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Section 6.3 The Matrix Inverse

Before Starting this Section, Review

1 Matrix multiplication (Section 6.2 , page 610)
2 Equality of matrices (Section 6.2 , page 606)
3 Gauss–Jordan elimination procedure (Section 6.1 , page 597)
4 Leontieff input–output model (Section 6.1 , page 599)

Objectives

1 Verify the multiplicative inverse of a matrix.
2 Find the inverse of a matrix.
3 Find the inverse of a 2×2 $2 \times 2$ matrix.
4 Use matrix inverses to solve systems of linear equations.
5 Use matrix inverses in applied problems.

Coding and Decoding Messages

Imagine you are a secret agent who is looking for spies in different parts of the world. Periodically, you communicate with your headquarters via your laptop or by satellite. You expect that your communications will be intercepted. So before you broadcast any message, you must write it using a secret code. A message written with a secret code is called a cryptogram.

First, you might replace each letter in the message with a number and send the message as a sequence of numbers. The problem with this approach is that it is rather easy to crack such a code. For example, if letters such as e are always represented by the same symbol, then a code breaker (hacker) can guess which symbol represents each letter and eventually translate your message. In Example 8, we show you a better way to encode and decode a message using matrix multiplication.

The Multiplicative Inverse of a Matrix

1 Verify the multiplicative inverse of a matrix.

In arithmetic, you know that if a is a nonzero real number, then a has a unique reciprocal, denoted by $\frac{1}{a},$ or $a^{- 1} .$ The number $a^{- 1}$ is called the multiplicative inverse of a, and

$a a^{- 1} = a^{- 1} a = 1 .$

Similarly, certain square matrices have multiplicative inverses, with an identity matrix acting as “1.”

Notice that if B is the inverse of A, then A is the inverse of B.

Recall that in general (even for square matrices), $A B \neq B A .$ However, if A and B are square matrices with $A B = I,$ then it can be shown that $B A = I .$ Therefore, to determine whether A and B are inverses, you need only check that either $A B = I$ or $B A = I .$

Arthur Cayley (1821–1895)

English mathematicians of the first third of the nineteenth century attempted to extend the basic ideas of algebra and to determine exactly how much one can generalize the properties of integers to other types of quantities. Although Sylvester (see page 607) coined the term matrix, it was Arthur Cayley who developed the algebra of matrices. Cayley studied mathematics at Trinity College, Cambridge, but because there was no suitable teaching job available, he decided to become a lawyer. Although he became skilled in legal work, he regarded the law as just a way to support himself. During his legal career, he was able to write more than 300 mathematical papers. In 1863, Cambridge University established a new post in mathematics and offered it to Cayley. He accepted the job eagerly even though it paid less money than he made as a lawyer.

Example 1 Verifying the Inverse of a Matrix

Show that B is the inverse of A.

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

Solution

You need to verify that $A B = I$ or $B A = I .$

$A B = [\begin{array}{r} 1 & 3 \\ 2 & 7 \end{array}] [\begin{array}{r} 7 & - 3 \\ - 2 & 1 \end{array}] = [\begin{array}{r} 1 (7) + 3 (- 2) & 1 (- 3) + 3 (1) \\ 2 (7) + 7 (- 2) & 2 (- 3) + 7 (1) \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] = I$

Because $A B = I,$ it follows that $B = A^{- 1} .$

Practice Problem 1

Show that B is the inverse of A.

$A = [\begin{array}{r} 3 & 2 \\ 2 & 1 \end{array}] and B = [\begin{array}{r} - 1 & 2 \\ 2 & - 3 \end{array}]$

In working with real numbers, you know that if $a = 0,$ then $a^{- 1}$ does not exist. So it should not come as a surprise that if A is a square zero matrix, then $A^{- 1}$ does not exist. Surprisingly, there are also nonzero matrices that do not have inverses.

Example 2 Proving That a Particular Nonzero Matrix Has No Inverse

Show that the matrix A does not have an inverse.

$A = [\begin{array}{r} 2 & 0 \\ 0 & 0 \end{array}]$

Solution

Suppose A has an inverse B, where

$B = [\begin{array}{l} x & y \\ z & w \end{array}] .$

Then you have

$\begin{aligned} [\begin{array}{r} 2 & 0 \\ 0 & 0 \end{array}] [\begin{array}{r} x & y \\ z & w \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] & A B = I \\ [\begin{array}{r} 2 x & 2 y \\ 0 & 0 \end{array}] = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] & Multiply the two matrices on the left side . \end{aligned}$

Because these two matrices are equal, you must have

$0 = 1 Equate the entries in the (2, 2) position .$

Because $0 = 1$ is a false statement, the matrix A does not have an inverse.

Practice Problem 2

Show that the matrix $A = [\begin{array}{r} 3 & 1 \\ 3 & 1 \end{array}]$ does not have an inverse.

Finding the Inverse of a Matrix

2 Find the inverse of a matrix.

It can be shown that if a square matrix A has an inverse, then the inverse is unique. This means that a square matrix cannot have more than one inverse. If a matrix A has an inverse, then A is called invertible or nonsingular. A square matrix that does not have an inverse is called a singular matrix. The following discussion derives a technique for finding the inverse of an invertible matrix.

Suppose we want to find the inverse of the matrix $A = [\begin{array}{r} 1 & 2 \\ 3 & 5 \end{array}] .$ We first let $B = [\begin{array}{l} x & y \\ z & w \end{array}]$ be the inverse of the matrix A. Then

$\begin{array}{rcll} A B & = & I & Definition of inverse \\ [\begin{array}{r} 1 & 2 \\ 3 & 5 \end{array}] [\begin{array}{l} x & y \\ z & w \end{array}] & = & [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] & Substitute for A, B, and I . \\ [\begin{array}{l} x + 2 z & y + 2 w \\ 3 x + 5 z & 3 y + 5 w \end{array}] & = & [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] & Multiply matrices A and B . \end{array}$

Equating corresponding entries in the last equation yields two systems of linear equations.

$\begin{array}{l} {\begin{array}{rcrcl} x & + & 2 z & = & 1 \\ 3 x & + & 5 z & = & 0 \end{array} & \begin{array}{l} Equate the entries in the (1, 1) \\ and (2, 1) positions . \end{array} \\ {\begin{array}{rcrcl} y & + & 2 w & = & 0 \\ 3 y & + & 5 w & = & 1 \end{array} & \begin{array}{l} Equate the entries in the (1, 2) \\ and (2, 2) positions . \end{array} \end{array}$

Let’s solve the two systems by Gauss–Jordan elimination.

