# 2.4: Solving Systems with Inverses

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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

The multiplicative inverse of a real number $$a$$ is $$a^{-1}$$, where the product is 1:

$aa^{-1} = a^{-1}a = \left( \frac{1}{a} \right)a = 1.\nonumber$

For example, $$2^{-1} = \frac{1}{2}$$ and $$\left( \frac{1}{2} \right)2 = 1$$. 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 $$I_n$$ where $$n$$ represents the dimension of the matrix.  Shown below are the identity matrices for a $$2 \times 2$$ matrix and a $$3 \times 3$$ matrix, respectively.

${I_2} = \left[\begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}\right] \quad {I_3} = \left[\begin{array}{*{20}{r}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}\right] \nonumber$

The identity matrix acts as a 1 in matrix algebra. For example, $$AI = IA = A. \nonumber$$

A matrix that has a multiplicative inverse has the properties

\begin{align*} A{A^{-1}} = I \\ {A^{-1}}A = I \end{align*} \nonumber

A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, $$A{A^{-1}} = {A^{-1}}A = I$$, is a requirement. Not all square matrices have an inverse, but if $$A$$  is invertible, then $$A^{-1}$$ is unique. We will look at two methods for finding the inverse of a $$2 \times 2$$ matrix and a third method that can be used on both $$2 \times 2$$ and $$3 \times 3$$ matrices.

Definition: The Identity Matrix and Multiplicative Inverse

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

\begin{align*} I_2 = \left[\begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}\right] \quad I_3 = \left[\begin{array}{*{20}{r}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}\right] \\ \quad \quad 2 \times 2 \quad \quad 3 \times 3 \end{align*} \nonumber

If $$A$$ is an $$n \times n$$ matrix and $$B$$ is an $$n \times n$$ matrix such that $$AB = BA = {I_n}$$, then $$B = A^{-1}$$, the multiplicative inverse of a matrix $$A.$$

Example $$\PageIndex{1}$$

Given matrix A, show that $$AI = IA = A.$$

$A = \left[\begin{array}{*{20}{r}} { 3}&4 \\ -2&5 \end{array}\right] \nonumber$

Solution

Use matrix multiplication to show that the product of $$A$$ and the identity is equal to the product of the identity and $$A$$.

$AI = \left[\begin{array}{*{20}{r}} 3&4 \\ -2&5 \end{array}\right]\left[\begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}\right] = \left[\begin{array}{*{20}{r}} {3 \cdot 1 + 4 \cdot 0}&{3 \cdot 0 + 4 \cdot 1} \\ {-2 \cdot 1 + 5 \cdot 0}&{-2 \cdot 0 + 5 \cdot 1} \end{array}\right] = \left[\begin{array}{*{20}{r}} { 3}&4 \\ -2&5 \end{array}\right] \nonumber$

$IA = \left[\begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}\right]\left[\begin{array}{*{20}{r}} { 3}&4 \\ -2&5 \end{array}\right] = \left[\begin{array}{*{20}{r}} {1 \cdot 3 + 0 \cdot (-2)}&{ 1 \cdot 4 + 0 \cdot 5} \\ {0 \cdot 3 + 1 \cdot (-2)}&{ 0 \cdot 4 + 1 \cdot 5} \end{array}\right] = \left[\begin{array}{*{20}{r}} { 3}&4 \\ -2&5 \end{array}\right] \nonumber$

Given two matrices, show that one is the multiplicative inverse of the other.

1. Given matrix $$A$$ of order $$n \times n$$ and matrix $$B$$ of order $$n \times n$$, multiply $$AB.$$
2. If $$AB = I$$, then find the product $$BA$$. If $$BA = I$$, then $$B = A^{-1}$$ and $$A = B^{-1}$$.

Example $$\PageIndex{2}$$

Show that the given matrices are multiplicative inverses of each other.

