatrix, then the product AB is the m × n matrix whose entries are determined asfollows. To find the entry in row i and column j of AB, single out row i from matrixA and column j from matrix B. Multiply the corresponding entries from the row andcolumn together, and then add up the resulting products.

In other words, if C = AB, then

cij = ai1b1j + ai2b2j + · · ·+ ainbnj

or, using summation notation,

cij =

n∑

k=1

aikbkj.

Example 1.12 Given the pairs of matrices A and B, find the products AB and BA, ifdefined.

(a) A =

[1 2 42 6 0

]

, B =

4 1 4 30 −1 3 12 7 5 2

(b) A =

1 2 34 5 67 8 9

, B =

xyz

Solution . . . . . . . . .

Note It should be apparent from Example 1.12 that multiplication of matrices “actdifferently” from multiplication of numbers. The latter has the property of commutativity(for all numbers a and b, ab = ba), but matrix multiplication does not.

One situation in which the products AB and BA are both defined (although theystill need not be equal) is when A and B have the same size and are square.


Partitioned Matrices

A matrix can be subdivided or partitioned into small matrices by inserting horizontaland vertical rules between selected rows and columns. For example, the following arethree possible partitions of a general 3× 4 matrix A:

• partition of A into four submatrices A11, A12, A21, and A22

A =

a11 a12 a13 a14a21 a22 a23 a24a31 a32 a33 a34

=

[A11 A12

A21 A22

]

• partition of A into its row matrices r1, r2, and r3

A =

a11 a12 a13 a14a21 a22 a23 a24a31 a32 a33 a34

=

r1r2r3

• partition of A into its column matrices c1, c2, c3, and c4

A =

a11 a12 a13 a14a21 a22 a23 a24a31 a32 a33 a34

=[c1 c2 c3 c4

]

Matrix Multiplication by Columns and by Rows

Sometimes it may be desirable to find a particular row or column of a matrix product ABwithout computing the entire product. The following results are useful for that purpose:

i th row matrix of AB =[i th row matrix of A

]B (1.6)

j th column of AB = A[j th column of B

](1.7)

Example 1.13 If A and B are the matrices in Example 1.12 (a), then find the first rowand second column of AB

Solution . . . . . . . . .

If a1, a2, . . ., am denote the row matrices of A and b1, b2, . . ., bn denote the columnmatrices of B, then it follows from Formula (1.6) and (1.7) that

AB =

a1

a2

...am

B =

a1Ba2B...

amB

(1.8)

(AB computed row by row)

AB = A[b1 b2 · · · bn

]=

[Ab1 Ab2 · · · Abn

](1.9)

(AB computed column by column)


Matrix Products as Linear Combinations

Suppose that

A =

a11 a12 · · · a1na21 a22 · · · a2n...

......


and x =

x1

x2...xn

Then

Ax =

a11x1 + a12x2 + · · ·+ a1nxn

a21x1 + a22x2 + · · ·+ a2nxn...

......

am1x1 + am2x2 + · · ·+ amnxn

= x1

a11a21...

am1

+ x2

a12a22...

am2

+ · · ·+ xn

a1na2n...

amn

(1.10)

In words, (1.10) tell us that the product Ax of the matrix A with a column matrix xis a linear combination of the column matrices of A with the coefficients coming fromthe matrix x. Moreover, the product yA of a 1×m matrix y with an m× n matrix A isa linear combination of the row matrices of A with scalar coefficients coming from y.

Example 1.14 The matrix product

2 −1 31 3 −2−3 2 1

−123

=

5−110

can be written as the linear combination of the column matrices

−1

21−3

+ 2

−132

+ 3

3−21

=

5−110

The matrix product

[5 −1 10

]

2 −1 31 3 −2−3 2 1

=[−21 12 27

]

can be written as the linear combination of the row matrices

5[2 −1 3

]− 1

[1 3 −2

]+ 10

[−3 2 1

]=

[−21 12 27

]♣

It follows from (1.8) and (1.10) that the j th column matrix of a product AB is alinear combination of the column matrices of A with the coefficients coming from the j thcolumn of B.


Example 1.15 We showed in Example 1.12 (a) that

AB =

3 −22 41 −3

[−2 1 34 1 6

]

=

−14 1 −312 6 30−14 −2 −15

The column matrices of AB can be expressed as linear combinations of the column ma-trices of A as follows:

−1412−14

= −2

321

+ 4

−24−3

16−2

= 1

321

+ 1

−24−3

−330−15

= 3

321

+ 6

−24−3

♣

Matrix Form of a Linear System

Matrix multiplication has an important application to system of linear equations Con-sider any system of m linear equations in n unknowns.

a11x1 + a12x2 + · · ·+ a1nxn = b1

a21x1 + a22x2 + · · ·+ a2nxn = b2 (1.11)...

......

...

am1x1 + am2x2 + · · ·+ amnxn = bm

Since two matrices are equal if and only if their corresponding entries are equal, wecan replace the m equations in this system by the single matrix equation

a11x1 + a12x2 + · · ·+ a1nxn

a21x1 + a22x2 + · · ·+ a2nxn...

......

am1x1 + am2x2 + · · ·+ amnxn

=

b1b2...bm

The m× 1 matrix on the left side of this equation can be written as a product to give

a11 a12 · · · a1na21 a22 · · · a2n...

......


x1

x2...xm

=

b1b2...bm

.

If we designate these matrices A, x, and b, respectively, then the original system of mequations in n unknowns has been replaced by the single matrix equation

Ax = b


The matrix A in this equation is called the coefficient matrix of the system. Theaugmented matrix for the system is obtained by adjoining b to A as the last column;thus the augmented matrix is

[A b

]=

a11 a12 · · · a1n b1a21 a22 · · · a2n b2...

......

...am1 am2 · · · amn bm

.

Transpose of a Matrix

Definition 1.10 Transpose of a matrix. The transpose of the m×n matrix A, writtenAT , is the n×m matrix B whose j th column is the j th row of A. Equivalently, B = AT

if and only if bij = aji for each i and j.

Example 1.16 Find the transpose of

(a) A =

[2 7−8 0

]

(b) B =

[1 2 −74 5 6

]

(c) C =[−1 3 4

]

Solution . . . . . . . . .

Definition 1.11 If A is a square matrix, then trace of A, denoted by tr(A), is definedto be the sum of the entries on the main diagonal of A. The trace of A is undefined if Ais not a square matrix.

Example 1.17 Find the trace of matrices A and B.

A =

a11 a12 a13a21 a22 a23a31 a32 a33

, B =

−1 2 7 03 5 −8 41 2 7 −34 −2 1 0

Solution . . . . . . . . .

Exercise 1.3

1. Suppose that A,B,C,D, and E are matrices with the following dimensions:

A B C D E(4× 5) (4× 5) (5× 2) (4× 2) (5× 4)

Determine which of the following matrix expression are defined. For those that aredefined, give the dimension of the resulting matrix.

(a) BA (b) AC +D (c) AE +B (d) AB +B

(e) E(A+B) (f) E(AC) (g) ETA (h) (AT + E)D

2. If a matrix A is 3× 5 and the product AB is 3× 7, what is the size of B?


3. Solve the following matrix equation for a, b, c, and d

[a− b b+ c3d+ c 2a− 4d

]

=

[8 17 6

]

4. Consider the matrices

A =

3 0−1 21 1

, B =

[4 −10 2

]

, C =

[1 4 23 1 5

]

,

D =

1 5 2−1 0 13 2 4

, E =

6 1 3−1 1 24 1 3

Compute the following (where possible).

(a) 2AT + C (b) DT − ET (c) (D − E)T (d) BT + 5CT

(e) 12CT − 1

4A (f) B −BT (g) 2ET − 3DT (h) (2ET − 3DT )T

5. Using the matrices in Exercise 4, compute the following (where possible).

(a) AB (b) BA (c) (3E)D (d) (AB)C

(e) A(BC) (f) CCT (g) (DA)T (h) (CTB)AT

(i) tr(DDT ) (j) tr(4ET −D) (k) tr(CTAT + 2ET )

6. Let A =

[1 −2−2 5

]

and AB =

[−1 2 −16 −9 3

]

. Find the first and second columns

of B.

7. Let

A =

1 2 13 3 52 4 1

and B =

1 0 22 1 15 4 1

Use the method of Example 1.13 to find

(a) the first row of AB (b) the third row of AB

(c) the second column of AB (d) the first column of BA

(e) the third row of AA (f) the third column of AA

8. Let A and B be the matrices Exercise7. Use the method of Example 1.15 to

(a) express each column matrix of AB as a linear combination of the columnmatrices of A

(b) express each column matrix of BA as a linear combination of the columnmatrices of B


9. In each part, express the given system of linear equations as a single matrix equationAx = b.

(a) 3x1 + 2x2 = 12x1 − 3x2 = 5

(b) x1 + x2 = 52x1 + x2 − x3 = 63x1 − 2x2 + 2x3 = 7

(c) 2x1 + x2 + x3 = 4x1 − x2 + 2x3 = 2

3x1 − 2x2 − x3 = 0

(d) 4x1 + −3x3 + x4 = 15x1 + x2 − 8x4 = 3

2x1 − 5x2 + 9x3 − x4 = 03x2 − x3 + 7x4 = 2

10. In each part, express the matrix equation as a system of linear equations.

(a)

3 −1 24 3 7−2 1 5

x1

x2

x3

=

2−14

(b)

3 −2 0 15 0 2 −23 1 4 7−2 5 1 6

wxyz

=

0000

11. In each part, find a 6× 6 matrix [aij ] that satisfies the state condition. Make youranswers as general as possible by using letters rather than specific numbers for thenonzero entries.

(a) aij = 0 if i 6= j (b) aij = 0 if i > j

(c) aij = 0 if i < j (d) aij = 0 if |i− j| > 1

12. Find the matrix A such that

A

xyz

=

x+ yx− y0

for all choices of x, y, and z?

13. Indicate whether the statement is always true or sometimes false. Justify youranswer with a counterexample.

(a) The expressions tr(AAT ) and tr(ATA) are always defined, regardless of thesize of A.

(b) tr(AAT ) = tr(ATA) for every matrix A.

(c) If the first column of A has all zeros, then so does the first column of everyproduct AB.

(d) If the first row of A has all zeros, then so does the first row of every productAB.

(e) If A is a square matrix with two identical rows, then AA has two identicalrows.

(f) If A is a square matrix and AA has a column of zeros, then A must have acolumn of zeros.


(g) If B is an n×n matrix whose entries are positive even integers, and if A is ann × n matrix whose entries are positive integers, then the entries of AB andBA are positive even integers.

(h) If the matrix sum AB +BA is defined, then A and B must be square.


1. (a) Undefined (b) 4× 2 (c) Undefined (d) Undefined (e) 5× 5 (f) 5× 2

(g) Undefined (h) 5× 2

2. 5× 7 3. a = 5, b = −3, c = 4, d = 1

4. (a)

[7 2 43 5 7

]

(b)

−5 0 −14 −1 1−1 −1 1

(c)

−5 0 −14 −1 1−1 −1 1

(d) Undefined

(e)

−14

32

94

034

94

(f)

[0 −11 0

]

(g)

9 1 −1−13 2 −40 1 −6

(h)

9 −13 01 2 1−1 −4 −6

5. (a)

12 −3−4 54 1

(b) Undefined (c)

42 108 7512 −3 2136 78 63

(d)

3 45 911 −11 177 17 13

(e)

3 45 911 −11 177 17 13

(f)

[21 1717 35

]

(g)

[0 −2 1112 1 8

]

(h)

12 6 948 −20 1424 8 16

(i) 61 (j) 35 (k) 28 6. b1 =

[74

]

, b2 =

[−8−5

]

7. (a)[10 6 5

](b)

[15 8 9

](c)

6238

(d)

5719

(e)[16 20 23

](f)

122323

8. (a)

103415

= 1

132

+ 2

234

+ 5

151

6238

= 0

132

+ 1

234

+ 4

151

5149

= 2

132

+ 1

234

+ 1

151

(b)

5719

= 1

125

+ 3

014

+ 2

211


101126

= 2

125

+ 3

014

+ 4

211

3826

= 1

125

+ 5

014

+ 1

211

9. (a)

[3 22 −3

] [x1

x2

]

=

[15

]

(b)

1 1 02 1 −13 −2 2

x1

x2

x3

=

567

(c)

2 1 11 −1 23 −2 −1

x1

x2

x3

=

420

(d)

4 0 −3 15 1 0 −82 −5 9 −10 3 −1 7

x1

x2

x3

x4

=

1302

10. (a) 3x1 − x2 + 2x3 = 24x1 + 3x2 + 7x3 = −1

−2x1 + x2 + 5x3 = 4

(b) 3w − 2x + z = 05w + 2y − 2z = 03w + x+ 4y + 7z = 0

−2w + 5x+ y + 6z = 0

11. (a)

a11 0 0 0 0 00 a22 0 0 0 00 0 a33 0 0 00 0 0 a44 0 00 0 0 0 a55 00 0 0 0 0 a66

(b)

a11 a12 a13 a14 a15 a160 a22 a23 a24 a25 a260 0 a33 a34 a35 a360 0 0 a44 a45 a460 0 0 0 a55 a560 0 0 0 0 a66

(c)

a11 0 0 0 0 0a21 a22 0 0 0 0a31 a32 a33 0 0 0a41 a42 a43 a44 0 0a51 a52 a53 a54 a55 0a61 a62 a63 a64 a65 a66

(d)

a11 a12 0 0 0 0a21 a22 a23 0 0 00 a32 a33 a34 0 00 0 a43 a44 a45 00 0 0 a54 a55 a560 0 0 0 a65 a66

12.

1 1 01 −1 00 0 0

13. (a) True (b) True (c) False; for example, A =

[0 20 4

]

and B =

[1 23 4

]

(d) True (e) True (f) False; for example, A =

[1 −11 −1

]

(g) True (h) True


1.4 Inverses; Rules of Matrix Arithmetic

In this section we shall discuss some properties of the arithmetic operations on matrices.We shall see that many of the basic rules of arithmetic for real numbers also hold formatrices, but a few do not.

Properties of Matrix Operations

For real numbers a and b, we always have ab = ba, which is called the commutative lawfor multiplication. For matrices, however, AB and BA need not be equal.