System of Equations	Matrix Form
${\begin{array}{rcrcl} x & + & 2 z & = & 1 \\ 3 x & + & 5 z & = & 0 \end{array}$	$[\begin{array}{r} 1 & 2 & 1 \\ 3 & 5 & 0 \end{array}]$
${\begin{array}{rcrcl} y & + & 2 w & = & 0 \\ 3 y & + & 5 w & = & 1 \end{array}$	$[\begin{array}{r} 1 & 2 & 0 \\ 3 & 5 & 1 \end{array}]$

Because the coefficient matrix for both systems of equations is the original matrix

$A = [\begin{array}{r} 1 & 2 \\ 3 & 5 \end{array}],$

the calculations involved in the Gauss–Jordan elimination on the coefficient matrix A will be the same for both systems. Therefore, you can solve both systems at once by considering the “doubly augmented” matrix

$C = [\begin{array}{r} 1 & 2 & 1 & 0 \\ 3 & 5 & 0 & 1 \end{array}] .$

Note that the matrix C is formed by adjoining the identity matrix I to matrix A:

$C = [A | I] .$

Now apply Gauss–Jordan elimination to the matrix C.

$\begin{array}{r} C = [\begin{array}{r} 1 & 2 & 1 & 0 \\ 3 & 5 & 0 & 1 \end{array}] \overset{- 3 R_{1} + R_{2} \to R_{2}}{\to} [\begin{array}{r} 1 & 2 & 1 & 0 \\ 0 & - 1 & - 3 & 1 \end{array}] \overset{(- 1) R_{2}}{\to} [\begin{array}{r} 1 & 2 & 1 & 0 \\ 0 & 1 & 3 & - 1 \end{array}] \\ \overset{- 2 R_{2} + R_{1} \to R_{1}}{\to} [\begin{array}{r} 1 & 0 & - 5 & 2 \\ 0 & 1 & 3 & - 1 \end{array}] \end{array}$

Converting the last matrix back into two systems of equations, we have

$[\begin{array}{r} 1 & 0 & - 5 \\ 0 & 1 & 3 \end{array}] represents {\begin{array}{rcl} 1 \cdot x + 0 \cdot z & = & - 5 \\ 0 \cdot x + 1 \cdot z & = & 3 \end{array} or {\begin{cases} x = - 5 \\ z = 3 \end{cases}$

Similarly,

$[\begin{array}{r} 1 & 0 & 2 \\ 0 & 1 & - 1 \end{array}] represents {\begin{array}{rcl} 1 \cdot y + 0 \cdot w & = & 2 \\ 0 \cdot y + 1 \cdot w & = & - 1 \end{array} or {\begin{cases} y = 2 \\ w = - 1 \end{cases}$

We conclude that $x = - 5, y = 2, z = 3,$ and $w = - 1 .$

Because $B = [\begin{array}{l} x & y \\ z & w \end{array}],$ it follows that

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

You should check that $A B = I .$

We have shown that to find the inverse of an invertible matrix A, we transform the matrix $[A | I]$ by a sequence of elementary row operations into $[I | B],$ where $B = A^{- 1} .$

Example 3 Finding the Inverse of a Matrix

Find the inverse (if it exists) of the matrix $A = [\begin{array}{r} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 2 & 4 & 6 \end{array}] .$

Solution

Step 1 Start by adjoining the identity matrix $I = [\begin{array}{r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ to the matrix A to form the augmented matrix $[A | I] .$

$[A | I] = [\begin{array}{r} 1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 5 & 7 & 0 & 1 & 0 \\ 2 & 4 & 6 & 0 & 0 & 1 \end{array}]$
Step 2 Perform row operations on $[A | I]$ to transform it to a matrix of the form $[I | B] .$

$[\begin{array}{r} 1 & 2 & 3 & 1 & 0 & 0 \\ 2 & 5 & 7 & 0 & 1 & 0 \\ 2 & 4 & 6 & 0 & 0 & 1 \end{array}] \begin{array}{l} \overset{- 2 R_{1} + R_{2} \to R_{2}}{\to} \\ \overset{- 2 R_{1} + R_{3} \to R_{3}}{\to} \end{array} [\begin{array}{r} 1 & 2 & 3 & 1 & 0 & 0 \\ 0 & 1 & 1 & - 2 & 1 & 0 \\ 0 & 0 & 0 & - 2 & 0 & 1 \end{array}] = [C | D]$

Because the third row of matrix C consists of zeros, it is impossible to transform A into I by row operations. Hence, A is not invertible.

Practice Problem 3

Find the inverse (if it exists) of the matrix $A = [\begin{array}{r} 1 & 4 & - 2 \\ - 1 & 1 & 2 \\ 3 & 7 & - 6 \end{array}] .$

Example 4 Finding the Inverse of a Matrix

Find the inverse (if it exists) of the matrix $A = [\begin{array}{r} 1 & 1 & 0 \\ 0 & 3 & 1 \\ 2 & 3 & 3 \end{array}] .$

Solution

Step 1 Start with the matrix $[A | I] :$

$[A | I] = [\begin{array}{r} 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 & 1 & 0 \\ 2 & 3 & 3 & 0 & 0 & 2 \end{array}] I = I_{3}$
Step 2 Use row operations on $[A | I] .$ Then $[A | I] \to$

