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3.5 Inverses of Matrices

Recall that the n×n $n \times n$ identity matrix is the diagonal matrix

I = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 100 ⋮ 0 010 ⋮ 0 001 ⋮ 0 \dots \dots \dots ⋱ \dots 000 ⋮ 1 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

$I = [\begin{array}{c} 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & 0 & \dots & 1 \end{array}]$ (1)

having ones on its main diagonal and zeros elsewhere. It is not difficult to deduce directly from the definition of the matrix product that I acts like an identity for matrix multiplication:

A I = A and I B = B

$AI = A and IB = B$ (2)

if the sizes of A and B are such that the products AI and IB are defined. It is, nevertheless, instructive to derive the identities in (2) formally from the two basic facts about matrix multiplication that we state below. First, recall that the notation

A = [a 1 a 2 a 3 \dots a n]

$A = [\begin{array}{l} a_{1} & a_{2} & a_{3} & \dots & a_{n} \end{array}]$ (3)

expresses the m×n $m \times n$ matrix A in terms of its column vectors a1,a2,a3,…,an $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ .

Fact 1 Ax in terms of columns of A

If A=[a1a2⋯an] $A = [\begin{array}{l} a_{1} & a_{2} & \dots & a_{n} \end{array}]$ and x=(x1,x2,…,xn) $x = (x_{1}, x_{2}, \dots, x_{n})$ is an n-vector, then

$A x = x 1 a 1 + x 2 a 2 + \dots + x n a n .$ $Ax = x_{1} a_{1} + x_{2} a_{2} + \dots + x_{n} a_{n} .$ (4)

The reason is that when each row vector of A is multiplied by the column vector x, its jth element is multiplied by xj $x_{j}$ .
Fact 2 AB in terms of columns of B

If A is an m×n $m \times n$ matrix and B=[b1b2⋯bp] $B = [\begin{array}{l} b_{1} & b_{2} & \dots & b_{p} \end{array}]$ is an n×p $n \times p$ matrix, then

$A B = [A b 1 A b 2 \dots A b p] .$ $AB = [\begin{array}{l} {Ab}_{1} & {Ab}_{2} & \dots & {Ab}_{p} \end{array}] .$ (5)

That is, the j th column of AB is the product of A and the j th column of B. The reason is that the elements of the j th column of AB are obtained by multiplying the individual rows of A by the j th column of B.

Example 1

The third column of the product AB of the matrices

A = [24 - 1 0 03] and B = ⎡ ⎣ ⎢ 3 - 2 5 761 53 - 2 - 4 6 - 1 ⎤ ⎦ ⎥

$A = [\begin{array}{r} 2 & - 1 & 0 \\ 4 & 0 & 3 \end{array}] and B = [\begin{array}{r} 3 & 7 & 5 & - 4 \\ - 2 & 6 & 3 & 6 \\ 5 & 1 & - 2 & - 1 \end{array}]$

A b 3 = [24 - 1 0 03] ⎡ ⎣ ⎢ 53 - 2 ⎤ ⎦ ⎥ = [714] .

${Ab}_{3} = [\begin{array}{r} 2 & - 1 & 0 \\ 4 & 0 & 3 \end{array}] [\begin{array}{r} 5 \\ 3 \\ - 2 \end{array}] = [\begin{array}{r} 7 \\ 14 \end{array}] .$

To prove that AI=A, $A I = A,$ note first that

I = [e 1 e 2 \dots e n],

$I = [\begin{array}{l} e_{1} & e_{2} & \dots & e_{n} \end{array}],$ (6)

where the jth column vector of I is the jth basic unit vector

e j = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 0 ⋮ 1 ⋮ 0 ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ \leftarrow j th entry:

$e_{j} = [\begin{array}{c} 0 \\ ⋮ \\ 1 \\ ⋮ \\ 0 \end{array}] \leftarrow j th entry:$ (7)

If A=[a1a2⋯an], $A = [\begin{array}{l} a_{1} & a_{2} & \dots & a_{n} \end{array}],$ then Fact 1 yields

A e j = 0 \cdot a 1 + \dots + 1 \cdot a j + \dots + 0 \cdot a n = a j .

${Ae}_{j} = 0 \cdot a_{1} + \dots + 1 \cdot a_{j} + \dots + 0 \cdot a_{n} = a_{j} .$ (8)

Hence Fact 2 gives

A I = = A [e 1 e 2 \dots e n] [A e 1 A e 2 \dots A e n] = [a 1 a 2 \dots a n];

$\begin{array}{rcl} AI & = & A [\begin{array}{l} e_{1} & e_{2} & \dots & e_{n} \end{array}] \\ = & [\begin{array}{l} {Ae}_{1} & {Ae}_{2} & \dots & {Ae}_{n} \end{array}] = [\begin{array}{l} a_{1} & a_{2} & \dots & a_{n} \end{array}]; \end{array}$

that is, AI=A. $A I = A .$ The proof that IB=B $I B = B$ is similar. (See Problems 41 and 42.)

The Inverse Matrix A^-1

If a≠0, $a \neq 0,$ then there is a number b=a−1=1/a $b = a^{- 1} = 1 / a$ such that ab=ba=1. $a b = b a = 1 .$ Given a nonzero matrix A, we therefore wonder whether there is a matrix B such that AB=BA=I. $AB = BA = I .$ The following two examples show that the answer to this question depends upon the particular matrix A.

Example 2

A = [4397] and B = [7 - 3 - 9 4],

$A = [\begin{array}{r} 4 & 9 \\ 3 & 7 \end{array}] and B = [\begin{array}{r} 7 & - 9 \\ - 3 & 4 \end{array}],$

then

A B = [4397] [7 - 3 - 9 4] = [1001] = I;

$AB = [\begin{array}{r} 4 & 9 \\ 3 & 7 \end{array}] [\begin{array}{r} 7 & - 9 \\ - 3 & 4 \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] = I;$

BA=I $BA = I$ by a similar computation.

Example 3

Let

A = [1 - 2 - 3 6] and B = [a c b d] .