Example 1.18 Consider the matrices

A =

[1 02 3

]

, B =

[1 23 4

]

Multiplying gives

AB =

[1 211 16

]

, BA =

[5 611 16

]

Thus, AB 6= BA ♣

Theorem 1.2 (Properties of Matrices Arithmetic) Let A, B, and C be matricesand a and b be scalars. Assume that the sizes of the matrices are such that each operationis defined. Then

(a) A+B = B + A (Commutative law for addition)

(b) A+ (B + C) = (A+B) + C (Associative law for addition)

(c) A(BC) = (AB)C (Associative law for multiplication)

(d) A(B + C) = AB + AC (Left distributive laws)

(e) (B + C)A = BA + CA (Right distributive laws)

(f) A(B − C) = AB −AC (g) (B − C)A = BA− CA

(h) a(B + C) = aB + aC (i) a(B − C) = aB − aC

(j) (a + b)C = aC + bC (k) (a− b)C = aC − bC

(l) (ab)C = a(bC) (m) a(BC) = (aB)C = B(aC)

Example 1.19 Let

A =

1 0 −12 1 −13 −2 0

, B =

−1 2 30 1 0−2 −1 0

, C =

−2 1 11 1 03 0 −3

Find A(BC), (AB)C, A(B + C), and AB + AC.

Solution . . . . . . . . .


Zero Matrices

Definition 1.12 The matrix analog of the number 0 is called a zero matrix. A zeromatrix, denoted by 0, is a matrix of any sizes consisting of only zero entries.

For example, the matrices

0 0 00 0 00 0 0

,

[0 0 00 0 0

]

,

000

, and[0 0 0 0

]

are all zero matrices.Since we already know that some of the rules of arithmetic for real number do not

carry over to matrix arithmetic. For example, consider the following two standard resultsin the arithmetic of real numbers.

• If ab = ac and a 6= 0, then b = c. (This is called the cancellation law.)

• If ad = 0, then at least one of the factors on the left is 0

As the next example shows, the corresponding results are not generally true in matrixarithmetic.


A =

[0 10 2

]

, B =

[1 13 4

]

, C =

[2 53 4

]

, D =

[3 70 0

]

Then

AB = AC =

[3 46 8

]

AD =

[0 00 0

]

Thus, although A 6= 0, it is incorrect to cancel the A from both sides of the equationAB = AC and write B = C. Also, AD = 0, yet A 6= 0 and D 6= 0. Thus, thecancellation law is not valid for matrix multiplication, and it is possible for a product ofmatrices to be zero without either factor being zero. ♣

Theorem 1.3 (Properties of Zero Matrices) Assuming that the sizes of the matricesare such that the indicated operations can be performed, the following rules of matrixarithmetic are valid.

(a) A+ 0 = 0+ A = A

(b) A− A = 0

(c) 0− A = −A

(d) A0 = 0; 0A = 0


Identity Matrices

Of special interest are square matrices with 1’s on the main diagonal and 0’s off the maindiagonal, such as

[1 00 1

]

,

1 0 00 1 00 0 1

,

1 0 0 00 1 0 00 0 1 00 0 0 1

.

Amatrix of this form is called an identity matrix and is denoted by I. If it is importantto emphasize the size, we shall write In for the n× n identity matrix.

The identity matrix is so named because it has the property that AIn = A andImA = A for any m × n matrix A. Thus it is the matrix analog of the number 1, themultiplicative identity for numbers.


A =

3 4 12 6 30 1 8

B =

b11 b12b21 b22b31 b32

Then

I3A =

1 0 00 1 00 0 1

3 4 12 6 30 1 8

=

3 4 12 6 30 1 8

= A

and

BI2 =

b11 b12b21 b22b31 b32

[1 00 1

]

=

b11 b12b21 b22b31 b32

= B ♣

Theorem 1.4 If R is the reduced row-echelon form of an n × n matrix A, then eitherR has a row of zeros or R is the identity matrix In.

Definition 1.13 Let A be a square matrix. If there exist a matrix B such that AB =BA = I, we say that A is invertible (or nonsingular) and that B is the inverse ofA. If A has no inverse, it is said to be noninvertible (or singular).

Example 1.22 Show that the matrix B =

[3 51 2

]

is an inverse of A =

[2 −5−1 3

]

.

Solution . . . . . . . . .

Properties of Inverses

Theorem 1.5 If B and C are both inverses of the matrix A, then B = C.

As a consequence of Theorem 1.5, we can say now speak of the inverse of an invertiblematrix. If A is invertible, then its inverse will be denoted by the symbol A−1. Thus,

AA−1 = I and A−1A = I


Theorem 1.6 The matrix

A =

[a bc d

]

is invertible if ad− bc 6= 0, in which case the inverse is given by the formula

A−1 =1

ad− bc

[d −b−c a

]

=

d

ad− bc− b

ad − bc

− c

ad− bc

a

ad− bc

.

Theorem 1.7 If A and B are invertible matrices of the same dimension, then AB isinvertible and

(AB)−1 = B−1A−1

Theorem 1.8 If A1, A2, . . . , An are invertible matrices of the same dimension, thenA1A2 · · ·An is invertible and

(A1A2 · · ·An)−1 = A−1

n A−1n−1 · · ·A−1

2 A−11

Example 1.23 Let A =

[7 −35 −2

]

and B =

[4 13 1

]

. Show that (AB)−1 = B−1A−1.

Solution . . . . . . . . .

Powers of a Matrix

Definition 1.14 If A is a square matrix, then we define the nonnegative integer powersof A to be

A0 = I An = AA · · ·A︸︷︷︸

n factors

(n > 0)

Moreover, if A is invertible, then we define the negative integer powers to be

A−n = (A−1)n = A−1A−1 · · ·A−1︸︷︷︸

n factors

Note In general, if A is a square matrix, and r and s are positive integers, we haveArAs = Ar+s. It is also true that under the same condition, (Ar)s = Ars.


[3 −2−1 1

]

and A−1 =

[1 21 3

]

. Find the matrices A3 and A−3.

Solution . . . . . . . . .

Theorem 1.9 (Laws of Exponents) If A is invertible matrix, then:

(a) A−1 is invertible and (A−1)−1 = A .

(b) An is invertible and (An)−1 = (A−1)n for n = 0, 1, 2, . . ..


(c) For any nonzero scalar k, the matrix kA is invertible and (kA)−1 =1

kA−1 .

Example 1.25 If

(1

3A

)−1

=

[2 −1−5 3

]

, then find A.

Solution . . . . . . . . .

Properties of the Transpose

Theorem 1.10 If the sizes of the matrices are such that the state operations can beperformed, then

(a) (AT )T = A

(b) (A +B)T = AT +BT and (A−B)T = AT − BT

(c) (kA)T = kAT , where k is any scalar

(d) (AB)T = BTAT

Invertibility of a Transpose

Theorem 1.11 If A is an invertible matrix, then AT is also invertible and

(AT )−1 = (A−1)T .


[−3 15 −2

]

. Show that (AT )−1 = (A−1)T .

Solution . . . . . . . . .

Exercise 1.4

1. Let

A =

1 −2 03 5 −12 3 4

, B =

6 0 24 1 −1−3 8 5

C =

4 −5 35 7 −2−3 2 −1

, a = 3, b = −5

Show that

(a) A + (B + C) = (A+B) + C (b) (AB)C = A(BC)

(c) (a+ b)C = aC + bC (d) a(B − C) = aB − aC

(e) a(BC) = (aB)C = B(aC) (f) A(B − C) = AB − AC

(g) (B + C)A = BA+ CA (h) a(bC) = (ab)C

(i) (AT )T = A (j) (A+B)T = AT +BT

(k) (aC)T = aCT (l) (AB)T = BTAT


2. Let 0 denote a 2× 2 matrix, each of whose entries is zero. Find

(a) 2× 2 matrix A such that A 6= 0 and AA = 0 and

(b) 2× 2 matrix A such that A 6= 0 and AA = A.

3. Use Theorem 1.6 to compute the inverses of the following matrices.

(a) A =

[1 32 −4

]

(b) B =

[2 31 1

]

(c) C =

[−4 −55 6

]

(d) D =

[3 −7−6 13

]

4. Use the matrices A, B, and C in Exercise 3 to verify that

(a) (A−1)−1 = A (b) (BT )−1 = (B−1)T

(c) (AB)−1 = B−1A−1 (d) (ABC)−1 = C−1B−1A−1

5. Let A =

[1 −1−1 2

]

. Find the power A2 and A3.

6. Let A =

1 0 02 1 03 2 1

. Find the power A2 and A3.

7. Let A =

[1 10 1

]

. Find a general formula for An.

8. In each part, use the given information to find A.

(a) A−1 =

[2 −13 5

]

(b) (7A)−1 =

[−3 71 −2

]

(c) (5AT )−1 =

[−3 −15 2

]

(d) (I + 2A)−1 =

[−1 24 5

]

9. Find the inverse of

[cos θ sin θ− sin θ cos θ

]

.

10. Find the inverse of

[12(ex + e−x) 1

2(ex − e−x)

12(ex − e−x) 1

2(ex + e−x)

]

.

11. Consider the matrix

A =

a11 0 · · · 00 a22 · · · 0...

......

0 0 · · · ann

where a11a22 · · · ann 6= 0. Show that A is invertible and find its inverse.


12. Let A, B, and 0 be 2× 2 matrices. Assuming that A is invertible, find a matrix Csuch that [

A−1 0C A−1

]

is the inverse of the partitioned matrix

[A 0B A

]

13. Use the result in Exercise 12 to find the inverses of the following matrices.

(a)

1 1 0 0−1 1 0 01 1 1 11 1 −1 1

(b)

1 1 0 00 1 0 00 0 1 10 0 0 1

14. (a) Find a nonzero 3× 3 matrix A such that A is symmetric.(b) Find a nonzero 3× 3 matrix A such that A is skew-symmetric.

15. A square matrix A is called symmetric if AT = A and skew-symmetric ifAT = −A. Show that if B is a square matrix, then

(a) BBT and B +BT are symmetric,

(b) B − BT is skew-symmetric.

16. Indicate whether the statement is always true or sometimes false. Justify youranswer with a counterexample.

(a) (AB)2 = A2B2, where A are B are n× n matrices.

(b) (A− B)2 = (B −A)2, where A are B are n× n matrices.

(c) (A+B)(A−B) = A2 − B2, where A are B are n× n matrices.

(d) (AB−1)(BA−1) = In, where A are B are n× n matrices.

(e) Let A and B be square matrices such that AB = 0. Show that if A isinvertible, then B = 0.

(f) If a square matrix A satisfies A2 − 3A+ I = 0, then A−1 = 3I − A.

(g) If A and B are m× n matrices, then ABT and ATB are defined.

(h) If AB = BA and if A is invertible, then A−1B = BA−1.

(i) If AT is singular, then A is singular.

(j) A matrix with a row of zeros cannot have an inverse.

(k) A matrix with a column of zeros cannot have an inverse.

(l) If A is invertible and AB = AC, then B = C.

(m) Let A be an n×n matrix and let x and y be n× 1 matrices. If Ax = Ay andx 6= y, then A is singular.


(n) If A is invertible and k is any nonzero scalar, then (kA)n = knAn for all integervalues of n.

(o) Sum of two invertible matrices is invertible.

(p) If AB is invertible, then B is invertible.

(q)

[a bb a

]

+

[x yy x

]

is symmetric.


2. (a)

[0 10 0

]

(b)

[1 00 0

]

3. (a) A−1 =

[25

310

15

− 110

]

(b) B−1 =

[−1 31 −2

]

(c) C−1 =

[6 5−5 −4

]

(d) D−1 =

[

−133

−73

−2 −1

]

5. A2 =

[2 −3−3 5

]

, A3 =

[5 −8−8 13

]

6. A2 =

1 0 04 1 010 4 1

, A3 =

1 0 06 1 021 6 1

7. An =

[1 n0 1

]

8. (a) A =

[513

113

− 313

213

]

(b) A =

[27

117

37

]

(c) A =

[

−25

1

−15

35

]

(d) A =

[

− 913

113

213

− 613

]

9.

[cos θ − sin θsin θ cos θ

]

10

[12(ex + e−x) −1

2(ex − e−x)

−12(ex − e−x) 1

2(ex + e−x)

]

11. A =

1a11

0 · · · 0

0 1a22

· · · 0...

......

0 0 · · · 1ann

12. C = −A−1BA−1

13. (a)

12

−12

0 012

12

0 0

0 0 12

−12

−1 0 12

12

(b)

1 −1 0 00 1 0 00 0 1 −10 0 0 1


14. (a)

1 2 32 1 43 4 5

(b)

0 −1 −11 0 −11 1 0

16. (a) False; for example, A =

[0 10 2

]

and B =

[1 23 4

]

(b) True

(c) False; for example, A =

[0 10 2

]

and B =

[1 23 4

]

(d) True (e) True

(f) True (g) True (h) True (i) True (j) True (k) True (l) True (m) True

(n) True (o) False; for example, A =

[1 00 1

]

and B =

[−1 00 −1

]

(p) True

(q) True

1.5 Elementary Matrices and a Method for Finding

A−1

In this section we shall develop an algorithm for finding the inverse of an invertiblematrix. We shall also discuss some of the basic properties of invertible matrices.

Definition 1.15 An n×n matrix is called an elementary matrix if it can be obtainedfrom the n× n identity matrix In by performing a single elementary row operation.

List below are three elementary matrices and the operations that produce them.

[1 00 −5

]

Multiply the second row of I2 by −5.

1 0 0 00 0 0 10 0 1 00 1 0 0

Interchange the second and fourth rows of I4.

1 0 −20 1 00 0 1

Add −2 times the third row of I3 to the first row.

Theorem 1.12 If the elementary matrix E results from performing a certain row oper-ation on Im and A is an m × n matrix, then the product EA is the matrix that resultswhen this same row operation is performed on A.

Example 1.27 Let

E1 =

1 0 00 1 0−4 0 1

, E2 =

0 1 01 0 00 0 1

, E3 =

1 0 00 1 00 0 5


and

A =

a b cd e fg h i

Find the product E1A, E2A, and E3A.

Solution . . . . . . . . .

If an elementary row operation is applied to an identity matrix I to produce anelementary matrix E, then there is a second row operation that, when applied to E,produces I back again. For example, if E is obtained by multiplying the ith row of Iby a nonzero constant k, then I can be recovered if the ith row of E is multiplied by1k. The various possibilities are listed in the following table. The operations on the right

side of this table are called the inverse operations of the corresponding operations onthe left.

Row Operation on I Row Operation on EThat Produces E That Reproduces IMultiply row i by k 6= 0 Multiply row i by 1

k

Interchange rows i and j Interchange rows i and jAdd k times row i to row j Add −k times row i to row j

Example 1.28 In each of the following, an elementary row operation is applied to the2×2 identity matrix to obtain an elementary matrix E, then E is restored to the identitymatrix by applying the inverse row operation.

[1 00 1

]

−→[1 00 3

]

−→[1 00 1

]

↑Multiply the

second row by 3

↑Multiply the

second row by 13

[1 00 1

]

−→[0 11 0

]

−→[1 00 1

]

↑Interchange the firstand second rows

↑Interchange the firstand second rows

[1 00 1

]

−→[1 −70 1

]

−→[1 00 1

]

↑Add −7 times thesecond row to

the first

↑Add −7 times thesecond row to

the first

The next theorem gives an important property of elementary matrices.

