We've updated our

TEXT

# Reading: Solving Systems with Inverses

Nancy plans to invest \$10,500 into two different bonds to spread out her risk. The first bond has an annual return of 10%, and the second bond has an annual return of 6%. In order to receive an 8.5% return from the two bonds, how much should Nancy invest in each bond? What is the best method to solve this problem? There are several ways we can solve this problem. As we have seen in previous sections, systems of equations and matrices are useful in solving real-world problems involving finance. After studying this section, we will have the tools to solve the bond problem using the inverse of a matrix.

## Finding the Inverse of a Matrix

We know that the multiplicative inverse of a real number is a–1, and $\displaystyle{a}{a}^{{-{1}}}={a}^{{-{{1}}}}{a}={(\frac{{1}}{{a}})}{a}={1}$. The multiplicative inverse of a matrix is similar in concept, except that the product of matrix A and its inverse A–1 equals the identity matrix. The identity matrix is a square matrix containing ones down the main diagonal and zeros everywhere else. We identify identity matrices by In where n represents the dimension of the matrix. The next two equations are the identity matrices for a 2 × 2 matrix and a 3 ×3 matrix, respectively. [latex-display]\displaystyle{I}_{{3}}=\left[\matrix{{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{1}}\right][/latex-display] The identity matrix acts as a 1 in matrix algebra. For example, AI = IA =A A matrix that has a multiplicative inverse has the properties [latex-display]\displaystyle\left[\matrix{{A}{A}^{{-{{1}}}}={I}\\{A}^{{-{{1}}}}{A}={I}}\right][/latex-display] A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, $\displaystyle{A}{A}^{{-{{1}}}}={A}^{{-{{1}}}}{A}={I}$, is a requirement. Not all square matrices have an inverse, but if Ais invertible, then A–1 is unique. We will look at two methods for finding the inverse of a 2 × 2matrix and a third method that can be used on both 2 × 2and 3 × 3matrices.

## The Identity Matrix and Multiplicative Inverse

The identity matrixIn,is a square matrix containing ones down the main diagonal and zeros everywhere else.

#### 2×2

[latex-display]\displaystyle{I}_{{2}}=\left[\matrix{{1}&{0}\\{0}&{1}}\right][/latex-display]

#### 3×3

[latex-display]\displaystyle{I}_{{3}}=\left[\matrix{{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{1}}\right][/latex-display] If is an n × n matrix and Bis an n × n matrix such that AB = BA = In, then B = A–1, the multiplicative inverse of a matrix A.

### Example 1

Given matrix A, show that AI = IA =A $\displaystyle{A}=\left[\matrix{{3}&{4}\\-{2}&{5}}\right]$

#### Solution

Use matrix multiplication to show that the product of A and the identity is equal to the product of the identity and A [latex-display]\displaystyle{A}{I}=\left[\matrix{{3}&{4}\\-{2}&{5}}\right]\left[\matrix{{1}&{0}\\{0}&{1}}\right]=\left[\matrix{{3}\cdot{1}+{4}\cdot{0}&{3}\cdot{0}+{4}\cdot{1}\\-{2}\cdot{1}+{5}\cdot{0}&-{2}\cdot{0}+{5}\cdot{1}}\right]=\left[\matrix{{3}&{4}\\-{2}&{5}}\right][/latex-display] [latex-display]\displaystyle{A}{I}={\left[\matrix{{1}&{0}\\{0}&{1}}\right]{\left[\matrix{{3}&{4}\\-{2}&{5}}\right]={\left[\matrix{{1}\cdot{3}+{0}\cdot-{2}&{1}\cdot{4}+{0}\cdot{5}\\{0}\cdot{3}+{1}\cdot-{2}&{0}\cdot{4}+{1}\cdot{5}}\right]={\left[\matrix{{3}&{4}\\-{2}&{5}}\right][/latex-display]

### How To

Given two matrices, show that one is the multiplicative inverse of the other.
1. Given matrix A of order n × n and matrix B of order n × n multiply AB
2. If AB = I, then find the product BA. If BA = I, then B = A–1 and A = B–1