$A = [\begin{array}{r} 1 & - 3 \\ - 2 & 6 \end{array}] and B = [\begin{array}{r} a & b \\ c & d \end{array}] .$

If the matrix B had the property that AB=BA=I, $AB = BA = I,$ then

A B = = [1 - 2 - 3 6] [a c b d] [a - 3 c - 2 a + 6 c b - 3 d - 2 b + 6 d] = [1001] .

$\begin{array}{rcl} AB & = & [\begin{array}{r} 1 & - 3 \\ - 2 & 6 \end{array}] [\begin{array}{r} a & b \\ c & d \end{array}] \\ = & [\begin{array}{r} a - 3 c & b - 3 d \\ - 2 a + 6 c & - 2 b + 6 d \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] . \end{array}$

But upon equating corresponding elements of AB and the 2×2 $2 \times 2$ identity matrix in the last line, we find that

a - 3 c - 2 a + 6 c = = 10 and b - 3 d - 2 b + 6 d = = 0 1.

$\begin{array}{l} \begin{array}{rcl} a - 3 c & = & 1 \\ - 2 a + 6 c & = & 0 \end{array} & and & \begin{array}{rcl} b - 3 d & = & 0 \\ - 2 b + 6 d & = & 1. \end{array} \end{array}$

It is clear that these equations are inconsistent. Thus there can exist no 2×2 $2 \times 2$ matrix B such that AB=I $AB = I$ .

Thus the matrix A of Example 2 is invertible, whereas the matrix A of Example 3 is not invertible.

A matrix B such that AB=BA=I $AB = BA = I$ is called an inverse matrix of the matrix A. The following theorem says that no matrix can have two different inverse matrices.

The unique inverse of an invertible matrix A is denoted by A−1. $A^{- 1} .$ Thus we say in Example 2 that

If A = [4397] then A - 1 = [7 - 3 - 9 4] .

$If A = [\begin{array}{r} 4 & 9 \\ 3 & 7 \end{array}] then A^{- 1} = [\begin{array}{r} 7 & - 9 \\ - 3 & 4 \end{array}] .$

In the case of a 2×2 $2 \times 2$ matrix A, it is easy to determine whether or not A is invertible and to find A−1 $A^{- 1}$ if it exists. In Problems 36 and 37 we ask you to verify the following result.

Equation (9) gives us the following prescription for writing the inverse of an invertible 2×2 $2 \times 2$ matrix:

First, interchange the two main diagonal entries.
Then, change the signs of the two off-diagonal elements.
Finally, divide each element of the resulting matrix by ad−bc $a d - b c$ .

You might check that this is how B=A−1 $B = A^{- 1}$ is obtained from A in Example 2 (in which ad−bc=1 $a d - b c = 1$ ).

Example 4

A = [4569],

$A = [\begin{array}{l} 4 & 6 \\ 5 & 9 \end{array}],$

then ad−bc=36−30=6≠0, $a d - b c = 36 - 30 = 6 \neq 0,$ so

A - 1 = 1 6 [9 - 5 - 6 4] = [3 2 - 5 6 - 1 2 3] .

$A^{- 1} = \frac{1}{6} [\begin{array}{r} 9 & - 6 \\ - 5 & 4 \end{array}] = [\begin{array}{r} \frac{3}{2} & - 1 \\ - \frac{5}{6} & \frac{2}{3} \end{array}] .$

Arbitrary integral powers of a square matrix A are defined as follows, though in the case of a negative exponent we must assume that A is also invertible. If n is a positive integer, we define

A 0 A n + 1 A - n = = = I and A 1 = A; A n A for n \geq 1; (A - 1) n .

$\begin{array}{rcl} A^{0} & = & I and A^{1} = A; \\ A^{n + 1} & = & A^{n} A for n \geq 1; \\ A^{- n} & = & {(A^{- 1})}^{n} . \end{array}$

In Problem 28 of Section 3.4, we asked you to verify the laws of exponents

A r A s = A r + s, (A r) s = A r s

$A^{r} A^{s} = A^{r + s}, {(A^{r})}^{s} = A^{r s}$ (10)

in the case of positive integral exponents, and Problem 31 of this section deals with the case of negative integral exponents. In Problem 29 we ask you to establish parts (a) and (b) of the following theorem.

In mathematics it is frequently important to note the surprises. The surprise in Eq. (11) is the reversal of the natural order of the factors in the right-hand side. You should now be able to show that

(A B C) - 1 = C - 1 B - 1 A - 1 .

${(ABC)}^{- 1} = C^{- 1} B^{- 1} A^{- 1} .$

In general, any product of invertible matrices of the same size is again invertible, and the inverse of a product of invertible matrices is the product in reverse order of their inverses.

Example 5

To solve the system

4 x 1 + 6 x 2 5 x 1 + 9 x 2 = = 6 18,

$\begin{array}{rcl} 4 x_{1} + 6 x_{2} & = & 6 \\ 5 x_{1} + 9 x_{2} & = & 18, \end{array}$

we use the inverse of the coefficient matrix

A = [4569]

$A = [\begin{array}{l} 4 & 6 \\ 5 & 9 \end{array}]$

that we found in Example 4. Then Eq. (13) yields

x = A - 1 b = = [4569] - 1 [618] [3 2 - 5 6 - 1 2 3] [618] = [- 9 7] .

$\begin{array}{rcl} x = A^{- 1} b & = & {[\begin{array}{r} 4 & 6 \\ 5 & 9 \end{array}]}^{- 1} [\begin{array}{r} 6 \\ 18 \end{array}] \\ = & [\begin{array}{r} \frac{3}{2} & - 1 \\ - \frac{5}{6} & \frac{2}{3} \end{array}] [\begin{array}{r} 6 \\ 18 \end{array}] = [\begin{array}{r} - 9 \\ 7 \end{array}] . \end{array}$

Thus x1=−9, x2=7 $x_{1} = - 9, x_{2} = 7$ is the unique solution.