Theorem 1.13 Every elementary matrix is invertible, and the inverse is also an ele-mentary matrix.


Theorem 1.14 If A is an n × n matrix, then the following statements are equivalent,that is, all true or all false.

(a) A is invertible.

(b) Ax = 0 has only the trivial solution.

(c) The reduced row-echelon form of A is In.

(d) A is expressible as a product of elementary matrices.

By Theorem 1.14 (c), we can find elementary matrices E1, E2, . . ., Ek−1, Ek such that

EkEk−1 · · ·E2E1A = In (1.12)

By Theorem 1.13, E1, E2, . . ., Ek−1, Ek are invertible. Multiplying both sides ofEquation (1.12) on the left successively by E−1

k , E−1k−1, . . ., E

−12 , E−1

1 we obtain

A = E−11 E−1

2 · · ·E−1k−1E

−1k In = E−1

1 E−12 · · ·E−1

k−1E−1k

By Theorem 1.13, this equation expresses A as a product of elementary matrices.

Row Equivalence

If a matrix B can be obtained from a matrix A by performing a finite sequence ofelementary row operations, then obviously we can get from B back to A by performingthe inverses of these elementary row operations in reverse order. Matrices that can beobtained from one another by a finite sequence of elementary row operations are said tobe row equivalent.

In other words, a matrix B is row equivalent to A if there exists a finite sequence E1,E2, . . ., Ek−1, Ek of elementary matrices such that

B = EkEk−1 · · ·E2E1A.

Example 1.29 Let

A =

1 2 −31 −2 15 −2 −3

.

Find the elementary matrices E1, E2, . . ., Ek−1, Ek such that

EkEk−1 · · ·E2E1A = Ar

where Ar is the reduced row-echelon form of A.

Solution . . . . . . . . .


A Method for Inverting Matrices

As our first application of Theorem 1.14, we shall establish a method for determiningthe inverse of an invertible matrix. Multiplying (1.12) on the right by A−1 yields

(EkEk−1 · · ·E2E1A)A−1 = InA

−1

EkEk−1 · · ·E2E1In = A−1

which tells us that A−1 can be obtained by multiplying In successively on the left by theelementary matrices E1, E2, . . ., Ek−1, Ek. Since each multiplication on the left by oneof these elementary matrices performs a row operation, it follow that the sequence of rowoperations that reduces A to In will reduce In to A−1. Thus we have the following result:

To find the inverse of an invertible matrix A we must find a sequence ofelementary row operations that reduces A to identity and then performthis sequence of operation on In to obtain A−1

To accomplish this we shall adjoint the identity matrix to the right side of A, therebyproducing a matrix of the form

[A | I].Then we shall apply row operations to this matrix until the left side is reduced to I;these operations will convert the right side to A−1, so the final matrix will have the form

[I |A−1]

Example 1.30 Find the inverse of

A =

1 −2 12 −5 23 2 −1

Solution . . . . . . . . .

Often it will not be known in advance whether a given matrix is invertible. If an n×nmatrix A is not invertible, then it cannot be reduced to In by elementary row operation(part (c) of Theorem 1.14

). Stated another way, the reduced row-echelon form of A has

at least one row of zeros. Thus, if the procedure in the last example is attempted on amatrix that is not invertible, then at some point in the computations a row of zeros willoccur on the left side. It can then be concluded that the given matrix is not invertible,and the computations can be stopped.

Example 1.31 Let A be the matrix

1 3 4−2 −5 −31 4 9

. Determine whether A is invertible.

Solution . . . . . . . . .


Example 1.32 In Example 1.30 we show that

A =

1 −2 12 −5 23 2 −1

is an invertible matrix. From Theorem 1.14, it follows that the homogeneous system

x1 − 2x2 + x3 = 0

2x1 − 5x2 + 2x3 = 0

3x1 + 2x2 − x3 = 0

has only the trivial solution. ♣

Exercise 1.5

1. Which of the following are elementary matrices?

(a)

[0 11 0

]

(b)

[−3 11 0

]

(c)

[1 0

0√2

]

(d)

1 0 00 1 05 0 1

(e)

1 1 00 0 10 0 0

(f)

0 1 01 0 00 0 1

(g)

2 0 0 20 1 0 00 0 1 00 0 0 1

2. Find a row operation that will restore the given elementary matrix to an identitymatrix.

(a)

[1 0−3 1

]

(b)

1 0 00 1 00 0 5

(c)

1 0 0 00 0 0 10 0 1 00 1 0 0

(d)

1 0 −14

00 1 0 00 0 1 00 0 0 1

3. For each of the following pairs of matrices, find an elementary matrix E such thatEA = B.


(a) A =

[2 −15 3

]

B =

[−4 25 3

]

(b) A =

2 1 3−2 4 53 1 4

B =

2 1 33 1 4−2 4 5

(c) A =

4 −2 31 0 2−2 3 1

B =

4 −2 31 0 20 3 5

(d) A =

3 4 12 −7 −12 −7 3

B =

3 4 12 −7 −18 1 5

4. Given

A =

1 2 42 1 31 0 2

, B =

1 2 42 1 32 2 6

, C =

1 2 40 −1 −32 2 6

(a) Find an elementary matrix E such that EA = B.

(b) Find an elementary matrixF such that FB = C.

(c) Is C row equivalent to A? Explain.

5. Given

A =

2 1 16 4 54 1 3

Find elementary matrices E1, E2, E3 such that

E3E2E1A = U

where U is an upper triangular matrix.

6. Use the method shown in Example 1.30 and Example 1.31 to find the inverse ofthe given matrix if the matrix is invertible.

(a)

[1 42 7

]

(b)

[−3 64 5

]

(c)

[6 −4−3 2

]

(d)

1 0 13 3 42 2 3

(e)

1 1 10 1 10 0 1

(f)

2 0 50 3 01 0 3

(g)

2 1 43 2 50 −1 1

(h)

15

15

−25

15

15

110

15

−45

110

(i)

√2 3

√2 0

−4√2

√2 0

0 0 1


(j)

1 0 0 01 3 0 01 3 5 01 3 5 7

(k)

−8 17 2 13

4 0 25

−90 0 0 0−1 13 4 2

(l)

0 0 2 01 0 0 10 −1 3 02 1 5 −3

7. Find all values of a such that the following matrices are invertible:

(a) A =

[2 a3 4

]

(b) A =

1 a 0−1 0 10 1 1

8. Given

A =

[3 15 2

]

B =

[1 23 4

]

compute A−1 and use it to:

(a) Find a 2× 2 matrix X such that AX = B.

(b) Find a 2× 2 matrix Y such that Y A = B.

9. Consider the matrix

A =

[1 0−5 2

]

(a) Find elementary matrices E1 and E2 such that E2E1A = I.

(b) Write A−1 as a product of two elementary matrices.

(c) Write A as a product of two elementary matrices.

10. Express the matrix

A =

0 1 7 81 3 3 8−2 −5 1 −8

in the form A = EFGR, where E, F , and G are elementary matrices and R is inrow-echelon form.

11. Consider the following given matrix A. Find elementary matrices E1, E2, . . ., Ek−1,Ek such that EkEk−1 · · ·E2E1A = Ar where Ar is the reduced row-echelon form ofA.

(a)

1 0 20 1 00 0 3

(b)

2 0 2 41 1 2 −10 0 2 4

(c)

0 0 1 11 2 0 03 0 0 1

(d)

1 1 −1 −12 0 0 20 0 0 32 2 −2 −2

12. Indicate whether the statement is always true or sometimes false.


(a) Every square matrix can be expressed as a product of elementary matrices.

(b) The product of two elementary matrices is an elementary matrix.

(c) If A is invertible and a multiple of the first row of A is added to the secondrow, then the resulting matrix is invertible.

(d) If A is invertible and AB = 0, then it must be true that B = 0.

(e) If A is a singular n× n matrix, then Ax = 0 has infinitely many solutions.

(f) If A is a singular n × n matrix, then the reduced row-echelon from of A hasat least one row of zeros.

(g) If A−1 is expressible as a product of elementary matrices, then the homoge-neous linear system Ax = 0 has only the trivial solution.


1. (a), (c), (d), (f) 2. (a) r2 + 3r1 (b) 5 r3 (c) r2 ↔ r4 (d) r1 +14r3

3. (a)

[−2 00 1

]

(b)

1 0 00 0 10 1 0

(c)

1 0 00 1 00 2 1

(d)

1 0 00 1 02 0 1

4. (a) E =

1 0 00 1 01 0 1

(b) F =

1 0 00 1 −10 0 1

5. E1 =

1 0 0−3 1 00 0 1

, E2 =

1 0 00 1 0−2 0 1

, E3 =

1 0 00 1 00 1 1

6. (a)

[−7 42 −1

]

(b)

[

− 539

213

439

113

]

(c) Not invertible (d)

1 2 −3−1 1 −10 −2 3

(e)

1 −1 00 1 −10 0 1

(f)

3 0 −50 1

30

−1 0 2

(g)

−7 5 33 −2 −23 −2 −1

(h)

1 3 10 1 −1−2 2 0

(i)

√2

26−3

√2

260

4√2

26

√2

260

0 0 1

(j)

1 0 0 0−1

313

0 00 −1

515

00 0 −1

717

(k) Not invertible (l)

−45

35

15

15

32

0 −1 012

0 0 045

25

−15

−15

7. (a) a 6= 83

(b) a 6= 1 8. (a)

[−1 04 2

]

(b)

[−8 5−14 9

]


9. (a) E1 =

[1 05 1

]

, E2 =

[1 00 1

2

]

(b) A−1 = E2E1 (c) A = E−11 E−1

2

10.

0 1 01 0 00 0 1

1 0 00 1 0−2 0 1

1 0 00 1 00 1 1

1 3 3 80 1 7 80 0 0 0

11. (a) E1 =

1 0 00 1 00 0 1

3

, E2 =

1 0 −20 1 00 0 1

(b) E1 =

12

0 00 1 00 1 0

, E2 =

1 0 0−1 1 00 0 1

, E3 =

1 0 00 1 00 0 1

2

,

E4 =

1 0 −10 1 00 0 1

, E5 =

1 0 00 1 −10 0 1

(c) E1 =

0 1 01 0 00 0 1

, E2 =

1 0 00 1 0−3 0 1

, E3 =

1 0 00 0 10 1 0

,

E4 =

1 0 00 −1

60

0 0 1

, E5 =

1 −2 00 1 00 0 1

(d) E1 =

1 0 0 0−2 1 0 00 0 1 00 0 0 1

, E2 =

1 0 0 00 1 0 00 0 1 0−2 0 0 1

, E3 =

1 0 0 00 −1

20 0

0 0 1 00 0 0 1

E4 =

1 −1 0 00 1 0 00 0 1 00 0 0 1

, E5 =

1 0 0 00 1 0 00 0 1

30

0 0 0 1

, E6 =

1 0 −1 00 1 0 00 0 1 00 0 0 1

,

E7 =

1 0 0 00 1 2 00 0 1 00 0 0 1

12. (a) False (b) True (c) True (d) True (e) True (f) True (g) True

1.6 Further Results on Systems of Equations and

Invertibility

A Basic Theorem

Theorem 1.15 Every system of linear equations has no solutions, or has exactly onesolutions, or has infinitely many solutions.


Solving Linear Systems by Matrix Inversion

So far, we have studied two methods for solving linear system: Gaussian elimination andGauss-Jordan elimination. The following theorem provides a new method for solvingcertain linear systems.

Theorem 1.16 If A is an invertible n × n matrix, then for each n × 1 matrix b, thesystem of equations Ax = b has exactly one solution, namely, x = A−1b.

Example 1.33 Solve the system

x1 − 2x2 + x3 = 7

2x1 − 5x2 + 2x3 = 6

3x1 + 2x2 − x3 = 1

Solution . . . . . . . . .

Remark Note that the method of Example 1.33 applies only when the system has asmany equations as unknowns, and the coefficient matrix is invertible. This method is lessefficient, computationally, than Gaussian elimination, but it is important in the analysisof equations involving matrices.

Linear Systems with a Common Coefficient Matrix

Frequently, one is concerned with solving a sequence of systems

Ax = b1, Ax = b2, Ax = b3, . . . , Ax = bk

each of which has the same square coefficient matrix A. If A is invertible, then thesolutions

x1 = A−1b1, x2 = A−1b2, x3 = A−1b3, . . . , xk = A−1bk

can be obtained with one matrix inversion and k matrix multiplications. Once again,however, a more efficient method is to form the matrix

[A |b1 |b2 | · · · |bk ] (1.13)

in which the coefficient matrix A is “augmented” by all k of the matrices and then reduce(1.13) to reduced row-echelon form by Gauss-Jordan elimination. In this way we cansolve all k systems at once. This method has the added advantage that it applies evenwhen A is not invertible.

Example 1.34 Solve the systems

(a) 2x1 − 4x2 = −10

x1 − 3x2 + x4 = −4

x1 − x3 + 2x4 = 4

3x1 − 4x2 + 3x3 − x4 = −11


(b) 2x1 − 4x2 = −8

x1 − 3x2 + x4 = −2

x1 − x3 + 2x4 = 9

3x1 − 4x2 + 3x3 − x4 = −15

Solution . . . . . . . . .

Properties of Invertible Matrices

Up to now, to show that an n × n matrix A is invertible, it has been necessary to findan n× n matrix B such that

AB = I and BA = I

The next theorem shows that if we produce an n×n matrix B satisfying either condition,then the other condition holds automatically.

Theorem 1.17 Let A be a square matrix.

(a) If B is a square matrix satisfying BA = I, then B = A−1.

(b) If B is a square matrix satisfying AB = I, then B = A−1.

Theorem 1.18 (Equivalent Statements) If A is an n×n matrix, then the followingare equivalent.




(d) A is expressible as a product of elementary matrices.

(e) Ax = b is consistent for every n× 1 matrix b.

(f) Ax = b has exactly one solution for every n× 1 matrix b.

Theorem 1.19 Let A and B be square matrices of the same size. If AB is invertible,then A and B must also be invertible.

The following fundamental problem will occur frequently in various contexts.

A Fundamental Problem: Let A be a fixed m× n matrix. Find allm× 1 matrices b such that the system of equations Ax = b is consistent.

If A is an invertible matrix, Theorem 1.16 completely solves this problem by assertingthat for every m × 1 matrix b, the linear system Ax = b has the unique solutionx = A−1b. If A is not square, or if A is square but not invertible, then Theorem 1.16does not apply. In this case the matrix b must usually satisfy certain conditions in orderfor Ax = b to be consistent.