### Example 2

1. Show that the given matrices are multiplicative inverses of each other. $\displaystyle{A}={\left[\matrix{{1}&{5}\\-{2}&-{9}}\right],{B}={\left[\matrix{-{9}&-{5}\\{2}&{1}}\right]$
2. Show that the following two matrices are inverses of each other $\displaystyle{A}={\left[\matrix{{1}&{4}\\-{1}&-{3}}\right],{B}={\left[\matrix{-{3}&-{4}\\{1}&{1}}\right]$

#### Solutions

1. Multiply A and B. If both products equal the identity, then the two matrices are inverses of each other. [latex-display]\displaystyle{A}{B}={\left[\matrix{{1}&{5}\\-{2}&{9}}\right]\cdot{\left[\matrix{-{9}&-{5}\\{2}&{1}}\right]={\left[\matrix{{1}{(-{9})}+{5}{({2})}&{1}{(-{5})}+{5}{({1})}\\-{2}{(-{9})}-{9}{({2})}&-{2}{(-{5})}-{9}{({1})}}\right]={\left[\matrix{{1}&{0}\\{0}&{1}}\right][/latex-display] A and B are inverses of each other.
2. $\displaystyle{A}{B}={\left[\matrix{{1}&{4}\\-{1}&-{3}}\right]\cdot{\left[\matrix{-{3}&-{4}\\{1}&{1}}\right]={\left[\matrix{{1}{(-{3})}+{4}{({1})}&{1}{(-{4})}+{4}{({1})}\\-{1}{(-{3})}+-{3}{({1})}&-{1}{(-{4})}+-{3}{({1})}}\right]={\left[\matrix{{1}&{0}\\{0}&{1}}\right]$

## Finding the Multiplicative Inverse Using Matrix Multiplication

We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication.

### Example 3

Use matrix multiplication to find the inverse of the given matrix. [latex-display]\displaystyle{A}={\left[\matrix{{1}&-{2}\\{2}&-{3}}\right][/latex-display]

#### Solution

For this method, we multiply by a matrix containing unknown constants and set it equal to the identity. [latex-display]\displaystyle{\left[\matrix{{1}&-{2}\\{2}&-{3}}\right]{\left[\matrix{{a}&{b}\\{c}&{d}}\right]={\left[\matrix{{1}&{0}\\{0}&{1}}\right][/latex-display] Find the product of the two matrices on the left side of the equal sign. [latex-display]\displaystyle{\left[\matrix{{1}&-{2}\\{2}&-{3}}\right]{\left[\matrix{{a}&{b}\\{c}&{d}}\right]={\left[\matrix{{1}{a}-{2}{c}&{1}{b}-{2}{d}\\{2}{a}-{3}{c}&{2}{b}-{3}{d}}\right][/latex-display] Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0. [latex-display]\displaystyle{1}{a}-{2}{c}={1} {R}_{{1}}[/latex-display] [latex-display]\displaystyle{2}{a}-{3}{c}={0} {R}_{{2}}[/latex-display] Using row operations, multiply and add as follows: $\displaystyle{(-{2})}{R}_{{1}}+{R}_{{2}}\rightarrow{R}_{{2}}$. Add the equations, and solve for c. [latex-display]\displaystyle{1}{a}-{2}{c}={1}[/latex-display] [latex-display]\displaystyle{0}+{1}{c}=-{2}[/latex-display] [latex-display]\displaystyle{c}=-{2}[/latex-display] Back-substitute to solve for a. [latex-display]\displaystyle{\left[\matrix{{a}-{2}{(-{2})}={1}\\{a}+{4}={1}\\{a}=-{3}}\right][/latex-display] Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity. [latex-display]\displaystyle{\left[\matrix{{1}{b}-{2}{d}={0}&{R}_{{1}}\\{2}{b}-{3}{d}={1}&{R}_{{2}}}\right][/latex-display] Using row operations, multiply and add as follows: $\displaystyle{(-{2})}{R}_{{1}}+{R}_{{2}}={R}_{{2}}$. Add the two equations and solve for [latex-display]\displaystyle{\left[\matrix{{1}{b}-{2}{d}={0}\\{0}+{1}{d}={1}\\{d}={1}}\right][/latex-display] Once more, back-substitute and solve for [latex-display]\displaystyle{\left[\matrix{{b}-{2}{({1})}={0}\\{b}-{2}={0}\\{b}={2}}\right][/latex-display] [latex-display]\displaystyle{A}^{{-{{1}}}}={\left[\matrix{-{3}&{2}\\-{2}&{1}}\right][/latex-display]