How to Find A^-1

Theorem 2 tells us only how to invert 2×2 $2 \times 2$ matrices. The development of a method for inverting larger matrices involves a special class of matrices, which we define next.

Example 6

We obtain some typical elementary matrices as follows.

[1001] ⎡ ⎣ ⎢ 100010001 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ 100010001 ⎤ ⎦ ⎥ - \to - - - - - - (3) R 1 - \to - - - - - - (2) R 1 + R 3 - \to - - - - - - SWAP (R 1, R 2) [3001] = E 1 ⎡ ⎣ ⎢ 102010001 ⎤ ⎦ ⎥ = E 2 ⎡ ⎣ ⎢ 010100001 ⎤ ⎦ ⎥ = E 3

$\begin{array}{rcl} [\begin{array}{l} 1 & 0 \\ 0 & 1 \end{array}] & \overset{(3) R_{1}}{\to} & [\begin{array}{l} 3 & 0 \\ 0 & 1 \end{array}] = E_{1} \\ [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] & \overset{(2) R_{1} + R_{3}}{\to} & [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{array}] = E_{2} \\ [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] & \overset{SWAP (R_{1}, R_{2})}{\to} & [\begin{array}{l} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}] = E_{3} \end{array}$

The three elementary matrices E1, E2, $E_{1}, E_{2},$ and E3 $E_{3}$ correspond to three typical elementary row operations.

Now, suppose that the m×m $m \times m$ elementary matrix E corresponds to a certain elementary row operation. It turns out that if we perform this same operation on an arbitrary m×n $m \times n$ matrix A, we get the product matrix EA that results upon multiplying A on the left by the matrix E. Thus we can carry out an elementary row operation by means of left multiplication by the corresponding elementary matrix. Problems 38–40 illustrate typical cases in the proof of the following theorem.

Elementary row operations are reversible. That is, to every elementary row operation there corresponds an inverse elementary row operation that cancels its effects (see Figure 3.5.1). It follows that every elementary matrix is invertible. To see why, let E be a given elementary matrix and let E1 $E_{1}$ be the elementary matrix corresponding to the inverse of the row operation that transforms I into E. Then the inverse operation transforms E to I, so Theorem 5 implies that E1E=I. $E_{1} E = I .$ We see similarly that EE1=I. $E E_{1} = I .$ Hence, the elementary matrix E is invertible with E−1=E1 $E^{- 1} = E_{1}$ .

FIGURE 3.5.1.

Inverse elementary row operations.

Elementary Row Operation	Inverse Operation
(c)Ri $(c) R_{i}$	1cRi $\frac{1}{c} R_{i}$
SWAP(Ri,Rj) $SWAP (R_{i}, R_{j})$	SWAP(Ri,Rj) $SWAP (R_{i}, R_{j})$
(c)Ri+Rj $(c) R_{i} + R_{j}$	(−c)Ri+Rj $(- c) R_{i} + R_{j}$

Elementary matrices are not ordinarily used for computational purposes; it is simpler to carry out row operations directly than to multiply by elementary matrices. Instead, their principal role is in the proof of the following theorem, which leads in turn to a practical method for inverting matrices.

Proof

Assume first that A is invertible. Then, by Theorem 4 (with b=0 $b = 0$ ), it follows that Ax=0 $Ax = 0$ has only the trivial solution x=0. $x = 0 .$ But Theorem 4 in Section 3.3 implies that this is so (if and) only if A is row equivalent to I.

Now assume, conversely, that A is row equivalent to I. That is, there is a finite sequence of elementary row operations that transforms A into I. According to Theorem 5, each of these operations can be performed by multiplying on the left by the corresponding elementary matrix. If E1,E2,…,Ek $E_{1}, E_{2}, \dots, E_{k}$ are the elementary matrices corresponding to these row operations, it follows that

E k E k - 1 \dots E 2 E 1 A = I .

$E_{k} E_{k - 1} \dots E_{2} E_{1} A = I .$ (14)

If we now multiply each side in Eq. (14) by the inverse matrices (Ek)−1, (Ek−1)−1, …, (E2)−1, (E1)−1 ${(E_{k})}^{- 1}, {(E_{k - 1})}^{- 1}, \dots, {(E_{2})}^{- 1}, {(E_{1})}^{- 1}$ in turn, we find that

A = (E 1) - 1 (E 2) - 1 \dots (E k - 1) - 1 (E k) - 1 .

$A = {(E_{1})}^{- 1} {(E_{2})}^{- 1} \dots {(E_{k - 1})}^{- 1} {(E_{k})}^{- 1} .$ (15)

Thus A is a product of invertible elementary matrices, and it follows from part (c) of Theorem 3 that A is invertible.

The proof of Theorem 6 actually tells us how to find the inverse matrix of A. If we invert each side in Eq. (15) (remembering to reverse the order on the right), we get

A - 1 = E k E k - 1 \dots E 2 E 1 I .

$A^{- 1} = E_{k} E_{k - 1} \dots E_{2} E_{1} I .$ (16)

Because each left multiplication by an elementary matrix is equivalent to performing the corresponding row operation, we see by comparison of Eqs. (14) and (16) that the same sequence of elementary row operations that transforms A into I also transforms I into A−1 $A^{- 1}$ .

As a practical matter, it generally is more convenient to carry out the two reductions—from A to I and from I to A−1 $A^{- 1}$ —in parallel, as illustrated in our next example.

Example 7

Find the inverse of the 3×3 $3 \times 3$ matrix

A = ⎡ ⎣ ⎢ 453365232 ⎤ ⎦ ⎥ .

$A = [\begin{array}{l} 4 & 3 & 2 \\ 5 & 6 & 3 \\ 3 & 5 & 2 \end{array}] .$

Solution

We want to reduce A to the 3×3 $3 \times 3$ identity matrix I while simultaneously performing the same sequence of row operations on I to obtain A−1. $A^{- 1} .$ In order to carry out this process efficiently, we adjoin I on the right of A to form the 3×6 $3 \times 6$ matrix

⎡ ⎣ ⎢ ⎢ 453365232100010001 ⎤ ⎦ ⎥ ⎥ .