Example 1.35 What conditions must b1, b2, and b3 satisfy in order for the system ofequations

x1 + x2 + 2x3 = b1

x1 + x3 = b2

2x1 + x2 + 3x3 = b3

to be consistent?

Solution . . . . . . . . .

Example 1.36 What conditions must b1, b2, and b3 satisfy in order for the system ofequations

x1 − 2x2 + x3 = b1

2x1 − 5x2 + 2x3 = b2

3x1 + 2x2 − x3 = b3

to be consistent?

Solution . . . . . . . . .

Exercise 1.6

1. Solve the following system by inverting the coefficient matrix and using Theo-rem 1.16.

(a) x1 + x2 = 25x1 + 6x2 = 9

(b) 2x1 − x2 = 86x1 − 5x2 = 32

(c) 5x1 + 3x2 + 2x3 = 43x1 + 3x2 + 2x3 = 2

x2 + x3 = 5

(d) y + z = 63x− y + z = −7x+ y − 3z = −13

(e) − x− 2y − 3z = 0w + x+ 4y + 4z = 7w + 3x+ 7y + 9z = 4

−w − 2x− 4y − 6z = 6

(f) x1 + 2x2 − 3x3 = 82x1 + 5x2 − 6x3 = 17

−x1 − 2x2 + x3 − x4 = −84x1 + 10x2 − 9x3 + x4 = 33

2. Solve the following general system by inverting the coefficient matrix and usingTheorem 1.16.

x1 + 2x2 + x3 = b1

x1 − x2 + x3 = b2

x1 + x2 = b3

Use the resulting formulas to find the solution if

(a) b1 = −1, b2 = 3, b3 = 4 (b) b1 = 5, b2 = 0, b3 = 0

(c) b1 = −1, b2 = −1, b3 = 3


3. Use the method of Example 1.34 to solve the systems in all part simultaneously.

(a) x1 − 5x2 = b13x1 + 2x2 = b2

i. b1 = 1, b2 = 4ii. b1 = −2, b2 = 5

(b) x1 − 2x2 = b13x1 + x2 = b2

i. b1 = −3, b2 = 12ii. b1 = −2, b2 = 1iii. b1 = −3, b2 = 9

(c) 4x1 − 7x2 = b1x1 + 2x2 = b2

i. b1 = 0, b2 = 1ii. b1 = −4, b2 = 6iii. b1 = −1, b2 = 3iv. b1 = −5, b2 = 1

(d) 2x1 − x2 + x3 = b1−2x1 + x2 − 3x3 = b24x1 − 3x2 + x3 = b3

i. b1 = 1, b2 = −5, b3 = 5ii. b1 = 1, b2 = −3, b3 = 1iii. b1 = 8, b2 = −14, b3 = 14

4. Solve the systems in both parts at the same time.

(a) x1 − 2x2 + 2x3 = −22x1 − 5x2 + x3 = 13x1 − 7x2 + 2x3 = −1

(b) x1 − 2x2 + 2x3 = 12x1 − 5x2 + x3 = −13x1 − 7x2 + 2x3 = 0

5. Find conditions that the b’s must satisfy for the system to be consistent.

(a) 6x1 − 4x2 = b13x1 − 2x2 = b2

(b) x1 − 2x2 + 5x3 = b14x1 − 5x2 + 8x3 = b2

−3x1 + 3x2 − 3x3 = b3

(c) x1 − x2 + 3x3 + 2x4 = b1−2x1 + x2 + 5x3 + x4 = b2−3x1 + 2x2 + 2x3 − x4 = b34x1 − 3x2 + 3x3 + 3x4 = b4

6. Solve the following matrix equation for X .

1 −1 12 3 00 2 −1

X =

2 −1 5 7 84 0 −3 0 13 5 −7 2 1

7. In each part, determine whether the homogeneous system has a nontrivial solution(without using pencil and paper); then state whether the given matrix is invertible.


(a) 3x1 − 2x2 + x3 + x4 = 04x2 + 5x3 − 2x4 = 0

x3 − 3x4 = 02x4 = 0

3 −2 1 10 4 5 −20 0 1 −30 0 0 2

(b) 4x1 + 3x2 − x3 + 2x4 = 03x3 − x4 = 0x3 + 5x4 = 0

9x4 = 0

4 3 −1 20 0 3 −10 0 1 50 0 0 9


1. (a) x1 = 3, x2 = −1 (b) x1 = 2, x2 = −4

(c) x1 = 1, x2 = −11, x3 = 16 (d) x = −3, y = 2, z = 4

(e) w = −6, x = 1, y = 10, z = −7

(f) x1 = 3, x2 = 1, x3 = −1, x4 = 2

2. (a) x1 =163, x2 = −4

3, x3 = −11

3(b) x1 = −5

3, x2 =

53, x3 =

103

(c) x1 = 3, x2 = 0, x3 = −4

3. (a) i. x1 =2217, x2 =

117

ii. x1 =2117, x2 =

1117

(b) i. x1 = −1, x2 = 1 ii. x1 = 2, x2 = −5 iii. x1 = 3, x2 = 0

(c) i. x1 =715, x2 =

415

ii. x1 =3415, x2 =

2815

iii. x1 =1915, x2 =

1315

iv. x1 = −15, x2 =

35

(d) i. x1 = −3, x2 = −5, x3 = 2 ii. x1 = 0, x2 = 0, x3 = 1

iii. x1 = 2, x2 = −1, x3 = 3

4. (a) x1 = −12− 3t, x2 = −5− t, x3 = t

(b) x1 = 7− 3t, x2 = 3− t, x3 = t

5. (a) b1 = 2b2 (b) b1 = b2 + b3 (c) b1 = b3 + b4, b2 = 2b3 + b4

6. X =

11 12 −3 27 26−6 −8 1 −18 −17−15 −21 9 −38 −35

7. (a) Only the trivial solution x1 = x2 = x3 = x4 = 0; invertible

(b) Infinitely many solutions; no invertible

1.7 LU-Decompositions

With Gaussian elimination and Gauss-Jordan elimination, a linear system is solved byoperating systematically on an augmented matrix. In this section we shall discuss adifferent organization of this approach, one based on factoring the coefficient matrix intoa product of lower and upper triangular matrices.


Solving Linear Systems by Factoring

We shall proceed in two stages. First, we shall show how a linear system Ax = b can besolved very easily once the coefficient matrix A is factored into a product of lower andupper triangular matrices. Second, we shall show how to construct such factorizations.

If an n× n matrix A can be factored into a product of n× n matrices as

A = LU

where L is lower triangular and U is upper triangular, then the linear system Ax = bcan be solved as follows:

Step 1. Rewrite the system Ax = b as

LUx = b (1.14)

Step 2. Define a new n× 1 matrix y by

Ux = y (1.15)

Step 3. Use (1.15) to rewrite (1.14) as Ly = b and solve this system for y.

Step 4. Substitute y in (1.15) and solve for x.

The following example illustrates this procedure.

Example 1.37 Given the factorization

2 1 34 1 7−6 −2 −12

=

2 0 04 −1 0−6 1 −2

1 12

32

0 1 −10 0 1

use this result and the method described above to solve the system

2 1 34 1 7−6 −2 −12

x1

x2

x3

=

−15−2

Solution . . . . . . . . .

LU-Decompositions

We now turn to the problem of constructing such factorizations. To motivate the method,suppose that an n× n matrix A has been reduced to a row-echelon form U by Gaussianelimination—that is, by a certain sequence of elementary row operations. By Theo-rem 1.12 each of these operations can be accomplished by multiplying on the left by anappropriate elementary matrix. Thus there are elementary matrices E1, E2, . . . , Ek suchthat

Ek · · ·E2E1A = U (1.16)


By Theorem 1.13, E1, E2, . . . , Ek are invertible, so we can multiply both sides of Equa-tion (1.16) on the left successively by

E−1k , . . . , E−1

2 , E−11

to obtainA = E−1

1 E−12 · · ·E−1

k U (1.17)

We can show that the matrix L defined by

L = E−11 E−1

2 · · ·E−1k (1.18)

is lower triangular provided that no row interchanges are used in reducing A to U .Substituting (1.18) into (1.17) yields

A = LU

which is a factorization of A into a product of a lower triangular matrix and an uppertriangular matrix.

The following theorem summarizes the above result.

Theorem 1.20 If A is a square matrix that can be reduced to a row-echelon form U byGaussian elimination without row interchange, then A can be factored as A = LU , whereL is a lower triangular matrix.

Definition 1.16 A factorization of a square matrix A as A = LU , where L is lowertriangular and U is upper triangular, is called an LU-decomposition or triangular

decomposition of the matrix A.

Example 1.38 Find an LU-decomposition of

A =

2 1 34 1 7−6 −2 −12

Solution To obtain an LU-decomposition, A = LU , we shall reduce A to a row-echelonform U using Gaussian elimination and then calculate L from (1.18). The steps are asfollows:

Elementary MatrixReduction to Corresponding to Inverse of the

Row-Echelon Form the Row Operation Elementary Matrix

2 1 34 1 7−6 −2 −12

Step 1 E1 =

12

0 00 1 00 0 1

E−11 =

2 0 00 1 00 0 1

1 12

32

4 1 7−6 −2 −12


Elementary MatrixReduction to Corresponding to Inverse of the

Row-Echelon Form the Row Operation Elementary Matrix

Step 2 E2 =

1 0 0−4 1 00 0 1

E−12 =

1 0 04 1 00 0 1

1 12

32

0 −1 1−6 −2 −12

Step 3 E3 =

1 0 00 1 06 0 1

E−13 =

1 0 00 1 0−6 0 1

1 12

32

0 −1 10 1 −3

Step 4 E4 =

1 0 00 −1 00 0 1

E−14 =

1 0 00 −1 00 0 1

1 12

32

0 1 −10 1 −3

Step 5 E5 =

1 0 00 1 00 −1 1

E−15 =

1 0 00 1 00 1 1

1 12

32

0 1 −10 0 −2

Step 6 E6 =

1 0 00 1 00 0 −1

2

E−16 =

1 0 00 1 00 0 −2

1 12

32

0 1 −10 0 1

Thus

U =

1 12

32

0 1 −10 0 1


and, from (1.18),

L =

2 0 00 1 00 0 1

1 0 04 1 00 0 1

1 0 00 1 0−6 0 1

1 0 00 −1 00 0 1

1 0 00 1 00 1 1

1 0 00 1 00 0 −2

=

2 0 04 −1 0−6 1 −2

so

2 1 34 1 7−6 −2 −12

=

2 0 04 −1 0−6 1 −2

1 12

32

0 1 −10 0 1

is an LU-decomposition of A. ♣

Procedure for Finding LU-Decompositions

As this example shows, most of the work in constructing an LU-decomposition is ex-pended in the calculation of L. However, all this work can be eliminated by some carefulbookkeeping of the operations used to reduce A to U . Because we are assuming that norow interchanges are required to reduce A to U , there are only two types of operationsinvolved:

• multiplying a row by a nonzero constant, and

• adding a multiple of one row to another.

The first operation is used to introduce the leading 1’s and the second to introduce zerosbelow the leading 1’s.

In Example 1.38, the multiplier needed to introduce the leading 1’s in successive rowswere as follows:

12

for the first row−1 for the second row−1

2for the third row

Note that in (1.19), the successive diagonal entries in L were precisely the reciprocals ofthese multipliers:

L =

2 0 04 −1 0−6 1 −2

(1.19)

Next, observe that to introduce zeros below the leading 1 in the first row, we used theoperations

add −4 times the first row to the secondadd 6 times the first row to the third

and to introduce the zero below the leading 1 in the second row, we used the operation

add −1 times the the second to the third


Now note in (1.20) that in each position below diagonal of L, the entry is the negativeof the multiplier in the operation that introduced the zero in that position in U :

L =

2 0 04 −1 0−6 1 −2

(1.20)

Therefore, we have the following procedure for constructing an LU-decomposition ofa square matrix A provided that A can be reduced to row-echelon form without rowinterchanges.

Step 1 Reduce A to a row-echelon form U by Gaussian elimination without row inter-changes, keeping track of the multiplier used to introduce the leading 1’s.

Step 2 In each position along the main diagonal of L, place the reciprocal of the mul-tiplier that introduced the leading 1 in that position of U .

Step 3 In each position below the main diagonal of L, place the negative of the multi-plier used to introduce the zero in that position in U .


A =

2 4 21 5 24 −1 9

Solution . . . . . . . . .


A =

6 −2 09 −1 13 7 5

Solution . . . . . . . . .

Exercise 1.7

1. Use the method of Example 1.37 and the LU-decomposition

[3 −6−2 5

]

=

[3 0−2 1

] [1 −20 1

]

to solve the system

3x1 − 6x2 = 0

−2x1 + 5x2 = 1


2. Use the method of Example 1.37 and the LU-decomposition

2 6 2−3 −8 04 9 2

=

2 0 0−3 1 04 −3 7

1 3 10 1 30 0 1

to solve the system

2x1 + 6x2 + 2x3 = 2

−3x1 − 8x2 = 2

4x1 + 9x2 + 2x3 = 3

3. Find an LU-decomposition of the following coefficient matrix; then use the methodof Example 1.37 to solve the system.

(a)

[2 8−1 −1

] [x1

x2

]

=

[−2−2

]

(b)

2 −2 −20 −2 2−1 5 2

x1

x2

x3

=

−4−26

(c)

5 5 10−8 −7 −90 4 26

x1

x2

x3

=

014

(d)

1 2 32 9 63 2 10

x1

x2

x3

=

0107

(e)

1 2 12 −1 −11 1 3

x1

x2

x3

=

410

(f)

1 −2 12 1 −1−3 1 −2

x1

x2

x3

=

21−5

(g)

−1 0 1 02 3 −2 60 −1 2 00 0 1 5

x1

x2

x3

x4

=

5−137

(h)

1 −2 0 3−2 3 1 −6−1 4 −4 35 −8 4 0

x1

x2

x3

x4

=

11−21−123

4. Let

A =

2 1 −1−2 −1 22 1 0

Find an LU-decomposition of A.


1. x1 = 2, x2 = 1 2. x1 = 2, x2 = −1, x3 = 2


3. (a) x1 = 3, x2 = −1 (b) x1 = −1, x2 = 1, x3 = 0 (c) x1 = −1, x2 = 1, x3 = 0

(d) x1 = −49, x2 = 2, x3 = 15 (e) x1 = 1, x2 = 2, x3 = −1

(f) x1 = −49, x2 = 0, x3 = 1 (g) x1 = −3, x2 = 1, x3 = 2, x4 = 1

(h) x1 = 3, x2 = −1, x3 = 0, x4 = 2

4. A = LU =

2 0 0−2 1 02 1 1

1 12

−12

0 0 10 0 0

Chapter 2

Determinants

2.1 Determinants by Cofactor Expansion

Recall from Theorem 1.6 that the 2× 2 matrix

A =

[a bc d

]

is invertible if ad − bc 6= 0. The expression ad − bc is called the determinant of thematrix A and is denoted by the symbol det(A) or |A|. With this notation, the formulafor A−1 given in Theorem 1.6 is

A−1 =1

det(A)

[d −b−c a

]

One of the goals of this chapter is to obtain analog of this formula to square matrix ofhigher order. This will require that we extend the concept of a determinant to squarematrices of all orders.

