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 variables 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 = 1

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

are linear. The equations√x+ 3y = 2, 3x− 2y − 5z + yz = 4, and y = cos x

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.

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.

1

MA131 (Section 750002): Prepared by Asst.Prof.Dr.Archara Pacheenburawana 2

Find the solution set of (a) 5x− 2y = 3 and (b) x1 − 3x2 + 4x3 = 2.

Example 1.1.

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


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:

• 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.

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 −√2 x3 = 1 (b) x1 + 3x2 + x1x3 = 2

(c) x1 = −7x2 + 3x3 (d) x−2

1 + x2 + 8x3 = 5

(e) x3/51 − 2x2 + x3 = 4 (f) πx1 −

√2 x2 +

1

3x3 = 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 = 1

2t− 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 = 3

7+ 5

7t, y = t (b) x1 =

5

3s− 4

3t− 1

8, x2 = s, x3 = t

(c) x1 =1

4r − 5

8s+ 3

4t− 1

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

(d) v = 8

3q − 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 = 1

2t− 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.

Solve the following system by using elementary row operations.

x+ y + 2z = 9

2x+ 4y − 3z = 1

3x+ 6y − 5z = 0

Example 1.2.

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 − 7

2z = −17

2

3y − 11z = −27

1 1 2 9

0 1 −7

2−17

2

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 − 7

2z = −17

2

−1

2z = −3

2

1 1 2 9

0 1 −7

2−17

2

0 0 −1

2−3

2

Multiply the third equationby −2 to obtain

Multiply the third rowby −2 to obtain

x+ y + 2z = 9

y − 7

2z = −17

2

z = 3

1 1 2 9

0 1 −7

2−17

2

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 + 11

2z = 35

2

y − 7

2z = −17

2

z = 3

1 0 11

2

35

2

0 1 −7

2−17

2

0 0 1 3

Add −11

2times the third equation

to the first and 7

2times the third

equation to the second to obtain

Add −11

2times the third row

to the first and 7

2times 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

Suppose that the augmented matrix for a linear equations has been reduced byrow 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

Example 1.3.

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.

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

Example 1.4.

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.

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

Example 1.5.

Solution


Solve

x+ y + 2z = 9

2x+ 4y − 3z = 1

3x+ 6y − 5z = 0

by Gaussian elimination and back-substitution.

Example 1.6.

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.


Solve the following homogeneous system of linear equations by using Gauss-Jordan elimination.

2x1 + 2x2 − x3 + x5 = 0

−x1 − x2 + 2x3 − 3x4 + x5 = 0

x1 + x2 − 2x3 − x5 = 0

x3 + x4 + x5 = 0

Example 1.7.

Solution


A homogeneous system of linear equations with more unknowns than equationshas infinitely many solutions.

Theorem 1.1.

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) −1

2x1 +

1

3x2 +

1

2x3 = 0

1

4x1 − 2

3x2 +

1

4x3 = 0

1

4x1 +

1

3x2 − 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 = −1

7− 3

7t, x2 =

1

7− 4

7t, x3 = t

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

(f) x1 =3

4− 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 = −4

3t, x2 = 0, x3 =

1

3t, x4 = t

(j) x = 14

5− 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 =5

3t, 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 = −1

2s, 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

A matrix is a rectangular array of numbers, each of which is called an entryof the matrix.

Definition 1.1.

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 A

will 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)

A square matrix in which all the entries above the main diagonal are zero iscalled lower triangular, and a square matrix in which all the entries belowthe main diagonal are zero is called upper triangular. A matrix that is eitherupper triangular or lower triangular is called triangular.

Definition 1.2.

Operations on Matrices

Given matrices A and B, the entries aij and bkp are corresponding entriesif 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 respectivematrices.

Definition 1.3.

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.

Two matrices A and B are equal, written A = B, if they have the same sizeand their corresponding entries are equal.In matrix notation, if A = [aij ] and B = [bij ] have the same size, then A = B ifand only if aij = bij for all i and j.

Definition 1.4 (Equality of matrices).

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

]

?

Example 1.8.

Solution

Let A and B be matrices with the same size. The sum of A and B, writtenA+B, is the matrix obtained by adding corresponding entries of A and B

In matrix notation, if A = [aij] and B = [bij ] have the same size, then C = A+B

if and only if cij = aij + bij for all i and j.