$[\begin{array}{l} 4 & 3 & 2 & 1 & 0 & 0 \\ 5 & 6 & 3 & 0 & 1 & 0 \\ 3 & 5 & 2 & 0 & 0 & 1 \end{array}] .$

We now apply the following sequence of elementary row operations to this 3×6 $3 \times 6$ matrix (designed to transform its left half into the 3×3 $3 \times 3$ identity matrix).

- \to - - - - - - (- 1) R 3 + R 1 - \to - - - - - - (- 1) R 3 + R 2 - \to - - - - - - (- 2) R 1 + R 2 - \to - - - - - - (- 3) R 1 + R 3 - \to - - - - - - (- 2) R 2 + R 3 - \to - - - - - - SWAP (R 2, R 3) - \to - - - - - - (2) R 2 + R 1 - \to - - - - - - (- 5) R 2 + R 3 ⎡ ⎣ ⎢ ⎢ 153 - 2 65 032100010 - 1 01 ⎤ ⎦ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ 123 - 2 15 012100010 - 1 - 1 1 ⎤ ⎦ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ 103 - 2 55 012 1 - 2 0 010 - 1 11 ⎤ ⎦ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ 100 - 2 511 012 1 - 2 - 3 010 - 1 14 ⎤ ⎦ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ 100 - 2 51 010 1 - 2 1 01 - 2 - 1 12 ⎤ ⎦ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ 100 - 2 15 001 11 - 2 0 - 2 1 - 1 21 ⎤ ⎦ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ 100015001 31 - 2 - 4 - 2 1 321 ⎤ ⎦ ⎥ ⎥ ⎡ ⎣ ⎢ ⎢ 100010001 31 - 7 - 4 - 2 11 32 - 9 ⎤ ⎦ ⎥ ⎥

$\begin{array}{l} \overset{(- 1) R_{3} + R_{1}}{\to} & [\begin{array}{r} 1 & - 2 & 0 & 1 & 0 & - 1 \\ 5 & 6 & 3 & 0 & 1 & 0 \\ 3 & 5 & 2 & 0 & 0 & 1 \end{array}] \\ \overset{(- 1) R_{3} + R_{2}}{\to} & [\begin{array}{r} 1 & - 2 & 0 & 1 & 0 & - 1 \\ 2 & 1 & 1 & 0 & 1 & - 1 \\ 3 & 5 & 2 & 0 & 0 & 1 \end{array}] \\ \overset{(- 2) R_{1} + R_{2}}{\to} & [\begin{array}{r} 1 & - 2 & 0 & 1 & 0 & - 1 \\ 0 & 5 & 1 & - 2 & 1 & 1 \\ 3 & 5 & 2 & 0 & 0 & 1 \end{array}] \\ \overset{(- 3) R_{1} + R_{3}}{\to} & [\begin{array}{r} 1 & - 2 & 0 & 1 & 0 & - 1 \\ 0 & 5 & 1 & - 2 & 1 & 1 \\ 0 & 11 & 2 & - 3 & 0 & 4 \end{array}] \\ \overset{(- 2) R_{2} + R_{3}}{\to} & [\begin{array}{r} 1 & - 2 & 0 & 1 & 0 & - 1 \\ 0 & 5 & 1 & - 2 & 1 & 1 \\ 0 & 1 & 0 & 1 & - 2 & 2 \end{array}] \\ \overset{SWAP (R_{2}, R_{3})}{\to} & [\begin{array}{r} 1 & - 2 & 0 & 1 & 0 & - 1 \\ 0 & 1 & 0 & 1 & - 2 & 2 \\ 0 & 5 & 1 & - 2 & 1 & 1 \end{array}] \\ \overset{(2) R_{2} + R_{1}}{\to} & [\begin{array}{r} 1 & 0 & 0 & 3 & - 4 & 3 \\ 0 & 1 & 0 & 1 & - 2 & 2 \\ 0 & 5 & 1 & - 2 & 1 & 1 \end{array}] \\ \overset{(- 5) R_{2} + R_{3}}{\to} & [\begin{array}{r} 1 & 0 & 0 & 3 & - 4 & 3 \\ 0 & 1 & 0 & 1 & - 2 & 2 \\ 0 & 0 & 1 & - 7 & 11 & - 9 \end{array}] \end{array}$

Now that we have reduced the left half of the 3×6 $3 \times 6$ matrix to I, we simply examine its right half to see that the inverse of A is

A - 1 = ⎡ ⎣ ⎢ 31 - 7 - 4 - 2 11 32 - 9 ⎤ ⎦ ⎥ .

$A^{- 1} = [\begin{array}{r} 3 & - 4 & 3 \\ 1 & - 2 & 2 \\ - 7 & 11 & - 9 \end{array}] .$

Remark

Ordinarily, we do not know in advance whether a given square matrix is invertible or not. To find out, we attempt to carry out the reduction process illustrated in Example 7. If we succeed in reducing A to I, then A is invertible and thereby we find A−1. $A^{- 1} .$ Otherwise—if, somewhere along the way, an all-zero row appears in the left half—we conclude that A is not row equivalent to I, and therefore (by Theorem 6) A is not invertible.

Matrix Equations

In certain applications, one needs to solve a system Ax=b $Ax = b$ of n equations in n unknowns several times in succession—with the same n×n $n \times n$ coefficient matrix A each time, but with different constant vectors b1,b2,…,bk $b_{1}, b_{2}, \dots, b_{k}$ on the right. Thus we want to find solution vectors x1,x2,…,xk $x_{1}, x_{2}, \dots, x_{k}$ such that

A x 1 = b 1, A x 2 = b 2, \dots, A x k = b k .

${Ax}_{1} = b_{1}, {Ax}_{2} = b_{2}, \dots, {Ax}_{k} = b_{k} .$ (17)

By Fact 2 at the beginning of this section,

[A x 1 A x 2 \dots A x k] = A [x 1 x 2 \dots x k] .