Minors and Cofactors

There are several ways in which we might proceed. The approach in this section isa recursive approach. It defined the determinant of an n × n matrix in terms of thedeterminants of certain (n− 1)× (n− 1) matrices. The (n− 1)× (n− 1) matrices thatwill appear in this definition are submatrices of the original matrix. These submatricesare given a special name:

Definition 2.1 If A is a square matrix, then the minor of entry aij is denoted by Mij

and is defined to be the determinant of the submatrix that remains after the i th row andj th column are deleted from A. The number (−1)i+jMij is denoted by Cij and is calledthe cofactor of entry aij.

Example 2.1 Let

A =

3 1 −42 5 61 4 8

Fine the minor and cofactor of a11 and a32.

57


Solution . . . . . . . . .

Note that the cofactor and the minor of an element aij differ only in sign; that is,Cij = ±Mij . A quick way to determine whether to use + or − is to use the fact that thesign relating Cij and Mij is in the i th row and j th column of the “checkerboard” array

+ − + − + · · ·− + − + − · · ·+ − + − + · · ·− + − + − · · ·...

......

......

For example, C11 = M11, C21 = −M21, C13 = M13, C32 = −M32, and so on.

Cofactor Expansions

The definition of a 3× 3 determinant in terms of minors and cofactors is

det(A) = a11M11 + a12(−M12) + a13M13

= a11C11 + a12C12 + a13C13 (2.1)

Equation (2.1) show that the determinant of A can be computed by multiplying theentries in the first row of A by their corresponding cofactors and adding the resultingproducts. More generally, we define the determinant of an n× n matrix to be

det(A) = a11C11 + a12C12 + · · ·+ a1nC1n

This method of evaluating det(A) is called cofactor expansion along the first row ofA.

Example 2.2 Let A =

3 1 0−2 −4 35 4 −2

. Evaluate det(A) by cofactor expansion along

the first row of A.

Solution . . . . . . . . .

If A is a 3× 3 matrix, then its determinant is

det(A) =

∣∣∣∣∣∣

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣

= a11

∣∣∣∣

a22 a23a32 a33

∣∣∣∣− a12

∣∣∣∣

a21 a23a31 a33

∣∣∣∣+ a13

∣∣∣∣

a21 a22a31 a32

∣∣∣∣

= a11(a22a33 − a23a32)− a12(a21a33 − a23a31)

+a13(a21a32 − a22a31) (2.2)

= a11a22a33 + a12a23a31 + a13a21a32 − a13a22a31

−a12a21a33 − a11a23a32 (2.3)


By rearranging the terms in (2.3) in various ways, it is possible to obtain other formulaslike (2.2). There should be no trouble checking that all of the following are correct:

det(A) = a11C11 + a12C12 + a13C13

= a11C11 + a21C21 + a31C31

= a21C21 + a22C22 + a23C23

= a12C12 + a22C22 + a32C32

= a31C31 + a32C32 + a33C33

= a13C13 + a23C23 + a33C33 (2.4)

Note that in each equation, the entries and cofactors all come from the same row orcolumn. These equations are called the cofactor expansions of det(A).

The results we have just given for 3× 3 matrices from a special case of the followinggeneral theorem.

Theorem 2.1 Expansions by Cofactors

The determinant of an n×n matrix A can be computed by multiplying the entries in anyrow (or column) by their cofactors and adding the resulting products; that is, for each1 ≤ i ≤ n and 1 ≤ j ≤ n,

det(A) = a1jC1j + a2jC2j + · · ·+ anjCnj

(cofactor expansion along the j th column)

and

det(A) = ai1Ci1 + ai2Ci2 + · · ·+ ainCin

(cofactor expansion along the i th row)

Note that we may choose any row or any column.

Example 2.3 Let A be the matrix in Example 2.2. Evaluate det(A) by cofactor expan-sion along the first column of A.

Solution . . . . . . . . .

Remark In general, the best strategy for evaluating a determinant by cofactor expansionis to expand along a row or column having the largest number of zeros.

Example 2.4 Let

A =

1 0 0 −13 1 2 21 0 −2 12 0 0 1

Find det(A) by a cofactor expansion along a row or column of your choice.

Solution . . . . . . . . .


Adjoint of a Matrix

Definition 2.2 If A is any n× n matrix and Cij is the cofactor of aij, then the matrix

C11 C12 · · · C1n

C21 C22 · · · C2n...

......

Cn1 Cn2 · · · Cnn

is call matrix of cofactors from A. The transpose of this matrix is called the adjointmatrix of A and is denoted by adj(A).

Example 2.5 Find adj(A) when

A =

3 2 −11 6 32 −4 0

.

Solution . . . . . . . . .

Theorem 2.2 Inverse of a Matrix Using Its Adjoint

If A is an invertible matrix, then

A−1 =1

det(A)adj(A) (2.5)

Example 2.6 Use (2.5) to find the inverse of the matrix A in Example 2.5.

Solution . . . . . . . . .

Theorem 2.3 If A is an n×n triangular matrix(upper triangular, lower triangular, or

diagonal), then det(A) is the product of the entries on the main diagonal of the matrix;

that is, det(A) = a11a22 · · · ann.

Example 2.7 Find det(A) when

A =

2 7 −3 8 30 −3 7 5 10 0 6 7 60 0 0 9 80 0 0 0 4

.

Solution . . . . . . . . .

Cramer’s Rule

The next theorem provides a formula for the solution of certain linear systems of nequations in n unknowns. This formula, known as Cramer’s rule.


Theorem 2.4 If Ax = b is a system of n linear equations in n unknowns such thatdet(A) 6= 0, then the system has a unique solution. This solution is

x1 =det(A1)

det(A), x2 =

det(A2)

det(A), . . . , xn =

det(An)

det(A)

where Aj is the matrix obtained by replacing the entries in the j th column of A by theentries in the matrix

b =

b1b2...bn

Example 2.8 Use Cramer’s rule to solve

x1 + 2x3 = 6

−3x1 + 4x2 + 6x3 = 30

−x1 − 2x2 + 3x3 = 8

Solution . . . . . . . . .

Exercise 2.1

1. Let

A =

1 −2 36 7 −1−3 1 4

(a) Find all the minors of A. (b) Find all the cofactors.

2. Evaluate the determinant of the matrix in Exercise 1 by a cofactor expansion along

(a) the first row (b) the first column (c) the second row

(d) the second column (e) the third row (f) the third column

3. For the matrix in Exercise 1, find

(a) adj(A) (b) A−1 using its adjoint

4. Evaluate det(A) by a cofactor expansion along a row or column of your choice.

(a) A =

3 3 11 0 −41 −3 5

(b) A =

k + 1 k − 1 72 k − 3 45 k + 1 k

5. Find A−1 using its adjoint.

(a) A =

2 5 5−1 −1 02 4 3

(b) A =

2 −3 50 1 −30 0 2


6. Let

A =

1 3 1 12 5 2 21 3 8 91 3 2 2

(a) Find A−1 using its adjoint.

(b) Find A−1 using the method of Example 1.30 in Section 1.5.

(c) Which method involves less computation?

7. Show that the matrix

A =

cos θ sin θ 0− sin θ cos θ 0

0 0 1

is invertible for all values of θ; then find A−1 using its adjoint.

8. (a) If A =

[A11 A12

0 A22

]

is an “upper triangular” block matrix, where A11 and A22

are square matrices, then det(A) = det(A11)det(A22). Use this result to evaluatedet(A) for

A =

2 −1 2 5 64 3 −1 3 40 0 1 3 50 0 −2 6 20 0 3 5 2

(b) Verify your answer in part (a) by using cofactor expansion to evaluate det(A).

9. Solve by Cramer’s rule, where it applies.

(a) 7x1 − 2x2 = 33x1 + x2 = 5

(b) x− 4y + z = 64x− y + 2z = −12x+ 2y − 3z = −20

(c) 3x1 − x2 + x3 = 4−x1 + 7x2 − 2x3 = 12x1 + 6x2 − x3 = 5

10. Solve the system

4x+ y + z + w = 6

3x+ 7y − z + w = 1

7x+ 3y − 5z + 8w = −3

x+ y + z + 2w = 3

by (a) Cramer’s rule (b) Guass-Jordan elimination.



1. (a) M11 = 29, M12 = 21, M13 = 27, M21 = −11, M22 = 13,

M23 = −5, M31 = −19, M32 = −19, M33 = 19

(b) C11 = 29, C12 = −21, C13 = 27, C21 = 11, C22 = 13, C23 = 5,

C31 = −19, C32 = 19, C33 = 19

2. 152

3. (a) adj(A) =

29 11 −19−21 13 1927 5 19

(b) A−1 =

29152

11152

− 19152

− 21152

13152

19152

27152

5152

19152

4. (a) −66 (b) k3 − 8k2 − 10k + 95

5. (a) A−1

3 −5 −5−3 4 52 −2 −3

(b) A−1 =

12

32

10 1 3

2

0 0 12

6. A−1 =

−4 3 0 −12 −1 0 0−7 0 −1 86 0 1 −7

7. A−1 =

cos θ − sin θ 0sin θ cos θ 00 0 1

8. det(A) = 10× (−108) = −1080

9. (a) x1 = 1, x2 = 2 (b) x = −14455, y = −61

55, z = 46

11

(c) Cramer’s rule do not apply.

10. x = 1, y = 0, z = 2, w = 0

2.2 Evaluating Determinants by Row Reduction

In this section we shall show that the determinant of a square matrix can be evaluatedby reducing the matrix to row-echelon form. This method is important since it is themost computationally efficient way to find the determinant of a general matrix.

We begin with a fundamental theorem that will lead us to an efficient procedure forevaluating the determinant of a matrix of any order n.

Theorem 2.5 Let A be a square matrix. If A has a row of zeros or a column of zeros,then det(A) = 0.

Theorem 2.6 Let A be a square matrix. Then det(A) = det(AT ).


Theorem 2.7 Let A be a square matrix.

1. If B is the matrix that results when a single row or single column of A is multipliedby a scalar k, then det(B) = k det(A).

2. If B is the matrix that results when two rows or two columns of A are interchanged,then det(B) = −det(A).

3. If B is the matrix that results when a multiple of one row of A is added to anotherrow or when a multiple of one column is added to another column, then det(B) =det(A).

The following table illustrates Theorem 2.7 for 3× 3 determinants.

Relationship Operation∣∣∣∣∣∣

k a11 k a12 k a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣

= k

∣∣∣∣∣∣

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣

The first row of Ais multiplied by k.

det(B) = k det(A)∣∣∣∣∣∣

a21 a22 a23a11 a12 a13a31 a32 a33

∣∣∣∣∣∣

= −

∣∣∣∣∣∣

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣

The first andsecond rows of Aare interchanged.

det(B) = − det(A)

∣∣∣∣∣∣

a11 + k a12 a12 + k a22 a13 + k a23a21 a22 a23a31 a32 a33

∣∣∣∣∣∣

=

∣∣∣∣∣∣

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣

A multiple of thesecond row of Ais added to thefirst row.

det(B) = det(A)

Elementary Matrices

If we let A = In in Theorem 2.7, then the matrix B is an elementary matrix, and thetheorem yields the following result about determinants of elementary matrices.

Theorem 2.8 Let E be an n× n elementary matrix.

1. If E results from multiplying a row of In by k, then det(E) = k.

2. If E results from interchanging two rows of In, then det(E) = −1.

3. If E results from adding a multiple of one row of In to another, then det(E) = 1.


Example 2.9 Find the determinant of the following matrices.

(a) A =

1 0 0 00 3 0 00 0 1 00 0 0 1

(b) B =

0 0 0 10 3 0 00 0 1 01 0 0 0

(c) C =

1 0 0 70 1 0 00 0 1 00 0 0 1

Matrices with Proportional Rows or Columns

Theorem 2.9 If A is a square matrix with two proportional rows or two proportionalcolumns, then det(A) = 0.

Each of the following matrices has two proportional rows or columns; thus, each hasa determinant of zero.

[−1 4−2 8

]

,

1 −2 7−4 8 52 −4 3

,

3 −1 4 −56 −2 5 25 8 1 4−9 3 −12 15

Evaluating Determinants by Row Reduction

We shall now give a method for evaluating determinants that involves substantially lesscomputation than the cofactor expansion method. The idea of this method is to reducethe given matrix to upper triangular from by elementary row operations, then computethe determinant of the upper triangular matrix (an easy computation), and then relatethat determinant to that of the original matrix. Here is an example:

Example 2.10 Evaluate det(A) where

A =

0 1 53 −6 92 6 1

Solution . . . . . . . . .

Example 2.11 Compute the determinant of

A =

1 2 0 70 7 6 30 0 3 13 6 0 −5


Solution


A =

1 2 0 −20 0 2 −10 −1 1 01 3 4 1

Solution . . . . . . . . .

Cofactor expansion and row operations can sometimes be used in combination to pro-vide an effective method for evaluating determinants. The following example illustratesthis idea.


A =

3 5 −2 61 2 −1 12 4 1 53 7 5 3

Solution . . . . . . . . .


A =

2 1 −3 1−3 −2 0 22 1 0 −11 0 1 2

Solution . . . . . . . . .


A =

13

0 34

25

−1 32

18

−34

54

Solution . . . . . . . . .