$\begin{array}{l} \begin{array}{r} \overset{\frac{1}{3} R_{2} \to R_{2}}{\to} \\ \overset{- 2 R_{1} + R_{3} \to R_{3}}{\to} \end{array} [\begin{array}{r} 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{3} & 0 & \frac{1}{3} & 0 \\ 0 & 1 & 3 & - 2 & 0 & 1 \end{array}] \begin{array}{r} \overset{(- 1) R_{2} + R_{3} \to R_{3}}{\to} \end{array} [\begin{array}{r} 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{3} & 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{8}{3} & - 2 & - \frac{1}{3} & 1 \end{array}] \\ \begin{array}{r} \overset{\frac{3}{8} R_{3}}{\to} \end{array} [\begin{array}{r} 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & \frac{1}{3} & 0 & \frac{1}{3} & 0 \\ 0 & 0 & 1 & - \frac{6}{8} & - \frac{1}{8} & \frac{3}{8} \end{array}] \begin{array}{r} \overset{- \frac{1}{3} R_{3} + R_{2} \to R_{2}}{\to} \end{array} [\begin{array}{r} 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac{2}{8} & \frac{3}{8} & - \frac{1}{8} \\ 0 & 0 & 1 & - \frac{6}{8} & - \frac{1}{8} & \frac{3}{8} \end{array}] \\ \begin{array}{r} \overset{(- 1) R_{2} - R_{1} \to R_{1}}{\to} \end{array} [\begin{array}{r} 1 & 0 & 0 & \frac{6}{8} & - \frac{3}{8} & \frac{1}{8} \\ 0 & 1 & 0 & \frac{2}{8} & \frac{3}{8} & - \frac{1}{8} \\ 0 & 0 & 1 & - \frac{6}{8} & - \frac{1}{8} & \frac{3}{8} \end{array}] . \end{array}$

Hence, A is invertible and

$A^{- 1} = [\begin{array}{r} \frac{6}{8} & - \frac{3}{8} & \frac{1}{8} \\ \frac{2}{8} & \frac{3}{8} & - \frac{1}{8} \\ - \frac{6}{8} & - \frac{1}{8} & \frac{3}{8} \end{array}] = \frac{1}{8} [\begin{array}{r} 6 & - 3 & 1 \\ 2 & 3 & - 1 \\ - 6 & - 1 & 3 \end{array}] .$
Step 3 You should verify that $A A^{- 1} = I .$

Practice Problem 4

Find the inverse (if it exists) of the matrix $A = [\begin{array}{r} 1 & 2 & 3 \\ - 2 & 3 & 1 \\ 4 & 5 & - 2 \end{array}] .$

A Rule for Finding the Inverse of a $2 \times 2$ Matrix

3 Find the inverse of a $2 \times 2$ matrix.

You can quickly determine whether any $2 \times 2$ matrix is invertible and, if so, what its inverse is.

Example 5 Finding the Inverse of a $2 \times 2$ Matrix

Find the inverse (if it exists) of each matrix.

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

Solution

For the matrix $A, a = 5, b = 2, c = 4,$ and $d = 3 .$ Here $a d - b c = (5) (3) - (2) (4) = 15 - 8 = 7 \neq 0;$ so A is invertible and

$\begin{array}{rcll} A^{- 1} & = & \frac{1}{7} [\begin{array}{r} 3 & - 2 \\ - 4 & 5 \end{array}] & Substitute for a, b, c, and d in \frac{1}{a d - b c} [\begin{array}{r} d & - b \\ - c & a \end{array}] . \\ = & [\begin{array}{r} \frac{3}{7} & - \frac{2}{7} \\ - \frac{4}{7} & \frac{5}{7} \end{array}] & Scalar multiplication \end{array}$

You should verify that $A A^{- 1} = I .$
For the matrix B, $a = 4, b = 6, c = 2,$ and $d = 3;$ so

$a d - b c = (4) (3) - (6) (2) = 12 - 12 = 0 .$

Therefore, the matrix B does not have an inverse.

Practice Problem 5

Find the inverse (if it exists) of each matrix.
1. $A = [\begin{array}{r} 8 & 2 \\ 4 & 1 \end{array}]$
2. $B = [\begin{array}{r} 8 & - 2 \\ 3 & 1 \end{array}]$

Solving Systems of Linear Equations by Using Matrix Inverses

4 Use matrix inverses to solve systems of linear equations.

Matrix multiplication can be used to write a system of linear equations in matrix form.

System of Equations	Matrix Form
${\begin{array}{r} 3 x - 2 y & = & 4 \\ 4 x - 3 y & = & 5 \end{array}$	$[\begin{array}{r} 3 & - 2 \\ 4 & - 3 \end{array}] [\begin{array}{r} x \\ y \end{array}] = [\begin{array}{r} 4 \\ 5 \end{array}]$
${\begin{array}{r} 2 x_{1} & + & 4 x_{2} & - & x_{3} & = & 9 \\ 3 x_{1} & + & x_{2} & + & 2 x_{3} & = & 7 \\ x_{1} & + & 3 x_{2} & - & 3 x_{3} & = & 4 \end{array}$	$[\begin{array}{r} 2 & 4 & - 1 \\ 3 & 1 & 2 \\ 1 & 3 & - 3 \end{array}] [\begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \end{array}] = [\begin{array}{r} 9 \\ 7 \\ 4 \end{array}]$

Solving a system of linear equations amounts to solving a corresponding matrix equation of the form

$A X = B,$

where A is the coefficient matrix of the system, X is the column matrix containing the variables, and B is the column matrix of the system’s constants.

If A is a square matrix and A is invertible, then the matrix equation $A X = B$ has a unique solution, $X = A^{- 1} B,$ as the following derivation shows.

$\begin{array}{rcll} A X & = & B & Given equation \\ A^{- 1} (A X) & = & A^{- 1} B & \begin{array}{l} Be careful to multiply by A^{- 1} on the left on both \\ sides of the equation because multiplication of \\ matrices is not commutative . \end{array} \\ (A^{- 1} A) X & = & A^{- 1} B & Associative property \\ I X & = & A^{- 1} B & A^{- 1} A = I \\ X & = & A^{- 1} B & I is the identity matrix, and I X = X . \end{array}$

Example 6 Solving a Linear System by Using an Inverse Matrix

Use a matrix inverse to solve the linear system: ${\begin{array}{rcrcrcl} x & + & y & = & 4 \\ 3 y & + & z & = & 7 \\ 2 x & + & 3 y & + & 3 z & = & 21 \end{array}$ .

Solution

Write the linear system in matrix form.