Definition 1.5 (Sum of matrices).

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

Example 1.9.

Solution


Let A be a matrix and c be a scalar. Then the product cA is the matrixobtained by multiplying each entry of A by scalar c. The matrix cA is said tobe a scalar multiple of A.In matrix notation, if A = [aij ], then B = cA if and only if bij = caij for all iand j.

Definition 1.6 (Multiplication by a scalar).

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

Definition 1.7.

Let A =

[1 −1 0 32 6 −4 8

]

. Find 3A and (−1)A.

Example 1.10.

Solution

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).

Definition 1.8 (Difference of matrices).

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

Find C = 2A− B where A =

5 4−1 79 −3

and B =

−3 04 25 −7

.

Example 1.11.

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 1

3.

If A is an m× r matrix and B is an r× n matrix, then the product AB is them×n matrix whose entries are determined as follows. To find the entry in row i

and column j of AB, single out row i from matrix A and column j from matrixB. Multiply the corresponding entries from the row and column together, andthen 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.

Definition 1.9 (Matrix multiplication).

Given the pairs of matrices A and B, find the products AB and BA, if defined.

(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 =

x

y

z

Example 1.12.

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 they

still 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 productAB without computing the entire product. The following results are useful for thatpurpose:

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)

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

Example 1.13.

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 =

a1B

a2B...

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 from

the matrix x. Moreover, the product yA of a 1×m matrix y with an m× n matrix A is

a linear combination of the row matrices of A with scalar coefficients coming from y.


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

]♣

Example 1.14.

It follows from (1.8) and (1.10) that the j th column matrix of a product AB is a

linear combination of the column matrices of A with the coefficients coming from the j th

column of B.


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 thecolumn matrices 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

♣

Example 1.15.

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

The transpose of the m×n matrix A, written AT , is the n×m matrix B whosej th column is the j th row of A. Equivalently, B = AT if and only if bij = ajifor each i and j.

Definition 1.10 (Transpose of a matrix).

Find the transpose of

(a) A =

[2 7−8 0

]

(b) B =

[1 2 −74 5 6

]

(c) C =[−1 3 4

]

Example 1.16.

Solution

If A is a square matrix, then trace of A, denoted by tr(A), is defined to be thesum of the entries on the main diagonal of A. The trace of A is undefined if Ais not a square matrix.

Definition 1.11.


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

Example 1.17.

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+ c

3d+ 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) 1

2CT − 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 equa-tion Ax = 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

w

x

y

z

=

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 the


nonzero 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

x

y

z

=

x+ y

x− y

0

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)

−1

4

3

29

40

3

4

9

4

(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 = 0


3w + 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 law

for multiplication. For matrices, however, AB and BA need not be equal.

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 ♣

Example 1.18.


Let A, B, and C be matrices and a and b be scalars. Assume that the sizes ofthe matrices are such that each operation is 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)

Theorem 1.2 (Properties of Matrices Arithmetic).

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.

Example 1.19.

Solution


Zero Matrices

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

Definition 1.12.

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 theequation AB = AC and write B = C. Also, AD = 0, yet A 6= 0 and D 6=0. Thus, the cancellation law is not valid for matrix multiplication, and it ispossible for a product of matrices to be zero without either factor being zero.♣

Example 1.20.


Assuming that the sizes of the matrices are such that the indicated operationscan be performed, the following rules of matrix arithmetic are valid.

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

(b) A−A = 0

(c) 0− A = −A

(d) A0 = 0; 0A = 0

Theorem 1.3 (Properties of Zero Matrices).

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 ♣

Example 1.21.


If R is the reduced row-echelon form of an n× n matrix A, then either R has arow of zeros or R is the identity matrix In.

Theorem 1.4.

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 of A.If A has no inverse, it is said to be noninvertible (or singular).

Definition 1.13.

Show that the matrix B =

[3 51 2

]

is an inverse of A =

[2 −5−1 3

]

.

Example 1.22.

Solution

Properties of Inverses

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

Theorem 1.5.

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


The matrix

A =

[a b

c 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 1.6.

If A and B are invertible matrices of the same dimension, then AB is invertibleand

(AB)−1 = B−1A−1

Theorem 1.7.