Exercise 2.2

1. Evaluate the following determinants.

(a)

∣∣∣∣∣∣

3 −17 40 5 10 0 −2

∣∣∣∣∣∣

(b)

∣∣∣∣∣∣∣∣

√2 0 0 0

−8 2 0 07 0 −1 09 5 6 1

∣∣∣∣∣∣∣∣

(c)

∣∣∣∣∣∣

−2 1 31 −7 4−2 1 3

∣∣∣∣∣∣

(d)

∣∣∣∣∣∣

1 −2 32 −4 65 −8 1

∣∣∣∣∣∣


2. Evaluate the determinant of the given matrix by reducing the matrix to row-echelonform.

(a)

3 6 90 0 −2−2 1 5

(b)

1 −3 0−2 4 15 −2 2

(c)

1 −2 3 15 −9 6 3−1 2 −6 −22 8 6 1

(d)

1 3 1 5 3−2 −7 0 −4 20 0 1 0 10 0 2 1 10 0 0 1 1

3. Given that

∣∣∣∣∣∣

a b cd e fg h i

∣∣∣∣∣∣

= −6, find

(a)

∣∣∣∣∣∣

d e fg h ia b c

∣∣∣∣∣∣

(b)

∣∣∣∣∣∣

3a 3b 3c−d −e −f4g 4h 4i

∣∣∣∣∣∣

(c)

∣∣∣∣∣∣

a + g b+ h c+ id e fg h i

∣∣∣∣∣∣

(d)

∣∣∣∣∣∣

−3a −3b −3cd e f

g − 4d h− 4e i− 4f

∣∣∣∣∣∣

4. Repeat Exercise 2 using a combination of row reduction and cofactor expansion,as in Example 2.13.

5. Solve the equation ∣∣∣∣∣∣

x 5 70 x+ 1 60 0 2x− 1

∣∣∣∣∣∣

= 0


1. (a) −30 (b) −2 (c) 0 (d) 0 2. (a) 30 (b) −17 (c) 39 (d) −2

3. (a) −6 (b) 72 (c) −6 (d) 18 5. x = 0,−1, 12

2.3 Properties of the Determinant Function

Basic Properties of Determinants

Suppose that A and B are n× n matrices and k is any scalar. We begin by consideringpossible relationships between det(A), det(B), and

det(kA), det(A+B), and det(AB)


Since a common factor of any row of a matrix can be moved though the det sign, andsince each of the n rows in kA has a common factor of k, we obtain

det(kA) = kndet(A) (2.6)

For example, ∣∣∣∣∣∣

ka11 ka12 ka13ka21 ka22 ka23ka31 ka32 ka33

∣∣∣∣∣∣

= k3

∣∣∣∣∣∣

a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣

Unfortunately, no simple relationship exists among det(A), det(B), and det(A+B).In particular, we emphasize that det(A+B) will usually not be equal to det(A)+det(B).For example, if

A =

[1 22 5

]

, B =

[3 11 3

]

, A+B =

[4 33 8

]

,

then det(A) = 1, det(B) = 8, and det(A+B) = 23; thus

det(A+B) 6= det(A) + det(B)

But if we consider two 2× 2 matrices that differ only in the second row:

A =

[a11 a12a21 a22

]

and B =

[a11 a12b21 b22

]

,

then we have

det(A) + det(B) = (a11a22 − a12a21) + (a11b22 − a12b21)

= a11(a22 + b22)− a12(a21 + b21)

= det

[a11 a12

a21 + b21 a22 + b22

]

Thus

det

[a11 a12a21 a22

]

+ det

[a11 a12b21 b22

]

= det

[a11 a12

a21 + b21 a22 + b22

]

This is a special case of the following theorem.

Theorem 2.10 Let A,B, and C be n × n matrices that differ only in the single row,say the r th, and assume that the r th row of C can be obtained by adding correspondingentries in the r th row of A and B. Then

det(C) = det(A) + det(B)

For example,

det

1 7 52 0 3

1 + 0 4 + 1 7 + (−1)

= det

1 7 52 0 31 4 7

+ det

1 7 52 0 30 1 −1


Determinant of a Matrix Product

Corollary 2.1 If B is an n× n matrix and E is an n× nelementary matrix, then

det(EB) = det(E)det(B)

Determinant Test for Invertibility

Theorem 2.11 A square matrix A is invertible if and only if det(A) 6= 0.

Example 2.16 Determine whether the matrix A =

1 2 31 0 12 4 6

is invertible?

Solution . . . . . . . . .

Theorem 2.12 If A and B are square matrices of the same size, then

det(AB) = det(A)det(B)

Example 2.17 Let

A =

[3 12 1

]

, B =

[−1 35 8

]

Show that det(AB) = det(A)det(B).

Solution . . . . . . . . .

Theorem 2.13 If A is invertible, then

det(A−1) =1

det(A).

We conclude this section by merging Theorem 2.11 with the list to produce thefollowing theorem that relates all the major topics we have studied thus far.

Theorem 2.14 (Equivalent Statements) If A is an n×n matrix, then the followingstatements are equivalent.




(d) A can be expressed as a product of elementary matrices.

(e) Ax = b is consistent for every n× 1 matrix b.

(f) Ax = b has exactly one solution for every n× 1 matrix b.

(g) det(A) 6= 0.


Exerise 2.3

1. Verify that det(kA) = kndet(A) for

(a) A =

[−1 23 4

]

; k = 2 (b) A =

2 −1 33 2 11 4 5

; k = −2

2. Verify that det(AB) = det(A)det(B) for

A =

2 1 03 4 00 0 2

and B =

1 −1 37 1 25 0 1

Is det(A+B) = det(A) + det(B)?

3. By inspection, explain why det(A) = 0.

A =

−2 8 1 43 2 5 11 10 6 54 −6 4 −3

4. Use Theorem 2.11 to determine which of the following matrices are invertible.

(a)

1 0 −19 −1 48 9 −1

(b)

4 2 8−2 1 −43 1 6

(c)

√2 −

√7 0

3√2 −3

√7 0

5 −9 0

(d)

−3 0 15 0 68 0 3

5. Let

A =

a b cd e fg h i

Assuming that det(A) = −7, find

(a) det(3A) (b) det(A−1) (c) det(2A−1)

(d) det(2A)−1 (e) det

a g db h ec i f

6. Without directly evaluating, show that x = 0 and x = 2 satisfy∣∣∣∣∣∣

x2 x 22 1 10 0 −5

∣∣∣∣∣∣

= 0


7. For which value(s) of k does A fail to be invertible?

(a) A =

[k − 3 −2−2 k − 2

]

(b) A =

1 2 43 1 6k 3 2

8. Indicate whether the statement is always true or sometimes false. Justify eachanswer by giving a logical argument or a counterexample.

(a) det(2A) = 2det(A)

(b) |A2| = |A|2

(c) det(I + A) = 1 + det(A)

(d) if det(A) = 0 then A is not expressible as a product of elementary matrices.

(e) If the reduced row-echelon form of A has a row of zeros, then det(A) = 0.

(f) There is no square matrix A such that det(AAT ) = −1.

(g) If det(A) = 0, then the homogeneous system Ax = 0 has infinitely manysolutions.


1. (a) det(2A) = −40 = 22det(A) (b) det(−2A) = −448 = (−2)3det(A)

2. det(AB) = −170 = 10 · (−17) = det(A)det(B)

3. The second and forth columns of A are proportional.

4. (a) Invertible (b) Not invertible (c) Not invertible (d) Not invertible

5. (a) −63 (b) −17

(c) −87

(d) − 156

(e) 7

6. If x = 0, the first and third rows are proportional.

If x = 2, the first and second rows are proportional.

7. (a) k = 5±√17

2(b) k = −1

8. (a) False (b) True (c) False (d) True (e) True (f) True (g) True

Chapter 3

Vector Spaces

3.1 Real Vector Spaces

Vector Space Axioms

Definition 3.1 Let V be an arbitrary nonempty on which the operations of additionsand scalar multiplication are defined. If the following axioms are satisfied by all objectsu, v, w in V and all scalars k and m, then we call V a vector space and we call theobjects in V vectors.

1. If u and v are objects in V , then u+ v is in V .

2. u+ v = v + u

3. u+ (v +w) = (u+ v) +w

4. There is an object 0 in V , called a zero vector for V , such that 0+u = u+0 = ufor all u in V .

5. For each u in V , there is an object −u in V , called a negative of u, such thatu+ (−u) = (−u) + u = 0.

6. If k is any scalar and u is any object in V , then ku is in V .

7. k(u+ v) = ku+ kv

8. (k +m)u = ku+mu

9. k(mu) = (km)u

10. 1u = u

Remark Vector spaces in which the scalars are complex numbers are called complex

vector spaces, and those in which the scalars must be real are called real vector

spaces. In this chapter, all of our scalar will be real numbers.

72


Examples of Vector Spaces

Example 3.1 Rn Is a Vector Space

Let u = (u1, u2, . . . , un) and v = (v1, v2, . . . , vn) are two vectors in Rn.

The sum u+ v is defined by

u+ v = (u1 + v1, u2 + v2, . . . , un + vn)

and if k is any scalar, the scalar multiple ku is defined by

ku = (ku1, ku2, . . . , kun)

The operations of addition and scalar multiplication in the above are called the standardoperations on R

n.We can show that the set V = R

n with these standard operations is a vector space.♣

The three most important special cases of Rn are R (the real numbers), R

2 (thevectors in the plane), and R

3 (the vectors in 3-space).

Example 3.2 A Vector Space of 2× 2 Matrices

Show that the set V of all 2 × 2 matrices with real entries is a vector space if additionis defined to be matrix addition and scalar multiplication is defined to be matrix scalarmultiplication.

Solution It convenient to verify the axioms in the following order: 1, 6, 2, 3, 7, 8, 9, 4,5, and 10. Let

u =

[u11 u12

u21 u22

]

and v =

[v11 v12v21 v22

]

To prove Axiom 1, we must show that u + v is an object in V ; that is, we must showthat u+ v is a 2× 2 matrix.

u+ v =

[u11 u12

u21 u22

]

+

[v11 v12v21 v22

]

=

[u11 + v11 u12 + v12u21 + v21 u22 + v22

]

Similarly, Axiom 6 holds because for any real scalar number k, we have

ku = k

[u11 u12

u21 u22

]

=

[ku11 ku12

ku21 ku22

]

so ku is a 2× 2 matrix and consequently is an object in V .Axiom 2 follows from Theorem 1.2 (a) since

u+ v =

[u11 u12

u21 u22

]

+

[v11 v12v21 v22

]

=

[v11 v12v21 v22

]

+

[u11 u12

u21 u22

]

= v + u

Similarly, Axiom 3 follows from part (b) of that theorem; and Axioms 7, 8, and 9 followfrom parts (f), (h), and (j), respectively.


To prove Axiom 4, we must find an object 0 in V such that 0 + u = u + 0 = u forall u in V . This can be done by defining 0 to be

0 =

[0 00 0

]

With this definition,

0+ u =

[0 00 0

]

+

[u11 u12

u21 u22

]

=

[u11 u12

u21 u22

]

= u

and similarly u+ 0 = u. To prove Axiom 5, we must show that each object u in V hasa negative −u such that u+ (−u) = 0 and (−u) +u = 0. This can be done by definingthe negative of u to be

−u =

[−u11 −u12

−u21 −u22

]

With this definition,

u+ (−u) =

[u11 u12

u21 u22

]

+

[−u11 −u12

−u21 −u22

]

=

[0 00 0

]

= 0

and similarly (−u) + u = 0. Finally, Axiom 10 is a simple computation:

1u = 1

[u11 u12

u21 u22

]

=

[u11 u12

u21 u22

]

= u ♣

Example 3.2 is a special case of a vector space V of all m × n matrices with realentries, together with the operations of matrix addition and scalar multiplication. Weshall denote this vector space by the symbol Mm×n.

Example 3.3 A Vector Space of Real-Valued Functions

Let V be the set of real-value functions defined on the entire real line (−∞,∞). Iff = f(x) and g = g(x) are two such functions and k is any real number, define the sumfunction f + g and the scalar multiple kf , respectively, by

(f + g)(x) = f(x) + g(x) and (kf)(x) = kf(x)

In the exercise we shall ask you to show that V is a vector space with respect to theseoperations. This vector space is denoted by F (−∞,∞). ♣


Example 3.4 The Vector Space Pn

Let Pn denote the set of all polynomials of degree less than or equal n. If p and q are twopolynomials in Pn and k is any real number, define the sum p+ q and the scalar multiplekp, respectively, by

(p+ q)(x) = p(x) + q(x) and (kp)(x) = kp(x)

for all real number x. In this case, the zero vector is the zero polynomial,

z(x) = 0xn + 0xn−1 + 0xn−2 + · · ·+ 0x+ 0

It is easily verified that all the vector space axioms holds. thus Pn, with the standardaddition and scalar multiplication of functions, is a vector space. ♣

Example 3.5 The Zero Vector Space

Let V consist of a single object, which we denote by 0, and define

0+ 0 = 0 and k0 = 0

for all scalar k. It is easy to check that all the vector space axioms are satisfied. We callthis the zero vector space.

Example 3.6 A Set That Is Not a Vector Space

Let V = R2 and define addition and scalar multiplication operations as follows: If u =

(u1, u2) and v = (v1, v2), then define

u+ v = (u1 + v1, u2 + v2)

and if k is any real number, then define

ku = (ku1, 0)

For example,if u = (2, 4),v = (−3, 5), and k = 7, then

u+ v = (2 + (−3), 4 + 5) = (−1, 8)ku = 7u = (7 · 2, 0) = (14, 0)

In the exercises we will ask you to show that the first nine vector space axioms aresatisfied; however, there are values of u for which Axiom 10 fails to hold. For example,if u = (u1, u2) is such that u2 6= 0, then

1u = 1(u1, u2) = (1 · u1, 0) = (u1, 0) 6= u

Thus V is not a vector space with the stated operations. ♣


Some Properties of Vectors

Theorem 3.1 Let V be a vector space, u a vector in V , and k a scalar; then:

(a) 0u = 0

(b) k0 = 0

(c) (−1)u = −u

(d) If ku = 0, then k = 0 or u = 0.

Exercise 3.1

1. A set of objects is given, together with operations of addition and scalar multipli-cation. Determine which sets are vector spaces under the given operations. Forthose that are not vector spaces, list all axioms that fail to hold.

(a) The set of all triples of real numbers (x, y, z) with the operations

(x, y, z) + (x′, y′, z′) = (x+ x′, y + y′, z + z′) and k(x, y, z) = (kx, y, z)

(b) The set of all pairs of real numbers (x, y) with the operations

(x, y) + (x′, y′) = (x+ x′, y + y′) and k(x, y) = (2kx, 2ky)

(c) The set of all pairs of real numbers of the form (x, 0) with the standardoperations on R

2.

(d) The set of all n-tuples of real numbers of the form (x, x, . . . , x) with thestandard operations on R

n.

(e) The set of 2× 2 matrices of the form

[a 11 b

]

with the standard matrix addition and scalar multiplication.

(f) The set of all real-valued functions f defined everywhere on the real line andsuch that f(1) = 0, with the operations defined in Example 3.3.

(g) The set of all pairs of real numbers of the form (1, x) with the operations

(1, y) + (1, y′) = (1, y + y′) and k(1, y) = (1, ky)

(h) The set of polynomials of the form a+ bx with the operations

(a0+a1x)+(b0+b1x) = (a0+b0)+(a1+b1)x and k(a0+a1x) = (ka0)+(ka1)x

2. Complete the unfinished details of Example 3.3.


3. Show that the first nine vector space axioms are satisfied if V = R2 has the addition

and scalar multiplication operations defined in Example 3.6.