## Finding the Multiplicative Inverse by Augmenting with the Identity

Another way to find the multiplicative inverse is by augmenting with the identity. When matrix A is transformed into I, the augmented matrix I transforms into A–1 For example, given [latex-display]\displaystyle{A}={\left[\matrix{{2}&{1}\\{5}&{3}}\right][/latex-display] augment with the identity [latex-display]\displaystyle{\left[\matrix{{2}&{1}&{\mid}&{1}&{0}\\{5}&{3}&{\mid}&{0}&{1}}\right][/latex-display] Perform row operations with the goal of turning into the identity.
1. Switch row 1 and row 2. $\displaystyle{\left[\matrix{{5}&{3}&{\mid}&{0}&{1}\\{2}&{1}&{\mid}&{1}&{0}}\right]$
2. Multiply row 2 by –2 and add to row 1. $\displaystyle{\left[\matrix{{1}&{1}&{\mid}&-{2}&{1}\\{2}&{1}&{\mid}&{1}&{0}}\right]$
3. Multiply row 1 by –2 and add to row 2. $\displaystyle{\left[\matrix{{1}&{1}&{\mid}&-{2}&{1}\\{0}&-{1}&{\mid}&{5}&-{2}}\right]$
4. Add row 2 to row 1. $\displaystyle{\left[\matrix{{1}&{0}&{\mid}&{3}&-{1}\\{0}&-{1}&{\mid}&{5}&-{2}}\right]$
5. Multiply row 2 by –1 $\displaystyle{\left[\matrix{{1}&{0}&{\mid}&{3}&-{1}\\{0}&{1}&{\mid}&-{5}&{2}}\right]$
The matrix we have found is A–1 [latex-display]\displaystyle{A}^{{-{{1}}}}={\left[\matrix{{3}&-{1}\\-{5}&{2}}]\right[/latex-display]

## Finding the Multiplicative Inverse of 2 × 2 Matrices Using a Formula

When we need to find the multiplicative inverse of a 2 × 2 matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity. If A is a 2 × 2 matrix, such as [latex-display]\displaystyle{A}={\left[\matrix{{a}&{b}\\{c}&{d}}\right][/latex-display] the multiplicative inverse of is given by the formula [latex-display]\displaystyle{A}^{{-{{1}}}}=\frac{{1}}{{{a}{d}-{b}{c}}}{\left[\matrix{{d}&-{b}\\-{c}&{a}}\right][/latex-display] where ad – bc ≠ 0. If ad – bc = 0, then A has no inverse.

### Example 4

1. Use the formula to find the multiplicative inverse of $\displaystyle{A}={\left[\matrix{{1}&-{2}\\{2}&-{3}}\right]$
2. Use the formula to find the inverse of matrix A. Verify your answer by augmenting with the identity matrix. $\displaystyle{A}={\left[\matrix{{1}&-{1}\\{2}&{3}}\right]$
3. Find the inverse, if it exists, of the given matrix. $\displaystyle{A}={\left[\matrix{{3}&{6}\\{1}&{2}}\right]$