$[\begin{array}{l} {Ax}_{1} & {Ax}_{2} & \dots & {Ax}_{k} \end{array}] = A [\begin{array}{l} x_{1} & x_{2} & \dots & x_{k} \end{array}] .$

So the k equations in (17) are equivalent to the single matrix equation

A X = B,

$AX = B,$ (18)

where

X = [x 1 x 2 \dots x k] and B = [b 1 b 2 \dots b k] .

$X = [\begin{array}{l} x_{1} & x_{2} & \dots & x_{k} \end{array}] and B = [\begin{array}{l} b_{1} & b_{2} & \dots & b_{k} \end{array}] .$

If A is invertible and we know A−1, $A^{- 1},$ we can find the n×k $n \times k$ matrix of “unknowns” by multiplying each term in Equation (18) on the left by A−1 $A^{- 1}$ :

X = A - 1 B .

$X = A^{- 1} B .$ (19)

Note that this equation is a generalization of Eq. (13) in Theorem 4. If k=1, $k = 1,$ it usually is simplest to solve the system by Gaussian elimination, but when several different solutions are sought, it may be simpler to find A−1 $A^{- 1}$ first and then to apply (19).

Example 8

Find a 3×4 $3 \times 4$ matrix X such that

⎡ ⎣ ⎢ 453365232 ⎤ ⎦ ⎥ X = ⎡ ⎣ ⎢ 375 - 1 42 214651 ⎤ ⎦ ⎥ .

$[\begin{array}{r} 4 & 3 & 2 \\ 5 & 6 & 3 \\ 3 & 5 & 2 \end{array}] X = [\begin{array}{r} 3 & - 1 & 2 & 6 \\ 7 & 4 & 1 & 5 \\ 5 & 2 & 4 & 1 \end{array}] .$

Solution

The coefficient matrix is the matrix A whose inverse we found in Example 7, so Eq. (19) yields

X = A - 1 B = ⎡ ⎣ ⎢ 31 - 7 - 4 - 2 11 32 - 9 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ 375 - 1 42 214651 ⎤ ⎦ ⎥,

$X = A^{- 1} B = [\begin{array}{r} 3 & - 4 & 3 \\ 1 & - 2 & 2 \\ - 7 & 11 & - 9 \end{array}] [\begin{array}{r} 3 & - 1 & 2 & 6 \\ 7 & 4 & 1 & 5 \\ 5 & 2 & 4 & 1 \end{array}],$

and hence

X = ⎡ ⎣ ⎢ - 4 - 1 11 - 13 - 5 33 148 - 39 1 - 2 4 ⎤ ⎦ ⎥ .

$X = [\begin{array}{r} - 4 & - 13 & 14 & 1 \\ - 1 & - 5 & 8 & - 2 \\ 11 & 33 & - 39 & 4 \end{array}] .$

By looking at the third columns of B and X, for instance, we see that the solution of

4 x 1 + 3 x 2 + 2 x 3 5 x 1 + 6 x 2 + 3 x 3 3 x 1 + 5 x 2 + 2 x 3 = = = 214

$\begin{array}{rcl} 4 x_{1} + 3 x_{2} + 2 x_{3} & = & 2 \\ 5 x_{1} + 6 x_{2} + 3 x_{3} & = & 1 \\ 3 x_{1} + 5 x_{2} + 2 x_{3} & = & 4 \end{array}$

is x1=14, x2=8, x3=−39 $x_{1} = 14, x_{2} = 8, x_{3} = - 39$ .

Nonsingular Matrices

Theorem 6 tells us that the square matrix A is invertible if and only if it is row equivalent to the identity matrix I, and Theorem 4 in Section 3.3 implies that the latter is true if and only if the system Ax=0 $Ax = 0$ has only the trivial solution x=0. $x = 0 .$ A square matrix having these equivalent properties is sometimes called a nonsingular matrix.

Proof

By the remarks preceding the statement of Theorem 7, we already know that properties (a), (b), and (c) are equivalent—if A has any one of these properties, then it also has the other two. We can, therefore, complete the proof by establishing the chain of logical implications

(c) \Rightarrow (d) \Rightarrow (e) \Rightarrow (a) .

$(c) \Rightarrow (d) \Rightarrow (e) \Rightarrow (a) .$

That is, we need to show that if A has property (c), then it has property (d), and, similarly, that (d) implies (e) and that (e) implies (a).

(c)⇒(d): $(c) \Rightarrow (d) :$ We already know that (c) implies (a), and Theorem 4 says that (a) implies (d). Therefore, (c) implies (d).

(d)⇒(e): $(d) \Rightarrow (e) :$ This is obvious, because if the system Ax=b $Ax = b$ has a unique solution, then it certainly has a solution, and thus is consistent.

(e)⇒(a): $(e) \Rightarrow (a) :$ Given the hypothesis that Ax=b $Ax = b$ is consistent for every b, we must prove that A is invertible. Let b=ej, $b = e_{j},$ the jth column vector of the identity matrix I. Then the consistency of Ax=ej $Ax = e_{j}$ yields an n-vector xj $x_{j}$ such that

A x j = e j .

${Ax}_{j} = e_{j} .$ (20)

Let the vectors x1,x2,…,xn $x_{1}, x_{2}, \dots, x_{n}$ be obtained in this way for j=1,2,…,n, $j = 1, 2, \dots, n,$ and let B be the n×n $n \times n$ matrix with these vectors as its columns:

B = [x 1 x 2 \dots x n] .

$B = [\begin{array}{l} x_{1} & x_{2} & \dots & x_{n} \end{array}] .$

Then

A B = = = A [x 1 x 2 \dots x n] [A x 1 A x 2 \dots A x n] [e 1 e 2 \dots e n] [by (20)] .

$\begin{array}{rcl} AB & = & A [\begin{array}{l} x_{1} & x_{2} & \dots & x_{n} \end{array}] \\ = & [\begin{array}{l} {Ax}_{1} & {Ax}_{2} & \dots & {Ax}_{n} \end{array}] \\ = & [\begin{array}{l} e_{1} & e_{2} & \dots & e_{n} \end{array}] [by (20)] . \end{array}$

Therefore, AB=I, $AB = I,$ and thus we have found a matrix B such that AB=I $AB = I$ .