1. (a) Not a vector space. Axiom 8 fails.

(b) Not a vector space. Axiom 9 and 10 fail.

(c) The set is a vector space under the given operations.

(d) The set is a vector space under the given operations.

(e) Not a vector space. Axiom 1, 4, 5, and 6 fail.

(f) The set is a vector space under the given operations.

(g) The set is a vector space under the given operations.

(h) The set is a vector space under the given operations.

3.2 Subspaces

Given a vector space V , it is often possible to form another vector space by taking asubset S of V and using the operations of V .

Definition 3.2 A subset S of a vector space V is called subspaces of V if S is itself avector space under the addition and scalar multiplication defined on V .

In general, one must verify the ten vector space axioms to show that a set S withaddition and scalar multiplication forms a vector space. However, if S is part of thelarger set V that is already known to be a vector space, then certain axioms need notbe verified for S because they are “inherited” from V . For example, there is no need tocheck that u + v = v + u (Axiom 2) for S because this holds for all vectors in V andconsequently for all vectors in S. Other axioms inherited by S from V are 3, 7, 8, 9, and10. Thus to show that the set S is a subspace of a vector space V , we need only verifyAxioms 1, 4, 5, and 6. The following theorem shows that even Axioms 4 and 5 can beomitted.

Theorem 3.2 If S is a nonempty subset of a vector space V , then S is a subspace of Vif and only if the following conditions hold.

(a) If u and v are vectors in S, then u+ v is in S.

(b) If k is any scalar and u in any vector in S, then ku is in S.

Remark1. A set S of one or more vectors form a vector space V is said to be closed

under addition if condition (a) in Theorem 3.2 holds and closed under scalar

multiplication if condition (b) holds. Thus Theorem 3.2 states that S is a subspace ofV if and only if S is closed under addition and closed under scalar multiplication.

2. Every nonempty vector space V has at least two subspaces: V itself is a subspace,and the set {0} consisting of just the zero vector in V is a subspace called the zerosubspace.


Example 3.7 Let S = {(x1, x2, x3) ∈ R3 | x1 = x2}. Show that S is a subspace of R3.

Solution . . . . . . . . .

Example 3.8 Let S = {A ∈ M2×2 | a12 = −a21}. Show that S is a subspace of M2×2.

Solution . . . . . . . . .

Example 3.9 Let S be the set of all polynomials of degree less than n with the propertyp(0) = 0. Show that S is a subspace of Pn.

Solution . . . . . . . . .

Example 3.10 Let Cn[a, b] be the set of all functions f that have a continuous n thderivative on [a, b]. We can verify that Cn[a, b] is a subspace of C[a, b]. ♣

Remark C[a, b] denote the set of all real-value functions that defined and continuous onthe closed interval [a, b].

Example 3.11 Let S be the set of all functions f in C2[a, b] such that

f ′′(x) + f(x) = 0

for all x in [a, b]. Show that S is a subspace of C2[a, b]

Solution . . . . . . . . .

Example 3.12 A Subset of R2 That Is Not a Subspace

Let W be the set of all points (x, y) in R2 such that x ≥ 0 and y ≥ 0. The set W

is not a subspace of R2 since it is not closed under scalar multiplication. For example,u = (1, 1) lies in W , but its negative (−1)u = −u = (−1,−1) does not. ♣

Example 3.13 Let W =

{[x1

] ∣∣∣∣x is a real number

}

. Determine whether W is a sub-

space of M2×1.

Solution . . . . . . . . .

Solution Spaces of Homogeneous Systems

If Ax = b is a system of linear equations, then each vector x that satisfies this equationis called solution vector of the system. The following theorem shows that the solutionvectors of a homogeneous linear system form a vector space, which we shall call solutionspace of the system.

Theorem 3.3 If Ax = 0 is a homogeneous linear system of m equations in n unknowns,then the set of solution vectors, denoted by N(A), is a subspace of Rn.

Note N(A) = {x ∈ Rn |Ax = 0}


Example 3.14 Determine N(A) if

A =

[1 1 1 02 1 0 1

]

Solution . . . . . . . . .

Definition 3.3 A vector w is called a linear combination of the vectors v1,v2, . . . ,vr

if it can be expressed in the form

w = k1v1 + k2v2 + · · ·+ krvr

where k1, k2, . . . , kr are scalars.

Remark If r = 1, then the equation in Definition 3.3 reduces to w = k1v1; that is, wis a linear combination of a single vector v1 if it is a scalar multiple of v1.

Example 3.15 Every vector v = (a, b, c) in R3 is expressible as a linear combination of

the standard basis vectors

i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)

sincev = (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = ai + bj + ck ♣

Example 3.16 Consider the vectors u = (1, 2,−1) and v = (6, 4, 2) in R3. Show that

(a) w = (9, 2, 7) is a linear combination of u and v and

(b) w′ = (4,−1, 8) is not a linear combination of u and v

Solution . . . . . . . . .

Spanning

If v1,v2, . . . ,vr are vectors in vector space V , then generally some vectors in V may belinear combinations of v1,v2, . . . ,vr and other may not. The following theorem showsthat if we construct a set W consisting of all those vectors that are expressible as linearcombinations of v1,v2, . . . ,vr, then W forms a subspace of V .

Theorem 3.4 If v1,v2, . . . ,vr are vectors in a vector space V , then

(a) The set W of all linear combinations of v1,v2, . . . ,vr is a subspace of V .

(b) W is the smallest subspace of V that contains v1,v2, . . . ,vr in the sense that everyother subspace of V that contains v1, v2, . . ., vr must contain W .

Definition 3.4 If S = {v1,v2, . . . ,vr} is the set of vectors in a vector space V , thenthe subspace W of V consisting of all linear combinations of the vector in S is called thespace spanned by v1, v2, . . ., vr, and we say that the vectors v1, v2, . . ., vr span W .To indicate that W is the space spanned by the vectors in the set S = {v1,v2, . . . ,vr},we write

W = Span(S) or W = Span{v1,v2, . . . ,vr}


Example 3.17 If v1 and v2 are noncollinear vectors in R3 with their initial points at

the origin, then Span{v1,v2}, which consists of all linear combinations k1v1 + k2v2, isthe plane determined by v1 and v2.

z

x

y

Span{v1,v2}

k1v1 + k2v2

v1

k1v1

v2

k2v2

Similarly, if v is a nonzero vector in R2 or R3, then Span{v}, which is the set of all

scalar multiples kv, is the line determined by v.

z

x

y

Span{v}

v

kv

Example 3.18 The polynomial 1, x, x2, . . . , xn span the vector space Pn defined inExample 3.4 since each polynomial p in Pn can be written as

p = a0 + a1x+ · · ·+ anxn

which is a linear combination of 1, x, x2, . . . , xn. We can denote this by writing

Pn = Span{1, x, x2, . . . , xn} ♣

Example 3.19 Determine whether v1 = (1, 1, 2), v2 = (1, 0, 1), and v3 = (2, 1, 3) spanthe vector space R

3.

Solution . . . . . . . . .

Example 3.20 Determine whether the vectors 1− x2, x+ 2, and x2 span P3.

Solution . . . . . . . . .


Theorem 3.5 If S = {v1,v2, . . . ,vr} and S ′ = {w1,w2, . . . ,wk} are two set of vectorsin a vector space V , then

Span{v1,v2, . . . ,vr} = Span{w1,w2, . . . ,wk}

if and only if each vector in S is a linear combination of those in S ′ and each vector inS ′ is a linear combination of those in S.

Exercise 3.2

1. Use Theorem 3.2 to determine which of the following sets form subspaces of R2.

(a) {(x1, x2)| x1 + x2 = 0} (b) {(x1, x2)| x1x2 = 0}(c) {(x1, x2)| x1 = 3x2} (d) {(x1, x2)| |x1| = |x2|}(e) {(x1, x2)| x2

1 = x22}

2. Use Theorem 3.2 to determine which of the following are subspaces of R3.

(a) all vectors of the form (a, 0, 0)

(b) all vectors of the form (a, 1, 1)

(c) all vectors of the form (a, b, c), where b = a+ c

(d) all vectors of the form (a, b, c), where b = a+ c + 1

(e) all vectors of the form (a, b, 0)

(f) all vectors of the form (a, b, c), where a = b = c

3. Use Theorem 3.2 to determine which of the following are subspaces of P3.

(a) all polynomials a0 + a1x+ a2x2 + a3x

3 for which a0 = 0

(b) all polynomials a0 + a1x+ a2x2 + a3x

3 for which a0 + a1 + a2 + a3 = 0

(c) all polynomials a0+a1x+a2x2+a3x

3 for which a0, a1, a2, and a3 are integers

(d) all polynomials p(x) of the form a0 + a1x+ a2x2 + a3x

3 such that p(0) = 0

4. Use Theorem 3.2 to determine which of the following sets form subspaces of M2×2.

(a) The set of all 2× 2 diagonal matrices

(b) The set of all 2× 2 triangular matrices

(c) The set of all 2× 2 lower triangular matrices

(d) The set of all 2× 2 matrices A such that a12 = 1

(e) The set of all 2× 2 matrices B such that b11 = 0

(f) The set of all symmetric 2× 2 matrices

(g) The set of all singular 2× 2 matrices

5. Use Theorem 3.2 to determine which of the following are subspaces of Mn×n


(a) all n× n matrices A such that tr(A) = 0

(b) all n× n matrices A such that AT = −A

(c) all n× n matrices A such that the linear system Ax = 0 has only the trivialsolution

(d) all n× n matrices A such that AB = BA for a fixed n× n matrix B

6. Use Theorem 3.2 to determine which of the following sets form subspaces ofC[−1, 1].

(a) The set of function f in C[−1, 1] such that f(−1) = f(1)

(b) The set of odd functions in C[−1, 1]

(c) The set of continuous nondecreasing functions on [−1, 1]

(d) The set of function f in C[−1, 1] such that f(−1) = 0 and f(1) = 0

(e) The set of function f in C[−1, 1] such that f(−1) = 0 or f(1) = 0

7. Determine the nullspace of each of the following matrices.

(a) A =

[2 13 2

]

(b) A =

[1 2 −3 −1−2 −4 6 3

]

(c) A =

1 3 −42 −1 −1−1 −3 4

(d) A =

1 1 −1 22 2 −3 1−1 −1 0 −5

8. Which of the following are linear combinations of u = (0,−2, 2) and v = (1, 3,−10)?

(a) (2, 2, 2) (b) (3, 1, 5) (c) (0, 4, 5) (d) (0, 0, 0)

9. Express the following as linear combinations of u = (2, 1, 4), v = (1,−1, 3), andw = (3, 2, 5).

(a) (−9,−7,−15) (b) (6, 11, 6) (c) (0, 0, 0) (d) (7, 8, 9)

10. Express the following as linear combinations of p1 = 2+x+4x2, p2 = 1−x+3x2,and p3 = 3 + 2x+ 5x2.

(a) −9− 7x− 15x2

(b) 6 + 11x+ 6x2

(c) 0

(d) 7 + 8x+ 9x2

11. In each part, determine whether the given vectors span R3.

(a) v1 = (2, 2, 2), v2 = (0, 0, 3), v3 = (0, 1, 1)

(b) v1 = (2,−1, 3), v2 = (4, 1, 2), v3 = (8,−1, 8)

(c) v1 = (3, 1, 4), v2 = (2,−3, 5), v3 = (5,−2, 9), v4 = (1, 4,−1)


(d) v1 = (1, 2, 6), v2 = (3, 4, 1), v3 = (4, 3, 1), v4 = (3, 3, 1)

12. Which of the following are spanning sets for P2? Justify your answer.

(a) {1, x2, x2 − 2} (b) {2, x2, x, 2x+ 3}(c) {x+ 2, x+ 1, x2 − 1} (d) {x+ 2, x2 − 1}

13. Let f = cos2 x and g = sin2 x. Which of the following lie in the space spanned byf and g?

(a) cos 2x (b) 3 + x2 (c) 1 (d) sin x (e) 0

14. In M2×2 let

E11 =

[1 00 0

]

, E12 =

[0 10 0

]

, E21 =

[0 01 0

]

, E22 =

[0 00 1

]

Show that E11, E12, E21, E22 span M2×2.


1. (a), (c) 2. (a), (c), (f) 3. (a), (b), (d) 4. (a), (c), (e), (f) 5.(a), (b), (d)

6. (a), (b), (d)

7. (a) {(0, 0)} (b) Span {(−2, 1, 0, 0), (3, 0, 1, 0)}(c) Span {(1, 1, 1)} (d) Span {(−5, 0,−3, 1), (−1, 1, 0, 0)}

8. (a), (b), (d)

9. (a) −2(2, 1, 4) + (1,−1, 3)− 2(3, 2, 5) (b) 4(2, 1, 4)− 5(1,−1, 3) + (3, 2, 5)

(c) 0(2, 1, 4) + 0(1,−1, 3) + 0(3, 2, 5) (d) 0(2, 1, 4)− 2(1,−1, 3) + 3(3, 2, 5)

10. (a) −2p1 + p2 − 2p3 (b) 4p1 − 5p2 + p3 (c) 0p1 + 0p2 + 0p3

(d) 0p1 − 2p2 + 3p3

11. (a) The vectors span. (b) The vectors do not span.

(c) The vectors do not span. (d) The vectors span.

12. (b), (c) 13. (a), (c), (e)

3.3 Linear Independence

In this section we look more closely at the structure of vector spaces. To begin with, werestrict ourselves to vector spaces that can be generated from a finite set of elements.Each vector in the vector space can be built up from the elements in this generatingset using only the operations of addition and scalar multiplication. This generating setis usually referred to as a spanning set. In particular, it is desirable to find a minimal


spanning set. By minimal we mean a spanning set with no unnecessary elements (i.e.all the elements in the set are needed in order to span the vector space). To see how tofind a minimal spanning set, it is necessary to consider how the vectors in the collectiondepend on each other. Consequently, we introduce the concepts of linear dependenceand linear independence. These simple concepts provide the key to understanding thestructure of vector spaces.

Consider the following vectors in R3:

x1 =

1−12

, x2 =

−231

, x3 =

−138

Let S be the subspace of R3 spanned by x1, x2, x3. Actually, S can be represented interms of two vectors x1 and x2, since the vector x3 is already in the span of x1 and x2.

x3 = 3x1 + 2x2 (3.1)

Any linear combination of x1, x2, x3 can be reduced to a linear combination of x1 andx2:

α1x1 + α2x2 + α3x3 = α1x1 + α2x2 + α3(3x1 + 2x2)

= (α1 + 3α3)x1 + (α2 + 2α3)x2

ThusS = Span{x1,x2,x3} = Span{x1,x2}

Equation (3.1) can be rewritten in the form

3x1 + 2x2 − 1x3 = 0 (3.2)

Since the three coefficients in (3.2) are nonzero, we could solve for any vector in termsof the other two.

x1 = −2

3x2 +

1

3x3, x2 = −3

2x1 +

1

2x3, x3 = 3x1 + 2x2

It follows that

Span{x1,x2,x3} = Span{x2,x3} = Span{x1,x3} = Span{x1,x2}

Because of the dependency relation (3.2), the subspace S can be represented as the spanof any two of the given vectors.