We need to solve the matrix equation $A X = B$ for X. Because the matrix A is invertible (see Example 4), the system has a unique solution $X = A^{- 1} B .$ Use the following:

$\begin{array}{rcll} A^{- 1} & = & \frac{1}{8} [\begin{array}{r} 6 & - 3 & 1 \\ 2 & 3 & - 1 \\ - 6 & - 1 & 3 \end{array}] & Computed in Example 4 \\ X & = & A^{- 1} B \\ = & \frac{1}{8} [\begin{array}{r} 6 & - 3 & 1 \\ 2 & 3 & - 1 \\ - 6 & - 1 & 3 \end{array}] [\begin{array}{r} 4 \\ 7 \\ 21 \end{array}] & Substitute for A^{- 1} and B . \\ = & \frac{1}{8} [\begin{array}{r} 6 (4) - 3 (7) + 1 (21) \\ 2 (4) + 3 (7) - 1 (21) \\ - 6 (4) - 1 (7) + 3 (21) \end{array}] & Matrix multiplication \\ = & \frac{1}{8} [\begin{array}{r} 24 \\ 8 \\ 32 \end{array}] = [\begin{array}{r} 3 \\ 1 \\ 4 \end{array}] & Scalar multiplication \end{array}$

The solution set is ${(3, 1, 4)},$ which you can check in the original system.

Practice Problem 6

Use a matrix inverse to solve the linear system.

${\begin{array}{rcrcrcl} 3 x & + & 2 y & + & 3 z & = & 9 \\ 3 x & + & y & = & 12 \\ x & + & z & = & 6 \end{array}$

Applications of Matrix Inverses

5 Use matrix inverses in applied problems.

The Leontief Input–Output Model

We can use the inverse of a matrix to analyze input–output models that we studied in Section 6.1.

Suppose a simplified economy depends on two products: energy (E) and food (F).

To produce 1 unit of E requires $\frac{1}{4}$ unit of E and $\frac{1}{2}$ unit of F. To produce 1 unit of F requires $\frac{1}{3}$ unit of E and $\frac{1}{4}$ unit of F. Then the interindustry consumption is given by the following matrix.

$\begin{array}{l} Input \\ E F \\ A = [\begin{array}{r} \frac{1}{4} & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{4} \end{array}] \begin{array}{l} E \\ F \end{array} Output \end{array}$

The matrix A is the interindustry technology input–output matrix, or simply the technology matrix, of the system. If the system is producing $x_{1}$ units of energy and $x_{2}$ units of food, then the column matrix $X = [\begin{array}{r} x_{1} \\ x_{2} \end{array}]$ is called the gross production matrix.

Consequently,

$A X = [\begin{array}{r} \frac{1}{4} & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{4} \end{array}] [\begin{array}{l} x_{1} \\ x_{2} \end{array}] = [\begin{array}{l} \frac{1}{4} x_{1} + \frac{1}{2} x_{2} \\ \frac{1}{3} x_{1} + \frac{1}{4} x_{2} \end{array}] \begin{array}{l} \leftarrow units consumed by E \\ \leftarrow units consumed by F \end{array}$

represents the interindustry consumptions. If the column matrix $D = [\begin{array}{l} d_{1} \\ d_{2} \end{array}]$ represents consumer demand, then

$\begin{array}{rcll} D & = & X - A X & Gross production minus \\ interindustry consumption \\ = & I X - A X & I is the identity matrix . \\ D & = & (I - A) X & Distributive property \\ {(I - A)}^{- 1} D & = & {(I - A)}^{- 1} (I - A) X & Multiply both sides by \\ {(I - A)}^{- 1} if I - A is invertible . \\ {(I - A)}^{- 1} D & = & I X & {(I - A)}^{- 1} (I - A) = I \\ = & X & I is the identity matrix . \end{array}$

So if $(I - A)$ is invertible, then the gross production matrix is $X = {(I - A)}^{- 1} D .$

Example 7 Using the Leontief Input–Output Model

In the preceding discussion, suppose the consumer demand for energy is 1000 units and for food is 3000 units. Find the level of production (X) that will meet interindustry and consumer demand.

Solution

For the matrix A, we have

$I - A = [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] - [\begin{array}{r} \frac{1}{4} & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{4} \end{array}] = [\begin{array}{r} \frac{3}{4} & - \frac{1}{2} \\ - \frac{1}{3} & \frac{3}{4} \end{array}] .$

Use the formula for the inverse of a $2 \times 2$ matrix.

$\begin{array}{rcll} {(I - A)}^{- 1} & = & \frac{1}{\frac{3}{4} \cdot \frac{3}{4} - (- \frac{1}{2}) (- \frac{1}{3})} [\begin{array}{r} \frac{3}{4} & \frac{1}{2} \\ \frac{1}{3} & \frac{3}{4} \end{array}] \\ = & \frac{48}{19} [\begin{array}{r} \frac{3}{4} & \frac{1}{2} \\ \frac{1}{3} & \frac{3}{4} \end{array}] = \frac{1}{19} [\begin{array}{r} 36 & 24 \\ 16 & 36 \end{array}] & Simplify . \end{array}$

The matrix $D = [\begin{array}{r} 1000 \\ 3000 \end{array}] .$ Therefore, the gross production matrix X is

$\begin{array}{rcll} X = {(I - A)}^{- 1} D & = & \frac{1}{19} [\begin{array}{l} 36 & 24 \\ 16 & 36 \end{array}] [\begin{array}{l} 1000 \\ 3000 \end{array}] & \begin{array}{l} Substitute for \\ {(I - A)}^{- 1} and D . \end{array} \\ = & \frac{1}{19} [\begin{array}{l} 108, 000 \\ 124, 000 \end{array}] = [\begin{array}{l} \frac{108, 000}{19} \\ \frac{124, 000}{19} \end{array}] \end{array}$

To meet the consumer demand for 1000 units of energy and 3000 units of food, the energy produced must be $\frac{108, 000}{19}$ units and food production must be $\frac{124, 000}{19}$ units.

Practice Problem 7

In Example 7 , find the level of production (X) that will meet both the interindustry and consumer demand if consumer demand for energy is 800 units and consumer demand for food is 3400 units.

Cryptography

To encode or decode a message, first associate each letter in the message with a number, in the obvious manner. The blank corresponds to 0.

$\begin{array}{l} blank & A & B & C & D & E & F & G & H & I & J & K & L & M \\ 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\ N & O & P & Q & R & S & T & U & V & W & X & Y & Z \\ 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 \end{array}$

For example, the message

JOHNSON IS IN DANGER

becomes

$\begin{array}{l} 10 & 15 & 8 & 14 & 19 & 15 & 14 & 0 & 9 & 19 & 0 & 9 & 14 & 0 & 4 & 1 & 14 & 7 & 5 & 18 \\ J & O & H & N & S & O & N & I & S & I & N & D & A & N & G & E & R \end{array}$

This message is easy to decode, so we will make the coding more complicated. First, partition these 20 numbers into groups of three, enclosed in brackets, and insert zeros at the end of the last bracket if necessary.