We next show that B is invertible by showing that Bx=0 $Bx = 0$ has only the trivial solution [and by using the fact that property (c) implies property (a)]. But if Bx=0, $Bx = 0,$ then

A (B x) = A 0 = 0,

$A (Bx) = A 0 = 0,$

which implies that Ix=0 $I x = 0$ and thus that x=0. $x = 0 .$ So B is indeed invertible. We can therefore multiply each term in the equation AB=I $AB = I$ on the right by B−1 $B^{- 1}$ to get

A B B - 1 = I B - 1,

${ABB}^{- 1} = {IB}^{- 1},$

so that A=B−1. $A = B^{- 1} .$ Thus A is the inverse of an invertible matrix, and so is itself invertible. This establishes that property (e) implies property (a), and therefore we have completed the proof.

The proof of Theorem 7 is a bit long, but it summarizes most of the basic theory of Chapter 1 and is therefore well worth the effort. Indeed, this theorem is one of the central theorems of elementary linear algebra, and we will need to refer to it repeatedly in subsequent chapters.

3.5 Problems

In Problems 1–8, first apply the formulas in (9) to find A−1. $A^{- 1} .$ Then use A−1 $A^{- 1}$ (as in Example 5) to solve the system Ax=b $Ax = b$ .

A=[3423],b=[56] $A = [\begin{array}{r} 3 & 2 \\ 4 & 3 \end{array}], b = [\begin{array}{r} 5 \\ 6 \end{array}]$
A=[3275],b=[−13] $A = [\begin{array}{r} 3 & 7 \\ 2 & 5 \end{array}], b = [\begin{array}{r} - 1 \\ 3 \end{array}]$
A=[6576],b=[2−3] $A = [\begin{array}{r} 6 & 7 \\ 5 & 6 \end{array}], b = [\begin{array}{r} 2 \\ - 3 \end{array}]$
A=[571217],b=[55] $A = [\begin{array}{r} 5 & 12 \\ 7 & 17 \end{array}], b = [\begin{array}{r} 5 \\ 5 \end{array}]$
A=[3524],b=[56] $A = [\begin{array}{r} 3 & 2 \\ 5 & 4 \end{array}], b = [\begin{array}{r} 5 \\ 6 \end{array}]$
A=[4376],b=[105] $A = [\begin{array}{r} 4 & 7 \\ 3 & 6 \end{array}], b = [\begin{array}{r} 10 \\ 5 \end{array}]$
A=[7597],b=[32] $A = [\begin{array}{r} 7 & 9 \\ 5 & 7 \end{array}], b = [\begin{array}{r} 3 \\ 2 \end{array}]$
A=[851510],b=[73] $A = [\begin{array}{r} 8 & 15 \\ 5 & 10 \end{array}], b = [\begin{array}{r} 7 \\ 3 \end{array}]$

In Problems 9–22, use the method of Example 7 to find the inverse A−1 $A^{- 1}$ of each given matrix A.

[5465] $[\begin{array}{r} 5 & 6 \\ 4 & 5 \end{array}]$
[5476] $[\begin{array}{r} 5 & 7 \\ 4 & 6 \end{array}]$
⎡⎣⎢122557101⎤⎦⎥ $[\begin{array}{r} 1 & 5 & 1 \\ 2 & 5 & 0 \\ 2 & 7 & 1 \end{array}]$
⎡⎣⎢1233810236⎤⎦⎥ $[\begin{array}{r} 1 & 3 & 2 \\ 2 & 8 & 3 \\ 3 & 10 & 6 \end{array}]$
⎡⎣⎢213737329⎤⎦⎥ $[\begin{array}{r} 2 & 7 & 3 \\ 1 & 3 & 2 \\ 3 & 7 & 9 \end{array}]$
⎡⎣⎢322543635⎤⎦⎥ $[\begin{array}{r} 3 & 5 & 6 \\ 2 & 4 & 3 \\ 2 & 3 & 5 \end{array}]$
⎡⎣⎢11314251312⎤⎦⎥ $[\begin{array}{r} 1 & 1 & 5 \\ 1 & 4 & 13 \\ 3 & 2 & 12 \end{array}]$
⎡⎣⎢1−12−31−3−32−3⎤⎦⎥ $[\begin{array}{r} 1 & - 3 & - 3 \\ - 1 & 1 & 2 \\ 2 & - 3 & - 3 \end{array}]$
⎡⎣⎢1−10−32−20−12⎤⎦⎥ $[\begin{array}{r} 1 & - 3 & 0 \\ - 1 & 2 & - 1 \\ 0 & - 2 & 2 \end{array}]$
⎡⎣⎢131−20−1212⎤⎦⎥ $[\begin{array}{r} 1 & - 2 & 2 \\ 3 & 0 & 1 \\ 1 & - 1 & 2 \end{array}]$
⎡⎣⎢112445351⎤⎦⎥ $[\begin{array}{r} 1 & 4 & 3 \\ 1 & 4 & 5 \\ 2 & 5 & 1 \end{array}]$
⎡⎣⎢211001−131⎤⎦⎥ $[\begin{array}{r} 2 & 0 & - 1 \\ 1 & 0 & 3 \\ 1 & 1 & 1 \end{array}]$
⎡⎣⎢⎢⎢⎢0103001010200001⎤⎦⎥⎥⎥⎥ $[\begin{array}{r} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 \\ 3 & 0 & 0 & 1 \end{array}]$
⎡⎣⎢⎢⎢⎢4303011213241101⎤⎦⎥⎥⎥⎥ $[\begin{array}{r} 4 & 0 & 1 & 1 \\ 3 & 1 & 3 & 1 \\ 0 & 1 & 2 & 0 \\ 3 & 2 & 4 & 1 \end{array}]$

In Problems 23–28, use the method of Example 8 to find a matrix X such that AX=B $AX = B$ .