On the other hand, no such dependency relationship exists between x1 and x2. In-deed, if there were scalars k1 and k2, not both 0, such that

k1x1 + k2x2 = 0 (3.3)

then we could solve for one of the vectors in terms of the other.

x1 = −k2k1

x2 (k1 6= 0) or x2 = −k1k2

x1 (k2 6= 0)


However, neither of the two vectors in question is a multiple of the other. Therefore,Span{x1} and Span{x2} are both proper subspaces of Span{x1,x2}, and the only waythat (3.3) can hold is if k1 = k2 = 0.

We can generalize this example by making the following observations.

(I) If {v1,v2, . . . ,vn} span a vector space V and one of these vectors can be writtenas a linear combination of the other n − 1 vectors, then those n − 1 vectors spanV .

(II) Given n vectors v1,v2, . . . ,vn, it is possible to write one of the vectors as a linearcombination of the other n−1 vectors if and only if there exist scalars k1, k2, . . . , knnot all zero such that

k1v1 + k2v2 + · · ·+ knvn = 0

Definition 3.5 If S = {v1,v2, . . . ,vr} is a nonempty set of vectors, then the vectorequation

k1v1 + k2v2 + · · ·+ krvr = 0

has at least one solution, namely

k1 = 0, k2 = 0, . . . , kr = 0

If this is the only solution, then S is called a linearly independent set. If there areother solutions, then S is called a linearly dependent set.

It follows from (I) and (II) that, if {v1,v2, . . . ,vr} is a minimal spanning set, thenv1,v2, . . . ,vr are linearly independent. Conversely, if v1,v2, . . . ,vr are linearly indepen-dent and span V , then {v1,v2, . . . ,vr} is a minimal spanning set for V .

Example 3.21 If v1 = (2,−1, 0, 3), v2 = (1, 2, 5,−1), and v3 = (7,−1, 5, 8), then theset of vectors S = {v1,v2,v3} is linearly dependent, since 3v1 + v2 − v3 = 0. ♣Example 3.22 The polynomial

p1 = 1− x, p2 = 5 + 3x− 2x2, p3 = 1 + 3x− x2

form a linear dependent set in P2 since 3p1 − p2 + 2p3 = 0. ♣Example 3.23 Consider the vectors i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) in R

3.In terms of components, the vector equation

k1i + k2j+ k3k = 0

becomesk1(1, 0, 0) + k2(0, 1, 0) + k3(0, 0, 1) = (0, 0, 0)

or, equivalently,(k1, k2, k2) = (0, 0, 0)

This implies that k1 = 0, k2 = 0, and k3 = 0, so the set S = {i, j,k} is linear independent.A similar argument can be used to show that the vectors

e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), . . . , en = (0, 0, 0, . . . , 1)

form a linear independent set in Rn. ♣


Example 3.24 Show that1, x, x2, . . . , xn

form a linearly independent set of vectors in Pn.

Solution . . . . . . . . .

Example 3.25 Determine whether the vectors

v1 = (1,−2, 3), v2 = (5, 6,−1), v3 = (3, 2, 1)

form a linear dependent set or a linear independent set.

Solution . . . . . . . . .

Theorem 3.6 Let v1,v2, . . . ,vn be n vectors in Rn and let X =

[v1 v2 · · · vn

]. The

vectors v1, v2, . . ., vn will be linearly dependent if and only if X is singular.

We can use Theorem 3.6 to test whether n vectors are linearly independent in Rn.

Simply form a matrix X whose columns are the vectors being tested. To determinewhether X is singular, calculate the value of det(X). If det(X) = 0, the vectors arelinearly dependent. If det(X) 6= 0, the vectors are linearly independent.

Example 3.26 Determine wether the vectors

v1 = (4, 2, 3), v2 = (2, 3, 1), v3 = (2,−5, 3)

are linearly dependent.

Solution . . . . . . . . .

To determine wether r vectors v1, v2, . . ., vr in Rn are linearly independent, we can

rewrite the equationk1v1 + k2v2 + · · ·+ krvr = 0

as a linear system Xk = 0, where X =[v1 v2 · · · vr

]. If r 6= n, then the matrix X

is not square, so we cannot use determinants to decide wether the vectors are linearlyindependent. The system is homogeneous, so it has the trivial solution k = 0. It willhave nontrivial solutions if and only if the row echelon forms of X involve free variables.If there are nontrivial solutions, then the vectors are linearly dependent. If there are nofree variables, then k = 0 is the only solution, and hence the vectors must be linearlyindependent.

Example 3.27 Given

v1 = (1,−1, 2, 3), v2 = (−2, 3, 1,−2), v3 = (1, 0, 7, 7)

To determine if the vectors are linearly independent, we reduce the system Xk = 0 torow echelon form

1 −2 1 0−1 3 0 02 1 7 03 −2 7 0

−−→

1 −2 1 00 1 1 00 0 0 00 0 0 0

Since the echelon form involves a free variables k3, there are nontrivial solutions andhence the vectors must be linearly dependent. ♣


To determine whether a set of vectors is linear independent in Rn, we must solve a

homogeneous linear system of equations. A similar situation holds for the vector spacePn.

Example 3.28 Determine whether the vectors

p1 = x2 − 2x+ 3, p2 = 2x2 + x+ 8, p3 = x2 + 8x+ 7

form a linear dependent set or a linear independent set.

Solution . . . . . . . . .

Next we consider a very important property of linearly independent vectors. Linearcombinations of linearly independent vectors are unique. More precisely, we have thefollowing theorem.

Theorem 3.7 Let v1, v2, . . ., vn be vectors in a vector space V . A vector v inSpan{v1,v2, . . . ,vn} can be written uniquely as a linear combination of v1, v2, . . ., vn

if and only if v1, v2, . . ., vn are linearly independent.

The following theorem gives two simple facts about linear independence that areimportant to know.

Theorem 3.8

(a) A finite set of vectors that contain the zero vector is linearly dependent.

(b) A set with exactly two vectors is linear independent if and only if neither vector isa scalar multiple of the other.

Geometric Interpretation of Linear Independence

Linear independence has some useful geometric interpretations in R2 and R

3.If x and y are linearly dependent in R

2, then

k1x+ k2y = 0

where k1 and k2 are not both 0. If, say, k1 6= 0, we can write

x = −k2k1

y

If two vectors in R2 are linearly dependent, one of the vectors can be written as a scalar

multiple of the other. Thus, if both vectors are placed at the origin, they will lie alongthe same line.


(x1, x2)

(y1, y2)

x and y linearly dependent

(x1, x2)(y1, y2)

xy

x and y linearly independent

If

x =

x1

x2

x3

and y =

y1y2y3

are linearly independent in R3, then the two points (x1, x2, x3) and (y1, y2, y3) will not lie

on the same line through the origin in 3-space. Since (0, 0, 0), (x1, x2, x3), and (y1, y2, y3)are not collinear, they determine the plane. If (z1, z2, z3) lies on this plane, the vectorz = (z1, z2, z3) can be written as a linear combination of x and y, and hence x, y, and zare linearly dependent. If (z1, z2, z3) does not lie on the plane, the three vectors will belinearly independent.

x

y z

x, y, and z linearly dependent

x

y

z

x, y, and z linearly independent

The next theorem shows that a linearly independent set in Rn can contain at most

n vectors.

Theorem 3.9 Let S = {v1,v2, . . . ,vn} be a set of vectors in Rn. If r > n, then S is

linearly dependent.

Remark Theorem 3.9 tells us that a set in R2 with more than two vectors is linearly

dependent and a set in R3 with more than three vectors is linearly dependent.

Linear Independence of Functions

In Example 3.26 a determinant was used to test wether three vectors were linearly inde-pendent in R

3. Determinants can also be used to help to decide if a set of n vectors islinearly independent in C(n−1)[a, b]. Indeed, let f1, f2, . . ., fn be elements of C(n−1)[a, b].


If these vectors are linearly dependent, then there exist scalars c1, c2, . . ., cn not all zerosuch that

c1f1(x) + c2f2(x) + · · ·+ cnfn(x) = 0 (3.4)

for each x ∈ [a, b]. Taking the derivative with respect to x of both sides of yields

c1f1(x) + c2f2(x) + · · ·+ cnfn(x) = 0

If we continue taking derivatives of both sides, we end up with the system

c1f1(x) + c2f2(x) + · · ·+ cnfn(x) = 0

c1f′1(x) + c2f

′2(x) + · · ·+ cnf

′n(x) = 0

......

......

c1f(n−1)1 (x) + c2f

(n−1)2 (x) + · · ·+ cnf

(n−1)n (x) = 0

For each fixed x ∈ [a, b], the matrix equation

f1(x) f2(x) · · · fn(x)f ′1(x) f ′

2(x) · · · f ′n(x)

......

...

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

c1c2...cn

=

00...0

(3.5)

will have the same nontrivial solution (c1, c2, . . . , cn)T . Thus, if f1, f2, . . ., fn are linearly

dependent in C(n−1)[a, b], then, for fixed x ∈ [a, b], the coefficient matrix of system (3.5)is singular, its determinant is zero.

Definition 3.6 Let f1, f2, . . ., fn be functions in C(n−1)[a, b] and define the functionW (x) on [a, b] by

W (x) =

∣∣∣∣∣∣∣∣∣

f1(x) f2(x) · · · fn(x)f ′1(x) f ′

2(x) · · · f ′n(x)

......

...

f(n−1)1 (x) f

(n−1)2 (x) · · · f

(n−1)n (x)

∣∣∣∣∣∣∣∣∣

The function W (x) is called the Wronskian of f1, f2, . . ., fn.

Theorem 3.10 Let f1, f2, . . ., fn be elements of C(n−1)[a, b]. If there exists a point x0

in [a, b] such that W (x0) 6= 0, then f1, f2, . . ., fn are linearly independent.

Example 3.29 Show that 1, ex, and e2x are linearly independent in C(−∞,∞).

Solution . . . . . . . . .

Example 3.30 Show that the vectors 1, x, x2, and x3 are linearly independent in P3.

Solution . . . . . . . . .

Example 3.31 Determine whether the functions x2 and x|x| are linearly independentin C[−1, 1].

Solution . . . . . . . . .


Exercise 3.3

1. Explain why the following are linearly dependent sets of vectors.

(a) v1 = (1,−2, 4) and v2 = (5,−10,−20) in R3

(b) v1 = (3,−1), v2 = (4, 5), v3 = (−4, 7) in R2

(c) p1 = 3− 2x+ x2 and p2 = 6− 4x+ 2x2 in P2

(d) A =

[−3 42 0

]

and B =

[3 −4−2 0

]

in M2×2

2. Determine whether the following vectors are linearly independent in R2.

(a) (2, 1), (3, 2) (b) (2, 3), (4, 6)

(c) (−2, 1), (1, 3), (2, 4) (d) (−1, 2), (1,−2), (2,−4)

(e) (1, 2), (−1, 1)


(a) (1, 0, 0), (0, 1, 1), (1, 0, 1) (b) (1, 0, 0), (0, 1, 1), (1, 0, 1), (1, 2, 3)

(c) (2, 1,−2), (3, 2,−2), (2, 2, 0) (d) (2, 1,−2), (−2,−1, 2), (4, 2,−4)

(e) (1, 1, 3), (0, 2, 1)


(a) (3, 8, 7,−3), (1, 5, 3,−1), (2,−1, 2, 6), (1, 4, 0, 3)

(b) (0, 0, 2, 2), (3, 3, 0, 0), (1, 1, 0,−1)

(c) (0, 3,−3,−6), (−2, 0, 0,−6), (0,−4,−2,−2), (0,−8, 4,−4)

(d) (3, 0,−3, 6), (0, 2, 3, 1), (0,−2,−2, 0), (−2, 1, 2, 1)

5. Determine whether the following vectors are linearly independent in M2×2.

(a)

[1 01 1

]

,

[0 10 0

]

(b)

[1 00 1

]

,

[0 10 0

]

,

[0 01 0

]

(c)

[1 00 1

]

,

[0 10 0

]

,

[2 30 2

]

6. Determine whether the following vectors are linearly independent in P3.

(a) 1, x2, x2 − 2 (b) 2, x2, x, 2x+ 3

(c) x+ 2, x+ 1, x2 − 1 (d) x+ 2, x2 − 1

7. (a) Show that the vectors v1 = (0, , 3, 1,−1),v2 = (6, 0, 5, 1), and v3 = (4,−7, 1, 3)form a linearly dependent set in R

4

(b) Express each vector as a linear combination of the other two.


8. For which real values of λ do the following vectors form a linear dependent set inR

3?v1 = (λ,−1

2,−1

2), v2 = (−1

2, λ,−1

2), v3 = (−1

2,−1

2, λ)

9. For what values of α will the two vectors cos(x+α) and sin x be linearly dependentin C[−π, π]?

10. Under what conditions is a set with one vector linearly independent?

11. Indicate whether each statement is always true or sometimes false. Justify youranswer by giving a logical argument or a counterexample.

(a) If {v1,v2} is a linearly dependent set, then each vector is a scalar multiple ofthe other.

(b) If {v1,v2,v3} is a linearly independent set, then so is the set{kv1, kv2, kv3} for every nonzero scalar k.

(c) The converse of Theorem 3.8 (a) is also true.

(d) If v1,v2, . . . ,vn span Rn, then they are linearly independent.

(e) If v1,v2, . . . ,vn span a vector space V , then there are linearly independent.

(f) If v1,v2, . . . ,vr are vectors in a vector space V and

Span{v1,v2, . . . ,vr} = Span{v1,v2, . . . ,vr−1}

then v1,v2, . . . ,vr are linearly dependent.


1. (a) v2 is a scalar multiple of v1.

(b) The vectors are linear dependent by Theorem 3.9.

(c) p2 is a scalar multiple of p1.

(d) B is a scalar multiple of A.

2. (a), (e) 3. (a), (e) 4. (a), (b), (c), (d) 5. (a), (b) 6. (c), (d)

7. (b) v1 =27v2 − 3

7v3, v2 =

72v1 +

32v3, v3 = −7

3v1 +

23v2

8. λ = −12, λ = 1 9. When α is an odd multiple of π/2.

10. If and only if the vector is not zero.

11. (a) False (b) True (c) False (d) True (e) False (f) True

chapter 1 systems of linear equations and...

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