$A = \left[\begin{array}{rr} 1&5 \\ -2&-9 \end{array}\right]\quad B = \left[\begin{array}{rr} -9&-5 \\ 2&1 \end{array}\right] \nonumber$

Solution

Multiply $$AB$$ and $$BA$$. If both products equal the identity, then the two matrices are inverses of each other.

\begin{align*} AB &= \left[\begin{array}{*{20}{r}} 1&5 \\ -2&-9 \end{array}\right] \cdot \left[\begin{array}{*{20}{r}} -9&-5 \\ 2&1 \end{array}\right] \\ &= \left[\begin{array}{*{20}{r}} {1\left( -9 \right) + 5\left( 2 \right)}&{1\left( -5 \right) + 5\left( 1 \right)} \\ {-2\left( -9 \right) - 9\left( 2 \right)}&{-2\left( -5 \right) - 9\left( 1 \right)} \end{array}\right] \\ &= \left[\begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}\right] \\ BA &= \left[\begin{array}{*{20}{r}} -9&-5 \\ 2&1 \end{array}\right] \cdot \left[\begin{array}{*{20}{r}} 1&5 \\ -2&-9 \end{array}\right] \\ &= \left[\begin{array}{*{20}{r}} { - 9\left( 1 \right) - 5\left( -2 \right)}&{ - 9\left( 5 \right) - 5\left( -9 \right)} \\ {2\left( 1 \right) + 1\left( -2 \right)}&{2\left( -5 \right) + 1\left( -9 \right)} \end{array}\right] \\ &= \left[\begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}\right] \end{align*} \nonumber

$$A$$ and $$B$$ are inverses of each other.

Exercise $$\PageIndex{1}$$

Show that the following two matrices are inverses of each other.

$A = \left[\begin{array}{*{20}{r}} 1&4 \\ -1&-3 \end{array}\right],\quad B = \left[\begin{array}{*{20}{r}} -3&-4 \\ 1&1 \end{array}\right] \nonumber$

\begin{align*} AB &= \left[\begin{array}{*{20}{r}} 1&4 \\ -1&-3 \end{array}\right]\left[\begin{array}{*{20}{r}} -3&-4 \\ 1&1 \end{array}\right] = \left[\begin{array}{*{20}{r}} {1\left( -3 \right) + 4\left( 1 \right)}&{1\left( -4 \right) + 4\left( 1 \right)} \\ {-1\left( -3 \right) + - 3\left( 1 \right)}&{-1\left( -4 \right) + - 3\left( 1 \right)} \end{array}\right] = \left[\begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}\right] \\ BA &= \left[\begin{array}{*{20}{r}} -3&-4 \\ 1&1 \end{array}\right]\left[\begin{array}{*{20}{r}} 1&4 \\ -1&-3 \end{array}\right] = \left[\begin{array}{*{20}{r}} { - 3(1) + - 4(-1)}&{ - 3(4) + - 4( - 3)} \\ {1(1) + 1(-1)}&{1(4) + 1( - 3)} \end{array}\right] = \left[\begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}\right] \end{align*} \nonumber

Finding the Multiplicative Inverse by Augmenting with the Identity

One 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

$A = \left[\begin{array}{*{20}{r}} 2&1 \\ 5&3 \end{array}\right] \nonumber$

augment $$A$$ with the identity

$\left[\begin{array}{*{20}{r}} 2&1 \\ 5&3 \end{array} \left| { \begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}} \right.\right] \nonumber$

Perform row operations with the goal of turning $$A$$ into the identity.

1.      Switch row 1 and row 2.

$\left[\begin{array}{*{20}{r}} 5&3 \\ 2&1 \end{array} \left| { \begin{array}{*{20}{r}} 0&1 \\ 1&0 \end{array}} \right.\right] \nonumber$

2.      Multiply row 2 by –2 and add to row 1.

$\left[\begin{array}{*{20}{r}} 1&1 \\ 2&1 \end{array} \left| { \begin{array}{*{20}{r}} -2&1 \\ 1&0 \end{array}} \right.\right] \nonumber$

3.      Multiply row 1 by –2 and add to row 2.

$\left[\begin{array}{*{20}{r}} 1&1 \\ 0&-1 \end{array} \left| { \begin{array}{*{20}{r}} -2&1 \\ 5&-2 \end{array}} \right.\right] \nonumber$

4.      Add row 2 to row 1.

$\left[\begin{array}{*{20}{r}} 1&0 \\ 0&-1 \end{array} \left| { \begin{array}{*{20}{r}} 3&-1 \\ 5&-2 \end{array}} \right.\right] \nonumber$

5.      Multiply row 2 by –1.

$\left[\begin{array}{*{20}{r}} 1&{ 0} \\ 0&1 \end{array} \left| { \begin{array}{*{20}{r}} 3&-1 \\ -5&2 \end{array}} \right.\right] \nonumber$

The matrix we have found is $$A^{-1}$$.

$A^{-1} = \left[\begin{array}{*{20}{r}} 3&-1 \\ -5&2 \end{array}\right] \nonumber$

Finding the Multiplicative Inverse of $$2 \times 2$$ Matrices Using a Formula

When we need to find the multiplicative inverse of a $$2 \times 2$$ matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.