$\begin{array}{l} [10 & 15 & 8] & [14 & 19 & 15] & [14 & 0 & 9] & [19 & 0 & 9] & [14 & 0 & 4] & [1 & 14 & 7] & [5 & 18 & 0] \\ J & O & H & N & S & O & N & I & S & I & N & D & A & N & G & E & R \end{array}$

Let’s write these seven groups of three numbers as the seven columns of a $3 \times 7$ matrix.

$M = [\begin{array}{r} 10 & 14 & 14 & 19 & 14 & 1 & 5 \\ 15 & 19 & 0 & 0 & 0 & 14 & 18 \\ 8 & 15 & 9 & 9 & 4 & 7 & 0 \end{array}]$

Now choose any invertible $3 \times 3$ matrix A, say,

$A = [\begin{array}{r} 1 & 2 & 3 \\ 1 & 3 & 3 \\ 1 & 2 & 4 \end{array}] .$

We can use the matrix A to convert the message into code and then use $A^{- 1}$ to decode the message, as in the next example. The matrix A is called the coding matrix.

Example 8 Encoding and Decoding a Message

Encode and decode the message

JOHNSON IS IN DANGER

by using the matrix A given above.

Solution

Step 1 Express the message numerically and partition the numbers into groups of three.

$[10 15 8] [14 19 15] [14 0 9] [19 0 9] [14 0 4] [1 14 7] [5 18 0]$
Step 2 Write each group of three numbers in Step 1 as a column of a matrix M.

$M = [\begin{array}{r} 10 & 14 & 14 & 19 & 14 & 1 & 5 \\ 15 & 19 & 0 & 0 & 0 & 14 & 18 \\ 8 & 15 & 9 & 9 & 4 & 7 & 0 \end{array}]$
Step 3 Select an invertible $3 \times 3$ matrix A, such as

$A = [\begin{array}{r} 1 & 2 & 3 \\ 1 & 3 & 3 \\ 1 & 2 & 4 \end{array}] .$
Step 4 Multiply the matrices in Steps 2 and 3 to form AM.

$\begin{array}{rcl} A M & = & [\begin{array}{r} 1 & 2 & 3 \\ 1 & 3 & 3 \\ 1 & 2 & 4 \end{array}] [\begin{array}{r} 10 & 14 & 14 & 19 & 14 & 1 & 5 \\ 15 & 19 & 0 & 0 & 0 & 14 & 18 \\ 8 & 15 & 9 & 9 & 4 & 7 & 0 \end{array}] \\ = & [\begin{array}{r} 64 & 97 & 41 & 46 & 26 & 50 & 41 \\ 79 & 116 & 41 & 46 & 26 & 64 & 59 \\ 72 & 112 & 50 & 55 & 30 & 57 & 41 \end{array}] \begin{array}{l} Matrix \\ multiplication \end{array} \end{array}$

The coded message sent will be the numbers from column 1 of AM, followed by the numbers from column 2, and so on.

$64 79 72 97 116 112 41 41 50 46 46 55 26 26 30 50 64 57 41 59 41$
Step 5 To decode the message in Step 4, write the numbers received in the message in groups of three and as columns of the matrix AM. Find the decoding matrix $A^{- 1}$ of the coding matrix A. You can verify that

$A^{- 1} = [\begin{array}{r} 6 & - 2 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array}] .$
Step 6 Compute the message matrix M.

$\begin{array}{rcl} M = A^{- 1} (A M) & = & [\begin{array}{r} 6 & - 2 & - 3 \\ - 1 & 1 & 0 \\ - 1 & 0 & 1 \end{array}] [\begin{array}{r} 64 & 97 & 41 & 46 & 26 & 50 & 41 \\ 79 & 116 & 41 & 46 & 26 & 64 & 59 \\ 72 & 112 & 50 & 55 & 30 & 57 & 41 \end{array}] \\ = & [\begin{array}{r} 10 & 14 & 14 & 19 & 14 & 1 & 5 \\ 15 & 19 & 0 & 0 & 0 & 14 & 18 \\ 8 & 15 & 9 & 9 & 4 & 7 & 0 \end{array}] \begin{array}{l} Matrix \\ multiplication \end{array} \end{array}$
Step 7 Finally, convert the numerical message from the matrix M in Step 6 back into English.

JOHNSON IS IN DANGER

Practice Problem 8

Rework Example 8 , replacing the message JOHNSON IS IN DANGER with the message JACK IS NOW SAFE.

Section 6.3 Exercises

Concepts and Vocabulary

For $n \times n$ matrices A and B, if $A B = I = B A,$ then B is called the of A.
An $n \times n$ matrix A is invertible if there is a matrix B such that $\underline{} = I .$
To find the inverse of an invertible matrix A, we transform $[A | I]$ by a sequence of row operations into $[I | B],$ where $B = \underline{} .$
$A = [\begin{array}{r} a & b \\ c & d \end{array}]$ is invertible if and only if
True or False. Every square matrix is invertible.
True or False. $A = [\begin{array}{r} 8 & 16 \\ 4 & 8 \end{array}]$ is invertible.
True or False. The inverse of $A = [\begin{array}{r} a & b \\ c & d \end{array}]$ is $\frac{1}{a d - b c} [\begin{array}{r} d & - b \\ - c & a \end{array}]$ if $a d - b c \neq 0$ .
True or False. If A is a square, invertible matrix and $A X = B$ , then $X = A^{- 1} B$ .

Building Skills

In Exercises 9–18, determine whether B is the inverse of A by computing AB and BA.