#### Solutions

1. Using the formula, we have
[latex-display]\displaystyle{{A}^{{-{{1}}}}=\frac{{1}}{{{({1})}{(-{3})}-{(-{2})}{({2})}}}{\left[\matrix{-{3}&{2}\\-{2}&{1}}\right][/latex-display] [latex-display]\displaystyle{A}^{{-{{1}}}}=\frac{{1}}{{-{3}+{4}}}{\left[\matrix{-{3}&{2}\\-{2}&{1}}\right][/latex-display] [latex-display]\displaystyle{A}^{{-{{1}}}}={\left[\matrix{-{3}&{2}\\-{2}&{1}}\right][/latex-display] Perform row operations with the goal of turning A into the identity. a) Multiply row 1 by –2 and add to row 2. [latex-display]\displaystyle{\left[\matrix{{1}&-{2}&{\mid}&{1}&{0}\\{0}&{1}&{\mid}&-{2}&{1}}\right][/latex-display] b) Multiply row 1 by 2 and add to row 1. [latex-display]\displaystyle{\left[\matrix{{1}&{0}&{\mid}&-{3}&{2}\\{0}&{1}&{\mid}&-{2}&{1}}\right][/latex-display] c) So, we have verified our original solution. [latex-display]\displaystyle{A}^{{-{{1}}}}={\left[\matrix{-{3}&{2}\\-{2}&{1}}]}[/latex-display] 2. $\displaystyle{A}^{{-{{1}}}}={\left[\matrix{\frac{{3}}{{5}}&\frac{{1}}{{5}}\-\frac{{2}}{{5}}&\frac{{1}}{{5}}}\right]$ We will use the method of augmenting with the identity. [latex-display]\displaystyle{\left[\matrix{{3}&{6}&{\mid}&{1}&{0}\\{1}&{3}&{\mid}&{0}&{1}}\right][/latex-display] a) Switch row 1 and row 2. [latex-display]\displaystyle{\left[\matrix{{1}&{3}&{\mid}&{0}&{1}\\{3}&{6}&{\mid}&{1}&{0}}\right][/latex-display] b) Multiply row 1 by −3 and add it to row 2. [latex-display]\displaystyle{\left[\matrix{{1}&{3}&{\mid}&{0}&{1}\\{0}&{0}&{\mid}&-{3}&{1}}\right][/latex-display] c) There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.

## Finding the Multiplicative Inverse of 3 × 3 Matrices

Unfortunately, we do not have a formula similar to the one for a 2 × 2 matrix to find the inverse of a 3 × 3 matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse. Given a 3 × 3 matrix [latex-display]\displaystyle{A}={\left[\matrix{{2}&{3}&{1}\\{3}&{3}&{1}\\{2}&{4}&{1}}\right][/latex-display] augment A with the identity matrix [latex-display]\displaystyle{A}{\mid}{I}={\left[\matrix{{2}&{3}&{1}&{\mid}&{1}&{0}&{0}\\{3}&{3}&{1}&{\mid}&{0}&{1}&{0}\\{2}&{4}&{1}&{\mid}&{0}&{0}&{1}}\right][/latex-display] To begin, we write the augmented matrix with the identity on the right and A on the left. Performing elementary row operations so that the identity matrix appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example.

### How To

Given a 3 × 3 matrix, find the inverse
1. Write the original matrix augmented with the identity matrix on the right.
2. Use elementary row operations so that
3. the identity appears on the left.
4. What is obtained on the right is the inverse of the original matrix.
5. Use matrix multiplication to show that $\displaystyle{A}{A}^{{-{{1}}}}={1}{\text{and}}{A}^{{-{{1}}}}{A}={I}$.

### Example 5

1. Given the matrix A, find the inverse. $\displaystyle{A}={\left[\matrix{{2}&{3}&{1}\\{3}&{3}&{1}\\{2}&{4}&{1}}\right]$
2. Find the inverse of the matrix. $\displaystyle{A}={\left[\matrix{{2}&-{17}&{11}\\-{1}&{11}&-{7}\\{0}&{3}&-{2}}\right]$

#### Solutions

1. Augment A with the identity matrix, and then begin row operations until the identity matrix replaces A. The matrix on the right will be the inverse of A. $\displaystyle{\left[\matrix{{2}&{3}&{1}&{\mid}&{1}&{0}&{0}\\{3}&{3}&{1}&{\mid}&{0}&{1}&{0}\\{2}&{4}&{1}&{\mid}&{0}&{0}&{1}}\right]{\stackrel{{\text{Interchange } {R}_{{2}}\text{ and } {R}_{{1}}\text{ } }}{\rightarrow}}{\left[\matrix{{3}&{3}&{1}&{\mid}&{0}&{1}&{0}\\{2}&{3}&{1}&{\mid}&{1}&{0}&{0}\\{2}&{4}&{1}&{\mid}&{0}&{0}&{1}}\right]$ Thus,$\displaystyle{A}^{{-{{1}}}}={B}={\left[\matrix{-{1}&{1}&{0}\\-{1}&{0}&{1}\\{6}&-{2}&-{3}}\right]$
2. $\displaystyle{A}^{{-{{1}}}}={\left[\matrix{{1}&{1}&{2}\\{2}&{4}&-{3}\\{3}&{6}&-{5}}\right]$

OpenStax, Precalculus, "Solving Systems with Inverses," licensed under a CC BY 3.0 license.