If A1, A2, . . . , An are invertible matrices of the same dimension, then A1A2 · · ·An

is invertible and

(A1A2 · · ·An)−1 = A−1

n A−1

n−1 · · ·A−1

2 A−1

1

Theorem 1.8.

Let A =

[7 −35 −2

]

and B =

[4 13 1

]

. Show that (AB)−1 = B−1A−1.

Example 1.23.

Solution


Powers of a Matrix

If A is a square matrix, then we define the nonnegative integer powers of A tobe

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

Definition 1.14.

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.

Let A =

[3 −2−1 1

]

and A−1 =

[1 21 3

]

. Find the matrices A3 and A−3.

Example 1.24.

Solution

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 .

Theorem 1.9 (Laws of Exponents).


If

(1

3A

)−1

=

[2 −1−5 3

]

, then find A.

Example 1.25.

Solution

Properties of the Transpose

If the sizes of the matrices are such that the state operations can be performed,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

Theorem 1.10.

Invertibility of a Transpose

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

(AT )−1 = (A−1)T .

Theorem 1.11.

Let A =

[−3 15 −2

]

. Show that (AT )−1 = (A−1)T .

Example 1.26.

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

[1

2(ex + e−x) 1

2(ex − e−x)

1

2(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 C

such 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 allinteger values of n.

(o) Sum of two invertible matrices is invertible.

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

(q)

[a b

b a

]

+

[x y

y x

]

is symmetric.


2. (a)

[0 10 0

]

(b)

[1 00 0

]

3. (a) A−1 =

[2

5

3

10

1

5− 1

10

]

(b) B−1 =

[−1 31 −2

]

(c) C−1 =

[6 5−5 −4

]

(d) D−1 =

[

−13

3−7

3

−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 n

0 1

]

8. (a) A =

[5

13

1

13

− 3

13

2

13

]

(b) A =

[2

71

1

7

3

7

]

(c) A =

[

−2

51

−1

5

3

5

]

(d) A =

[

− 9

13

1

13

2

13− 6

13

]


9.

[cos θ − sin θsin θ cos θ

]

10

[1

2(ex + e−x) −1

2(ex − e−x)

−1

2(ex − e−x) 1

2(ex + e−x)

]

11. A =

1

a110 · · · 0

0 1

a22· · · 0

......

...0 0 · · · 1

ann

12. C = −A−1BA−1

13. (a)

1

2−1

20 0

1

2

1

20 0

0 0 1

2−1

2

−1 0 1

2

1

2

(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.

An n × n matrix is called an elementary matrix if it can be obtained fromthe n× n identity matrix In by performing a single elementary row operation.

Definition 1.15.


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.

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

Theorem 1.12.

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 c

d e f

g h i

Find the product E1A, E2A, and E3A.

Example 1.27.

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 by 1

k.

The various possibilities are listed in the following table. The operations on the rightside of this table are called the inverse operations of the corresponding operations onthe left.

Row Operation on I Row Operation on E

That 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 j

Add k times row i to row j Add −k times row i to row j

In each of the following, an elementary row operation is applied to the 2 × 2identity matrix to obtain an elementary matrix E, then E is restored to theidentity matrix 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 1

3

[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

Example 1.28.

The next theorem gives an important property of elementary matrices.

Every elementary matrix is invertible, and the inverse is also an elementarymatrix.

Theorem 1.13.


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.

Theorem 1.14.

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

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−1

k−1, . . ., E−1

2 , E−1

1 we obtain

A = E−1

1 E−1

2 · · ·E−1

k−1E−1

k In = E−1

1 E−1

2 · · ·E−1

k−1E−1

k

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.

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.

Example 1.29.

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 row

operations 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]

Find the inverse of A =

1 −2 12 −5 23 2 −1

.

Example 1.30.

Solution


Often it will not be known in advance whether a given matrix is invertible. If an n×n

matrix 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.

Let A be the matrix

1 3 4−2 −5 −31 4 9

. Determine whether A is invertible.

Example 1.31.

Solution


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. ♣

Example 1.32.

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 −1

40

0 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)

1

5

1

5−2

51

5

1

5

1

101

5−4

5

1

10

(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 1

3

4 0 2

5−9

0 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 a

3 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 +1

4r3

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)

[

− 5

39

2

13

4

39

1

13

]

(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

3

1

30 0

0 −1

5

1

50

0 0 −1

7

1

7

(k) Not invertible (l)

−4

5

3

5

1

5

1

53

20 −1 0

1

20 0 0

4

5

2

5−1

5−1

5

7. (a) a 6= 8

3(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−1

1 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 =

1

20 0

0 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

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

Theorem 1.15.