$A = [\begin{array}{r} 1 & 2 \\ 1 & 3 \end{array}], B = [\begin{array}{r} 3 & - 2 \\ - 1 & 1 \end{array}]$
$A = [\begin{array}{r} 3 & 2 \\ 4 & 3 \end{array}], B = [\begin{array}{r} 3 & - 2 \\ - 4 & 3 \end{array}]$
$A = [\begin{array}{r} 3 & 2 \\ 1 & 4 \end{array}], B = [\begin{array}{r} \frac{2}{5} & - \frac{1}{5} \\ - \frac{1}{10} & \frac{3}{10} \end{array}]$
$A = [\begin{array}{r} 2 & - 3 \\ 4 & - 3 \end{array}], B = [\begin{array}{r} - \frac{1}{2} & \frac{1}{2} \\ - \frac{2}{3} & \frac{1}{3} \end{array}]$
$A = [\begin{array}{r} 1 & 0 & - 1 \\ - 1 & 1 & 1 \end{array}], B = [\begin{array}{r} \begin{matrix} 2 \\ 1 \\ 1 \end{matrix} & \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \end{array}]$
$A = [\begin{array}{r} - 2 & 1 & 3 \\ 0 & - 1 & 1 \\ 1 & 2 & 0 \end{array}], B = \frac{1}{8} [\begin{array}{r} - 2 & 6 & 4 \\ 1 & - 3 & 2 \\ 1 & 5 & 2 \end{array}]$
$A = [\begin{array}{r} 1 & - 2 & 1 \\ - 8 & 6 & - 2 \\ 5 & - 3 & 1 \end{array}], B = \frac{1}{2} [\begin{array}{r} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}]$
$A = [\begin{array}{r} 2 & 3 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 2 \end{array}], B = \frac{1}{18} [\begin{array}{r} 1 & - 5 & 7 \\ 7 & 1 & - 5 \\ - 5 & 7 & 1 \end{array}]$
$A = [\begin{array}{r} 1 & 1 & 1 \\ 1 & 2 & 3 \\ - 1 & 1 & - 1 \end{array}], B = \frac{1}{4} [\begin{array}{r} 5 & - 2 & - 1 \\ 2 & 0 & 2 \\ - 3 & 2 & - 1 \end{array}]$
18. $A = [\begin{array}{r} 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ 1 & 0 & 1 \end{array}], B = \frac{1}{2} [\begin{array}{r} 1 & 1 & 1 \\ - 1 & 1 & 1 \\ - 1 & - 1 & 1 \end{array}]$

In Exercises 19–28, find $A^{- 1}$ (if it exists) by forming the matrix $[A | I]$ and using row operations to obtain $[I | B],$ where $B = A^{- 1} .$

$A = [\begin{array}{r} 2 & 0 \\ 1 & 3 \end{array}]$
$A = [\begin{array}{r} 4 & 3 \\ 1 & 0 \end{array}]$
$A = [\begin{array}{r} 2 & 4 \\ 3 & 6 \end{array}]$
$A = [\begin{array}{r} 9 & 6 \\ 6 & 4 \end{array}]$
$A = [\begin{array}{r} 1 & 6 & 4 \\ 0 & 2 & 3 \\ 0 & 1 & 2 \end{array}]$
$A = [\begin{array}{r} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}]$
$A = [\begin{array}{r} 2 & 3 & 1 \\ 2 & 4 & 1 \\ 3 & 7 & 2 \end{array}]$
$A = [\begin{array}{r} 1 & 1 & 5 \\ 4 & 3 & - 5 \\ 1 & 1 & 0 \end{array}]$
$A = [\begin{array}{r} 3 & - 3 & 4 \\ 2 & - 3 & 4 \\ 0 & - 1 & 1 \end{array}]$
$A = [\begin{array}{r} 1 & 2 & - 2 \\ - 1 & 3 & 0 \\ 0 & - 2 & 1 \end{array}]$

In Exercises 29–34, find the inverse (if it exists) of $A = [\begin{array}{r} a & b \\ c & d \end{array}]$ by using the formula $A^{- 1} = \frac{1}{a d - b c} [\begin{array}{r} d & - b \\ - c & a \end{array}] .$ Check that $A^{- 1} A = I .$

$A = [\begin{array}{r} 1 & 0 \\ 3 & 2 \end{array}]$
$A = [\begin{array}{r} 3 & 4 \\ 5 & 6 \end{array}]$
$A = [\begin{array}{r} 2 & - 3 \\ - 3 & 5 \end{array}]$
$A = [\begin{array}{r} 3 & - 4 \\ 2 & - 2 \end{array}]$
$A = [\begin{array}{r} a & - b \\ b & - a \end{array}], a^{2} \neq b^{2}$
$A = [\begin{array}{r} 2 & - 1 \\ 1 & - 1 \end{array}]$

In Exercises 35–38, write the given linear system of equations as a matrix equation $A X = B .$

${\begin{array}{rc} 2 x & + & 3 y & = & - 9 \\ x & - & 3 y & = & 13 \end{array}$
${\begin{array}{rcl} 5 x & - & 4 y & = & 7 \\ 4 x & - & 3 y & = & 5 \end{array}$
${\begin{array}{rcl} 3 x & + & 2 y & + & z & = & 8 \\ 2 x & + & y & + & 3 z & = & 7 \\ x & + & 3 y & + & 2 z & = & 9 \end{array}$
${\begin{array}{rcl} x & + & 3 y & + & z & = & 4 \\ x & - & 5 y & + & 2 z & = & 7 \\ 3 x & + & y & - & 4 z & = & - 9 \end{array}$

In Exercises 39–42, write each matrix equation as a linear system of equations.

$[\begin{array}{r} 1 & - 2 \\ 2 & 1 \end{array}] [\begin{array}{r} x \\ y \end{array}] = [\begin{array}{r} 0 \\ 5 \end{array}]$
$[\begin{array}{r} 2 & 3 \\ 3 & - 1 \end{array}] [\begin{array}{r} x_{1} \\ x_{2} \end{array}] = [\begin{array}{r} 0 \\ 11 \end{array}]$
$[\begin{array}{r} 2 & 3 & 1 \\ 5 & 7 & - 1 \\ 4 & 3 & 0 \end{array}] [\begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \end{array}] = [\begin{array}{r} - 1 \\ 5 \\ 5 \end{array}]$
$[\begin{array}{r} 3 & - 2 & 3 \\ 5 & 0 & 4 \\ 2 & 7 & 0 \end{array}] [\begin{array}{r} r \\ s \\ t \end{array}] = [\begin{array}{r} 4 \\ 3 \\ - 8 \end{array}]$
Let $A = [\begin{array}{r} 1 & 2 & 5 \\ 2 & 3 & 8 \\ - 1 & 1 & 2 \end{array}] .$ Show that $A^{- 1} = [\begin{array}{r} 2 & - 1 & - 1 \\ 12 & - 7 & - 2 \\ - 5 & 3 & 1 \end{array}] .$

In Exercises 44–46, use the result of Exercise 43 to solve the system of equations.