Formula for the Inverse of a 2 by 2 Matrix

If $$A$$ is a $$2 \times 2$$ matrix, such as $A = \left[\begin{array}{*{20}{r}} a&b \\ c&d \end{array}\right] \nonumber$

the multiplicative inverse of $$A$$ is given by the formula

$A^{-1} = \frac{1}{ad - bc}\left[\begin{array}{*{20}{r}} d&{ - b} \\ { - c}&a \end{array}\right] \nonumber$

where $$ad - bc \ne 0.$$ If $$ad - bc = 0,$$ then $$A$$ has no inverse.

Example $$\PageIndex{3}$$

Use the formula to find the multiplicative inverse of

$A = \left[\begin{array}{*{20}{r}} 2&1 \\ 5&3 \end{array}\right] \nonumber$

Solution

Using the formula, we have

\begin{align*} A^{-1} &= \frac{1}{\left( 2 \right)\left( 3 \right) - \left( 5 \right)\left( 1 \right)}\left[\begin{array}{*{20}{r}} 3&-1 \\ -5&2 \end{array}\right] \\ &= \frac{1}{1}\left[\begin{array}{*{20}{r}} 3&-1 \\ -5&2 \end{array}\right] \\ &= \left[\begin{array}{*{20}{r}} 3&-1 \\ -5&2 \end{array}\right] \end{align*} \nonumber

Notice this matches the inverse we found above by augmenting with the identity matrix.

Exercise $$\PageIndex{2}$$

Use the formula to find the inverse of matrix $A = \left[\begin{array}{*{20}{r}} 1&-1 \\ 2&3 \end{array}\right] \nonumber$

$A^{-1} = \left[\begin{array}{*{20}{r}} {\frac{3}{5}}&{\frac{1}{5}} \\ { - \frac{2}{5}}&{\frac{1}{5}} \end{array}\right]$

Example $$\PageIndex{4}$$

Find the inverse, if it exists, of the given matrix:  $A = \left[\begin{array}{*{20}{r}} 3&6 \\ 1&2 \end{array}\right] \nonumber$

Solution

We will use the method of augmenting with the identity.

$\left[\begin{array}{*{20}{r}} 3&6 \\ 1&2 \end{array} \left| { \begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}} \right.\right] \nonumber$

Switch row 1 and row 2.

$\left[\begin{array}{*{20}{r}} 1&2 \\ 3&{6 } \end{array} \left| { \begin{array}{*{20}{r}} 0&1 \\ 1&0 \end{array}} \right.\right] \nonumber$

Multiply row 1 by $$- 3$$ and add it to row 2.

$\left[\begin{array}{*{20}{r}} 1&2 \\ 0&0 \end{array} \left| { \begin{array}{*{20}{r}} 0&0 \\ 1&-3 \end{array}} \right.\right] \nonumber$

There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.

Finding the Multiplicative Inverse of $$3 \times 3$$ Matrices

Unfortunately, we do not have a formula similar to the one for a $$2 \times 2$$ matrix to find the inverse of a $$3 \times 3$$ matrix. Instead, we will augment the original matrix with the identity matrix and use row operations to obtain the inverse.

Given a $$3 \times 3$$ matrix

$A = \left[\begin{array}{*{20}{r}} 2&3&1 \\ 3&3&1 \\ 2&4&1 \end{array}\right] \nonumber$

augment $$A$$ with the identity matrix

$A\left| I \right. = \left[\begin{array}{*{20}{r}} 2&3&1 \\ 3&3&1 \\ 2&4&1 \end{array} \left| { \begin{array}{*{20}{r}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right.\right] \nonumber$

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.

Given a 3 by 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 the identity appears on the left.
3. What is obtained on the right is the inverse of the original matrix.
4. Use matrix multiplication to show that $$AA^{-1} = I$$ and $$A^{-1}A = I.$$

Example $$\PageIndex{5}$$

Given the $$3 \times 3$$ matrix $$A$$, find the inverse.

$A = \left[\begin{array}{*{20}{r}} 2&3&1 \\ 3&3&1 \\ 2&4&1 \end{array}\right] \nonumber$

Solution

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 $$A$$.