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 solving


certain linear systems.

If A is an invertible n× n matrix, then for each n× 1 matrix b, the system ofequations Ax = b has exactly one solution, namely, x = A−1b.

Theorem 1.16.

Solve the system

x1 − 2x2 + x3 = 7

2x1 − 5x2 + 2x3 = 6

3x1 + 2x2 − x3 = 1

Example 1.33.

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.

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

Example 1.34.

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.

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 1.17.

If A is an n× n matrix, then the following are equivalent.

(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.

(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 1.18 (Equivalent Statements).

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

and B must also be invertible.

Theorem 1.19.

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.


What conditions must b1, b2, and b3 satisfy in order for the system of equations

x1 + x2 + 2x3 = b1

x1 + x3 = b2

2x1 + x2 + 3x3 = b3

to be consistent?

Example 1.35.

Solution


What conditions must b1, b2, and b3 satisfy in order for the system of equations

x1 − 2x2 + x3 = b1

2x1 − 5x2 + 2x3 = b2

3x1 + 2x2 − x3 = b3

to be consistent?

Example 1.36.

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 =16

3, x2 = −4

3, x3 = −11

3(b) x1 = −5

3, x2 =

5

3, x3 =

10

3

(c) x1 = 3, x2 = 0, x3 = −4

3. (a) i. x1 =22

17, x2 =

1

17ii. x1 =

21

17, x2 =

11

17

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

(c) i. x1 =7

15, x2 =

4

15ii. x1 =

34

15, x2 =

28

15iii. x1 =

19

15, x2 =

13

15

iv. x1 = −1

5, x2 =

3

5

(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.


Given the factorization

2 1 34 1 7−6 −2 −12

=

2 0 04 −1 0−6 1 −2

1 1

2

3

2

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

Example 1.37.

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−1

k , . . . , E−1

2 , E−1

1

to obtainA = E−1

1 E−1

2 · · ·E−1

k U (1.17)

We can show that the matrix L defined by

L = E−1

1 E−1

2 · · ·E−1

k (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.

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

Theorem 1.20.

A factorization of a square matrix A as A = LU , where L is lower triangularand U is upper triangular, is called an LU-decomposition or triangulardecomposition of the matrix A.

Definition 1.16.


Find an LU-decomposition of

A =

2 1 34 1 7−6 −2 −12

Example 1.38.

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 =

1

20 0

0 1 00 0 1

E−1

1 =

2 0 00 1 00 0 1

1 1

2

3

2

4 1 7−6 −2 −12

Step 2 E2 =

1 0 0−4 1 00 0 1

E−1

2 =

1 0 04 1 00 0 1

1 1

2

3

2

0 −1 1−6 −2 −12

Step 3 E3 =

1 0 00 1 06 0 1

E−1

3 =

1 0 00 1 0−6 0 1

1 1

2

3

2

0 −1 10 1 −3

Step 4 E4 =

1 0 00 −1 00 0 1

E−1

4 =

1 0 00 −1 00 0 1

1 1

2

3

2

0 1 −10 1 −3


Elementary MatrixReduction to Corresponding to Inverse of the

Row-Echelon Form the Row Operation Elementary Matrix

Step 5 E5 =

1 0 00 1 00 −1 1

E−1

5 =

1 0 00 1 00 1 1

1 1

2

3

2

0 1 −10 0 −2

Step 6 E6 =

1 0 00 1 00 0 −1

2

E−1

6 =

1 0 00 1 00 0 −2

1 1

2

3

2

0 1 −10 0 1

Thus

U =

1 1

2

3

2

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 1

2

3

2

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 somecareful bookkeeping of the operations used to reduce A to U . Because we are assumingthat no row interchanges are required to reduce A to U , there are only two types ofoperations involved:

• 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:

1

2for 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 negative

of 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 .

Find an LU-decomposition of A =

2 4 21 5 24 −1 9

.

Example 1.39.

Solution


Find an LU-decomposition of A =

6 −2 09 −1 13 7 5

.

Example 1.40.

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 1

2−1

2

0 0 10 0 0

chapter 1 systems of linear equations and...

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