${\begin{array}{rcl} x & + & 2 y & + & 5 z & = & 4 \\ 2 x & + & 3 y & + & 8 z & = & 6 \\ - x & + & y & + & 2 z & = & 3 \end{array}$
${\begin{array}{rcl} x_{1} & + & 2 x_{2} & + & 5 x_{3} & = & 1 \\ 2 x_{1} & + & 3 x_{2} & + & 8 x_{3} & = & 3 \\ - x_{1} & + & x_{2} & + & 2 x_{3} & = & - 3 \end{array}$
${\begin{array}{rcl} x & + & 2 y & + & 5 z & = & - 4 \\ 2 x & + & 3 y & + & 8 z & = & - 6 \\ - x & + & y & + & 2 z & = & - \frac{5}{2} \end{array}$
1. Find the inverse of the matrix $A = [\begin{array}{r} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{array}] .$
2. Use the result of part (a) to solve the linear system.
  
  ${\begin{array}{rcl} x & + & y & + & z & = & 6 \\ x & + & 2 y & + & 3 z & = & 14 \\ x & + & 4 y & + & 9 z & = & 36 \end{array}$
1. Find the inverse of the matrix $A = [\begin{array}{r} 2 & 4 & - 1 \\ 3 & 1 & 2 \\ 1 & 3 & - 3 \end{array}] .$
2. Use the result of part (a) to solve the linear system.
  
  ${\begin{array}{rcr} 2 x & + & 4 y & - & z & = & 9 \\ 3 x & + & y & + & 2 z & = & 7 \\ x & + & 3 y & - & 3 z & = & 4 \end{array}$

Applying the Concepts

In Exercises 49–54, solve each linear system by using an inverse matrix.

${\begin{aligned} 3 x & + & 7 y & = & 11 \\ - 5 x & + & 4 y & = & 13 \end{aligned}$

${\begin{array}{rcr} x & - & 7 y & = & 3 \\ 2 x & + & 3 y & = & 23 \end{array}$
${\begin{array}{rcr} x & + & y & + & 2 z & = & 7 \\ x & - & y & - & 3 z & = & - 6 \\ 2 x & + & 3 y & + & z & = & 4 \end{array}$
${\begin{array}{rcr} x & + & y & + & z & = & 6 \\ 2 x & - & 3 y & + & 3 z & = & 5 \\ 3 x & - & 2 y & - & z & = & - 4 \end{array}$
${\begin{array}{rcr} 2 x & + & 2 y & + & 3 z & = & 7 \\ 5 x & + & 3 y & + & 5 z & = & 3 \\ 3 x & + & 5 y & + & z & = & - 5 \end{array}$
${\begin{array}{rcr} 9 x & + & 7 y & + & 4 z & = & 12 \\ 6 x & + & 5 y & + & 4 z & = & 5 \\ 4 x & + & 3 y & + & z & = & 7 \end{array}$

In Exercises 55–58, use the method presented in this section to solve the problem.

Investment. At the beginning of 2016, Liz decided to invest $90,000, splitting it three different ways. Part of the money she invested in a treasury bill that yields 3% annual interest, she invested another part in bonds with an annual yield 7%, and the rest she invested in a mutual fund that estimated an annual interest rate of 8%. Liz expected her net income for 2016 to be $5920. Unfortunately, the mutual fund lost 11% of its assets during the year so that Liz received only $980 income from all three investments combined. How much money did Liz put in each investment?
Sandwich sales. The Fresh Sandwich specializes in chicken, fish, and ham sandwiches. In April, May, and June 2016, they sold the following quantities.

Sales Volume in Hundreds

April May June

Chicken 8 7 4

Fish 11 8 6

Ham 6 7 3

Monthly revenue, in hundreds, from the sales of these sandwiches $164 $141 $86

Find the selling price of each of the three types of sandwiches.
Euler’s formula. A plane graph connects points called vertices with nonintersecting lines called edges and divides the plane into regions. The accompanying figure has seven vertices, nine edges, and four regions. Euler’s theorem states that in a plane graph,

$V - E + R = 2,$

where
- $\begin{array}{l} V = the number of vertices, \\ E = the number of edges, and \\ R = the number of regions . \end{array}$
In a certain plane graph, twice the number of edges is three times the number of vertices and twice the number of regions is one less than the number of edges. Find the number of vertices, edges, and regions for this graph. Sketch a plane graph that satisfies the conditions of this exercise.
Projectile. On a planet, the height h of a projectile fired up from an initial height $h_{0}$ with initial vertical velocity of $v_{0} ft / sec$ after t seconds is given by $h = \frac{1}{2} a t^{2} + v_{0} t + h_{0},$ where a is the gravitational constant. The following measurements were recorded.

Time t in seconds 1 2 3

Height h in feet 100 92 76
1. Find the initial height.
2. Find the height of the projectile after 2.5 seconds.

	Sales Volume in Hundreds
Chicken	8	7	4
Fish	11	8	6
Ham	6	7	3
Monthly revenue, in hundreds, from the sales of these sandwiches	$164	$141	$86

In Exercises 59–62, technology matrix A, representing interindustry demand, and matrix D, representing consumer demand for the sectors in a two- or three-sector economy, are given. Use the matrix equation $X = {(I - A)}^{- 1} D$ to find the production level X that will satisfy both demands.

Two-sector economy. In a two-sector economy,

$A = [\begin{array}{r} 0.1 & 0.4 \\ 0.5 & 0.2 \end{array}], D = [\begin{array}{r} 50 \\ 30 \end{array}] .$
Two-sector economy. In a two-sector economy,

$A = [\begin{array}{r} 0.2 & 0.5 \\ 0.1 & 0.6 \end{array}], D = [\begin{array}{r} 11 \\ 18 \end{array}] .$
Three-sector economy. In a three-sector economy,

$A = [\begin{array}{l} 0.2 & 0.2 & 0 \\ 0.1 & 0.1 & 0.3 \\ 0.1 & 0 & 0.2 \end{array}], D = [\begin{array}{r} 400 \\ 600 \\ 800 \end{array}] .$
Three-sector economy. In a three-sector economy,

$A = [\begin{array}{r} 0.5 & 0.4 & 0.2 \\ 0.2 & 0.3 & 0.1 \\ 0.1 & 0.1 & 0.3 \end{array}], D = [\begin{array}{r} 500 \\ 300 \\ 200 \end{array}] .$

In Exercises 63 and 64, use a $2 \times 2$ invertible coding matrix A of your choice. Remember to partition the numerical message into groups of two. (a) Write a cryptogram for each message. (b) Check by decoding.