A=[4534],B=[1−13−2−55] $A = [\begin{array}{r} 4 & 3 \\ 5 & 4 \end{array}], B = [\begin{array}{r} 1 & 3 & - 5 \\ - 1 & - 2 & 5 \end{array}]$
A=[7867],B=[20054−3] $A = [\begin{array}{r} 7 & 6 \\ 8 & 7 \end{array}], B = [\begin{array}{r} 2 & 0 & 4 \\ 0 & 5 & - 3 \end{array}]$
A=⎡⎣⎢122487134⎤⎦⎥,B=⎡⎣⎢10−1021320⎤⎦⎥ $A = [\begin{array}{r} 1 & 4 & 1 \\ 2 & 8 & 3 \\ 2 & 7 & 4 \end{array}], B = [\begin{array}{r} 1 & 0 & 3 \\ 0 & 2 & 2 \\ - 1 & 1 & 0 \end{array}]$
A=⎡⎣⎢1215171−22⎤⎦⎥,B=⎡⎣⎢201030102⎤⎦⎥ $A = [\begin{array}{r} 1 & 5 & 1 \\ 2 & 1 & - 2 \\ 1 & 7 & 2 \end{array}], B = [\begin{array}{r} 2 & 0 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & 2 \end{array}]$
A=⎡⎣⎢122−212377⎤⎦⎥,B=⎡⎣⎢001010101110⎤⎦⎥ $A = [\begin{array}{r} 1 & - 2 & 3 \\ 2 & 1 & 7 \\ 2 & 2 & 7 \end{array}], B = [\begin{array}{r} 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \end{array}]$
A=⎡⎣⎢653534322⎤⎦⎥,B=⎡⎣⎢2−11131050205⎤⎦⎥ $A = [\begin{array}{r} 6 & 5 & 3 \\ 5 & 3 & 2 \\ 3 & 4 & 2 \end{array}], B = [\begin{array}{r} 2 & 1 & 0 & 2 \\ - 1 & 3 & 5 & 0 \\ 1 & 1 & 0 & 5 \end{array}]$
Verify parts (a) and (b) of Theorem 3.

Problems 30 through 37 explore the properties of matrix inverses.

Suppose that A, B, and C are invertible matrices of the same size. Show that the product ABC is invertible and that (ABC)−1=C−1B−1A−1 ${(ABC)}^{- 1} = C^{- 1} B^{- 1} A^{- 1}$ .
Suppose that A is an invertible matrix and that r and s are negative integers. Verify that ArAs=Ar+s $A^{r} A^{s} = A^{r + s}$ and that (Ar)s=Ars ${(A^{r})}^{s} = A^{r s}$ .
Prove that if A is an invertible matrix and AB=AC, $AB = AC,$ then B=C. $B = C .$ Thus invertible matrices can be canceled.
Let A be an n×n $n \times n$ matrix such that Ax=x $Ax = x$ for every n-vector x. Show that A=I $A = I$ .
Show that a diagonal matrix is invertible if and only if each diagonal element is nonzero. In this case, state concisely how the inverse matrix is obtained.
Let A be an n×n $n \times n$ matrix with either a row or a column consisting only of zeros. Show that A is not invertible.
Show that A=[acbd] $A = [\begin{array}{l} a & b \\ c & d \end{array}]$ is not invertible if ad−bc=0 $a d - b c = 0$ .
Suppose that ad−bc≠0 $a d - b c \neq 0$ and A−1 $A^{- 1}$ is defined as in Equation (9). Verify directly that AA−1=A−1A=I $A A^{- 1} = A^{- 1} A = I$ .

Problems 38 through 40 explore the effect of multiplying by an elementary matrix.

Let E be the elementary matrix E1 $E_{1}$ of Example 6. If A is a 2×2 $2 \times 2$ matrix, show that EA is the result of multiplying the first row of A by 3.
Let E be the elementary matrix E2 $E_{2}$ of Example 6 and suppose that A is a 3×3 $3 \times 3$ matrix. Show that EA is the result upon adding twice the first row of A to its third row.
Let E be the elementary matrix E3 $E_{3}$ of Example 6. Show that EA is the result of interchanging the first two rows of the matrix A.

Problems 41 and 42 complete the proof of Eq. (2).

Show that the ith row of the product AB is AiB, $A_{i} B,$ where Ai $A_{i}$ is the ith row of the matrix A.
Apply the result of Problem 41 to show that if B is an m×n $m \times n$ matrix and I is the m×m $m \times m$ identity matrix, then IB=B $I B = B$ .
Suppose that the matrices A and B are row equivalent. Use Theorem 5 to prove that B=GA, $B = GA,$ where G is a product of elementary matrices.
Show that every invertible matrix is a product of elementary matrices.
Extract from the proof of Theorem 7 a self-contained proof of the following fact: If A and B are square matrices such that AB=I, $AB = I,$ then A and B are invertible.
Deduce from the result of Problem 45 that if A and B are square matrices whose product AB is invertible, then A and B are themselves invertible.

3.5 Application Automated Solution of Linear Systems

Linear systems with more than two or three equations are most frequently solved with the aid of calculators or computers. If an n×n $n \times n$ linear system is written in the matrix form Ax=b, $Ax = b,$ then we need to calculate first the inverse matrix A−1 $A^{- 1}$ and then the matrix product x=A−1b. $x = A^{- 1} b .$ Suppose the n×n $n \times n$ matrix A and the column vector b have been entered (as illustrated in the 3.2 Application). If A is invertible, then the inverse matrix A−1 $A^{- 1}$ is calculated by the Maple command with(linalg): inverse(A), the Mathematica command Inverse[A], or the Matlab command inv(A). Consequently, the solution vector x is calculated by the Maple command


with(linalg):   x := multiply(inverse(A),b);

or the Mathematica command


x = Inverse[A].b

or the Matlab command


x = inv(A)*b

Figure 3.5.2 illustrates a similar calculator solution of the linear system

3 x 1 2 x 1 5 x 1 4 x 1 - + + - 2 x 2 4 x 2 x 2 6 x 2 + - + - 7 x 3 x 3 7 x 3 8 x 3 + + - + 5 x 4 6 x 4 3 x 4 9 x 4 = = = = 505435286445

$\begin{array}{rcrcrcrcl} 3 x_{1} & - & 2 x_{2} & + & 7 x_{3} & + & 5 x_{4} & = & 505 \\ 2 x_{1} & + & 4 x_{2} & - & x_{3} & + & 6 x_{4} & = & 435 \\ 5 x_{1} & + & x_{2} & + & 7 x_{3} & - & 3 x_{4} & = & 286 \\ 4 x_{1} & - & 6 x_{2} & - & 8 x_{3} & + & 9 x_{4} & = & 445 \end{array}$

FIGURE 3.5.2.