$\left[\begin{array}{ccc|ccc} 2&3&1&1&0&0 \\ 3&3&1&0&1&0 \\ 2&4&1&0&0&1\end{array} \right]\xrightarrow{\text{ Interchange } R_2 \text{ and } R_1}\left[\begin{array}{ccc|ccc} 3&3&1& 0&1&0 \\ 2&3&1& 1&0&0 \\ 2&4&1& 0&0&1 \end{array}\right] \nonumber$

\begin{align*} - {R_2} + {R_1} = {R_1} &\to \left[\begin{array}{*{20}{r}} 1&0&0 \\ 2&3&1 \\ 2&4&1 \end{array} \left| { \begin{array}{*{20}{r}} -1&1&0 \\ 1&0&0 \\ 0&0&1 \end{array}} \right.\right] \\ - {R_2} + {R_3} = {R_3} &\to \left[\begin{array}{*{20}{r}} 1&0&0 \\ 2&3&1 \\ 0&1&0 \end{array}\left| { \begin{array}{*{20}{r}} -1&1&0 \\ 1&0&0 \\ -1&0&1 \end{array}} \right.\right] \\ {R_3} \leftrightarrow {R_2} &\to \left[\begin{array}{*{20}{r}} 1&0&0 \\ 0&1&0 \\ 2&3&1 \end{array} \left| { \begin{array}{*{20}{r}} -1&1&0 \\ -1&0&1 \\ 1&0&0 \end{array}} \right.\right] \\ -2{R_1} + {R_3} = {R_3} &\to \left[\begin{array}{*{20}{r}} 1&0&0 \\ 0&1&0 \\ 0&3&1 \end{array} \left| { \begin{array}{*{20}{r}} -1&1&0 \\ -1&0&1 \\ 3&-2&0 \end{array}} \right.\right] \\ - 3{R_2} + {R_3} = {R_3} &\to \left[\begin{array}{*{20}{r}} 1&0&0 \\ 0&1&0 \\ 0&0&{ 1 } \end{array}\left| { \begin{array}{*{20}{r}} -1&1&{ 0} \\ -1&{ 0}&1 \\ { 6}&-2&-3 \end{array}} \right.\right] \end{align*}

Thus,

$A^{-1} = B = \left[\begin{array}{*{20}{r}} -1&{ 1}&{ 0} \\ -1&{ 0}&1 \\ { 6}&-2&-3 \end{array} \right]. \nonumber$

To prove that $$B=A^{-1}$$, multiply the two matrices together to see if the product equals the identity, if $$AA^{-1}=I$$ and $$A^{-1}A=I$$.

\begin{align*} AA^{-1} &= \left[\begin{array}{*{20}{r}} 2&3&1 \\ 3&3&1 \\ 2&4&1 \end{array}\right]\left[ \begin{array}{*{20}{r}} -1&1&0 \\ -1&{ 0}&{ 1} \\ 6&-2&-3 \end{array}\right] \\ &= \left[\begin{array}{*{20}{r}} 2( -1 ) + 3( -1 ) + 1( 6 )&2( 1 ) + 3( 0 ) + 1( -2 )&2( 0 ) + 3( 1 ) + 1( -3 ) \\ 3( -1 ) + 3( -1 ) + 1( 6 )&3( 1 ) + 3( 0 ) + 1( -2 )&3( 0 ) + 3( 1 ) + 1( -3 ) \\ 2( -1 ) + 4( -1 ) + 1( 6 )&2( 1 ) + 4( 0 ) + 1( -2 )&2( 0 ) + 4( 1 ) + 1( -3 ) \end{array}\right] \\ &= \left[\begin{array}{*{20}{r}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}\right] \end{align*}

\begin{align*} A^{-1}A &= \left[ \begin{array}{*{20}{r}} -1&1&0 \\ -1&0&1 \\ 6&-2&-3 \end{array}\right]\left[\begin{array}{*{20}{c}} 2&3&1 \\ 3&3&1 \\ 2&4&1 \end{array}\right] \\ &= \left[\begin{array}{*{20}{r}} -1(2) + 1(3) + 0(2)&-1(3) + 1(3) + 0(4)&-1(1) + 1(1) + 0(1) \\ -1(2) + 0(3) + 1(2)&-1(3) + 0(3) + 1(4)&-1(1) + 0(1) + 1(1) \\ 6(2) + -2(3) + -3(2)&6(3) + -2(3) + -3(4)&6(1) + -2(1) + -3(1) \end{array}\right] \\ &= \left[\begin{array}{*{20}{r}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}\right] \end{align*}

Exercise $$\PageIndex{3}$$

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

$A = \left[\begin{array}{*{20}{r}} { 2}&{-17}&{11} \\ -1&{ 11}&-7 \\ { 0}&{ 3}&-2 \end{array}\right] \nonumber$