Coding and decoding. CANNOT FIND COLOMBO.
Coding and decoding. FOUND HEAD OF TERRORIST NETWORK.

In Exercises 65 and 66, use a $3 \times 3$ invertible coding matrix A of your choice.

Code and then decode the message of Exercise 63.
Code and then decode the message of Exercise 64.

Beyond the Basics

Suppose A is an invertible square matrix.
1. Is $A^{- 1}$ invertible? Why or why not?
2. If your answer to part (a) is yes, then what is ${(A^{- 1})}^{- 1} ?$
Let $A^{- 1} = [\begin{array}{r} 2 & 1 \\ 3 & - 4 \end{array}] .$ Find A.
Let A be a square matrix. Show that if $A^{2}$ is invertible, then A must be invertible. [Hint: $A^{2} B = A (A B) .$ ]
Show that if A is invertible, then $A^{2}$ is invertible.
Show that if A and B are $n \times n$ invertible matrices, then AB is invertible and ${(A B)}^{- 1} = B^{- 1} A^{- 1} .$ (Note the order.)
Verify the result of Exercise 71 for

$A = [\begin{array}{r} 1 & 2 \\ 3 & 4 \end{array}], B = [\begin{array}{r} - 1 & 0 \\ 3 & - 2 \end{array}] .$
Let A, B, and C be $n \times n$ matrices such that $A B = A C .$ Show that if A is invertible, then $B = C .$
Is the result of Exercise 73 true if A is not invertible?
Let A and B be $n \times n$ matrices such that $A B = A .$ Show that if A is invertible, then $B = I .$
Give an example to show that the result of Exercise 75 is not true if A is not invertible.
Let $A = [\begin{array}{r} 3 & 4 \\ 2 & 3 \end{array}] .$
1. Show that the matrix A satisfies the equation
  
  $A^{2} - 6 A + I = 0 .$
2. Use the equation in part (a) to show that $A^{- 1} = 6 I - A .$
3. Use the equation in part (b) to find $A^{- 1} .$
Let $A = [\begin{array}{r} 2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2 \end{array}] .$
1. Show that A satisfies the equation $A^{3} - 6 A^{2} + 9 A - 4 I = 0 .$
2. Use the equation in part (a) to show that
  
  $A^{- 1} = \frac{1}{4} (A^{2} - 6 A + 9 I) .$
  
  [Hint: Consider the product $\frac{1}{4} (A^{2} - 6 A + 9 I) A .$ ]
3. Use part (b) to find $A^{- 1} .$

In Exercises 79 and 80, use your graphing utility to find the inverse of the coefficient matrix. Then use the inverse to solve each linear system.

${\begin{array}{r} x & - & y & - & z & - & w & = & - 4 \\ x & + & y & + & z & - & w & = & 2 \\ 2 x & + & y & + & z & - & w & = & 3 \\ x & - & y & + & z & - & w & = & - 2 \end{array}$
${\begin{array}{r} x & + & y & + & z & - & w & = & 1 \\ x & - & y & - & z & + & w & = & 1 \\ x & - & y & + & z & + & 3 w & = & 1 \\ x & + & y & + & z & + & w & = & 3 \end{array}$
Suppose A and B are $2 \times 2$ matrices and $B^{- 1} = [\begin{array}{r} 1 & 2 \\ 3 & 4 \end{array}] .$

Find A.
1. If $A B = [\begin{array}{r} - 1 & 3 \\ 0 & 2 \end{array}] .$
2. If $B A = [\begin{array}{r} - 1 & 3 \\ 0 & 2 \end{array}] .$
3. If $B A B^{- 1} = [\begin{array}{r} 1 & 1 \\ 1 & 2 \end{array}] .$
4. If $B^{- 1} A B = [\begin{array}{r} 1 & 1 \\ 1 & 2 \end{array}] .$
1. Is the matrix $[\begin{array}{c} x & - 1 \\ 2 & x + 2 \end{array}]$ invertible for all real numbers x?
2. Is the matrix $[\begin{array}{c} x - 1 & - 1 \\ 3 & x - 5 \end{array}]$ invertible for all real numbers x?

Critical Thinking/Discussion/Writing

True or False. Justify your answers.
1. If A and B are matrices for which $A B = B A,$ then the formula $A^{2} B = B A^{2}$ must hold.
2. If A and B are invertible matrices, then $A + B$ must be invertible.
Give an example of a nonzero $2 \times 2$ matrix A such that $A^{2} = 0 .$ Show that the matrix $I + A$ is invertible and that ${(I + A)}^{- 1} = I - A .$
Is it true that every nonsingular matrix is a square matrix? Why or why not?

Getting Ready for the Next Section

In Exercises 86–89, state whether each expression is equal to 1 or $- 1 .$

${(- 1)}^{0}$
${(- 1)}^{10}$
${(- 1)}^{17}$
${(- 1)}^{6 + 9}$

In Exercises 90–93, perform the indicated operation.

$[- 1, 4] [\begin{array}{r} 2 \\ 3 \end{array}]$
$[\begin{array}{r} - 1 \\ - 3 \end{array}] [5, 0]$
$3 [\begin{array}{r} 7 & 1 \\ 5 & - 2 \end{array}]$
$(- 1) [\begin{array}{r} - 6 & 7 \\ 4 & - 1 \end{array}]$

In Exercises 94–97, solve each system of equations.

${\begin{aligned} 2 x & - & 3 y & = & 16 \\ x & - & y & = & 7 \end{aligned}$
${\begin{aligned} 16 x & - & 9 y & = & - 5 \\ 10 x & + & 18 y & = & - 11 \end{aligned}$
${\begin{aligned} x - y + 5 z & = & - 6 \\ 3 x + 3 y - z & = & 10 \\ x + 3 y + 2 z & = & 5 \end{aligned}$
${\begin{aligned} 2 x - y + 2 z & = & 3 \\ 2 x + 2 y - z & = & 0 \\ - x + 2 y + 2 z & = & - 12 \end{aligned}$

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Time `t` in seconds	1	2	3
Height `h` in feet	100	92	76

Table of Contents for Section 6.3 The Matrix Inverse

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Table of Contents for
Section 6.3 The Matrix Inverse