TI-89 solution of a linear system Ax=b $Ax = b$ .

for the solution x1=59, x2=13, x3=17, x4=47. $x_{1} = 59, x_{2} = 13, x_{3} = 17, x_{4} = 47 .$ This solution is also given by the Wolfram|Alpha $Wolfram | Alpha$ query


A = ((3, −2, 7, 5), (2, 4, −1, 6), (5, 1, 7, −3),
     (4, −6, −8, 9)),
b = (505, 435, 286, 445),
inv(A).b

Remark

Whereas the preceding commands illustrate the handy use of conveniently available inverse matrices to solve linear systems, it might be mentioned that modern computer systems employ direct methods—involving Gaussian elimination and still more sophisticated techniques—that are more efficient and numerically reliable to solve a linear system Ax=b $Ax = b$ without first calculating the inverse matrix A−1 $A^{- 1}$ .

Use an available calculator or computer system to solve the linear systems in Problems 1–6 of the 3.3 Application. The applied problems below are elementary in character—resembling the “word problems” of high school algebra—but might illustrate the practical advantages of automated solutions.

You are walking down the street minding your own business when you spot a small but heavy leather bag lying on the sidewalk. It turns out to contain U.S. Mint American Eagle gold coins of the following types:
- One-half ounce gold coins that sell for $285 each,
- One-quarter ounce gold coins that sell for $150 each, and
- One-tenth ounce gold coins that sell for $70 each.
A bank receipt found in the bag certifies that it contains 258 such coins with a total weight of 67 ounces and a total value of exactly $40,145. How many coins of each type are there?
Now you really strike it rich! You find a bag containing one-ounce U.S. American Eagle gold coins valued at $550 each, together with half-ounce and quarter-ounce coins valued as in the preceding problem. If this bag contains a total of 365 coins with a total weight of exactly 11 pounds and a total value of $100,130, how many gold coins of each type are there?
A commercial customer orders 81 gallons of paint that contains equal amounts of red paint, green paint, and blue paint—and, hence, could be prepared by mixing 27 gallons of each. However, the store wishes to prepare this order by mixing three types of paint that are already available in large quantity:
- a reddish paint that is a mixture of 50% red, 25% green, and 25% blue paint;
- a greenish paint that is 12.5% red, 75% green, and 12.5% blue paint; and
- a bluish paint that is 20% red, 20% green, and 60% blue paint.
How many gallons of each are needed to prepare the customer’s order?
Now the paint store receives a really big order—for 244 gallons of paint that is 1/2 red paint, 1/4 green paint, and 1/4 blue paint. The store has three already-mixed types of paint available in large quantity—the greenish paint and the bluish paint of the preceding problem, plus a reddish paint that is 2/3 red paint, 1/6 green paint, and 1/6 blue paint. How many gallons of each must be mixed in order to fill this order?
A tour busload of 45 people attended two Florida theme parks on successive days. On Day 1 the entrance fee was $15 per adult, $8 per child, $12 per senior citizen and the total charge was $558. On Day 2 the entrance fee was $20 per adult, $12 per child, $17 per senior citizen and the total charge was $771. How many adults, children, and senior citizens were on this tour bus?
For some crazy reason, the lunches bought at the first theme park were totaled separately for the adults, children, and seniors. The adults ordered 34 hot dogs, 15 French fries, and 24 soft drinks for a total bill of $70.85. The children ordered 20 hot dogs, 14 French fries, and 15 soft drinks for a total bill of $46.65. The senior citizens ordered 11 hot dogs, 10 French fries, and 12 soft drinks for a total bill of $30.05. What were the prices of a hot dog, an order of French fries, and a soft drink?
A fast-food restaurant sells four types of sandwiches—hamburgers, cheeseburgers, roast beef, and chicken—and has four cash registers. At the end of each day, each cash register tallies the number of each type of sandwich sold, and the total sandwich receipts for the day. The four cash register operators work at different speeds, and one day’s totals were as follows:

Hamburgers Cheeseburgers Roast Beef Chicken Receipts

Register 1 37 44 17 23 $232.99

Register 2 28 35 13 17 $178.97

Register 3 32 39 19 21 $215.99

Register 4 47 51 25 29 $294.38

What was the price of each of the four types of sandwiches?

	Hamburgers	Cheeseburgers	Roast Beef	Chicken	Receipts
Register 1	37	44	17	23	$232.99
Register 2	28	35	13	17	$178.97
Register 3	32	39	19	21	$215.99
Register 4	47	51	25	29	$294.38

The fast-food restaurant of the preceding problem adds a ham sandwich to its menu and, because of increased business, it also adds a fifth cash register and reduces prices. After this expansion, one day’s totals were as follows:

	Hamburgers	Cheeseburgers	Roast Beef	Chicken	Ham	Receipts
Register 1	41	49	22	26	19	$292.79
Register 2	34	39	18	20	16	$236.73
Register 3	36	43	23	24	18	$270.70
Register 4	49	52	26	31	24	$340.19
Register 5	52	55	24	28	25	$341.64

What were the new prices of the five types of sandwiches?

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Table of Contents for 3.5 Inverses of Matrices

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Table of Contents for
3.5 Inverses of Matrices