$A^{ - 1} = \left[\begin{array}{*{20}{r}} 1&1&2 \\ 2&4&-3 \\ 3&6&{ - 5} \end{array}\right]\nonumber$

Solving a System of Linear Equations Using the Inverse of a Matrix

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: $$X$$ is the matrix representing the variables of the system, and $$B$$ is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as

$AX = B \nonumber$

To solve a system of linear equations using an inverse matrix, let $$A$$ be the coefficient matrix, let $$X$$ be the variable matrix, and let $$B$$ be the constant matrix. Thus, we want to solve a system $$AX = B$$. For example, look at the following system of equations.

\begin{align*} a_1x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{align*} \nonumber

From this system, the coefficient matrix is

$A = \left[\begin{array}{*{20}{r}} a_1&b_1 \\ a_2&b_2 \end{array}\right] \nonumber$

The variable matrix is

$X = \left[\begin{array}{*{20}{r}} x \\ y \end{array}\right] \nonumber$

And the constant matrix is

$B = \left[\begin{array}{*{20}{r}} c_1 \\ c_2 \end{array}\right] \nonumber$

Then $$AX = B$$ looks like

$\left[\begin{array}{*{20}{r}} a_1&b_1 \\ a_2&b_2 \end{array}\right]\left[\begin{array}{*{20}{r}} x \\ y \end{array}\right] = \left[\begin{array}{*{20}{r}} c_1 \\ c_2 \end{array}\right] \nonumber$

Solving this system of equations written in matrix form is similar to solving a linear equation, except there is no matrix “division” so instead we need to use the inverse. The goal is the same—to isolate the variable.  We will investigate this idea in detail, but it is helpful to begin with a $$2 \times 2$$ system and then move on to a $$3 \times 3$$ system.

Solving a System of Equations Using the Inverse of a Matrix

Given a system of equations, write the coefficient matrix $$A$$, the variable matrix $$X$$, and the constant matrix $$B$$. Then

$AX = B \nonumber$

Multiply both sides by the inverse of $$A$$ to obtain the solution.

\begin{align*} \left( A^{-1} \right)AX &= \left( A^{-1} \right)B \\ \left[\left( A^{-1} \right)A\right]X &= \left( A^{-1} \right)B \\ IX &= \left( A^{-1} \right)B \\ X &= \left( A^{-1} \right)B \end{align*}

Important:  If the coefficient matrix does not have an inverse, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.

Example $$\PageIndex{6}$$

Solve the given system of equations using the inverse of a matrix.

\begin{align*} 3x + 8y = 5 \\ 4x + 11y = 7 \end{align*} \nonumber

Solution

Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.

$A = \left[\begin{array}{*{20}{r}} 3&8 \\ 4&{11} \end{array}\right],\quad X = \left[\begin{array}{*{20}{r}} x \\ y \end{array}\right],\quad B = \left[\begin{array}{*{20}{r}} 5 \\ 7 \end{array}\right] \nonumber$

Then

$\left[\begin{array}{*{20}{r}} 3&8 \\ 4&{11} \end{array}\right]\left[\begin{array}{*{20}{r}} x \\ y \end{array}\right] = \left[\begin{array}{*{20}{r}} 5 \\ 7 \end{array}\right] \nonumber$

Using the formula to calculate $$A^{-1}$$, we have:

\begin{align*} A^{-1} &= \frac{1}{ad - bc}\left[\begin{array}{*{20}{r}} d&-b \\ -c&a \end{array}\right] \\ &= \frac{1}{3(11) - 8(4)}\left[\begin{array}{*{20}{r}} {11}&-8 \\ -4&3 \end{array}\right] \\ &= \frac{1}{1}\left[\begin{array}{*{20}{r}} {11}&-8 \\ -4&3 \end{array}\right] \\ A^{-1} &= \left[\begin{array}{*{20}{r}} 11&-8 \\ -4&3 \end{array}\right] \end{align*}

Now we are ready to solve. Multiply both sides of the equation by $$A^{-1}$$.

\begin{align*} \left( A^{-1} \right)AX &= \left( A^{-1} \right)B \\ \left[\begin{array}{*{20}{r}} 11&-8 \\ -4&3 \end{array}\right]\left[\begin{array}{*{20}{r}} 3&8 \\ 4&11 \end{array}\right]\left[\begin{array}{*{20}{r}} x \\ y \end{array}\right] &= \left[\begin{array}{*{20}{r}} {11}&-8 \\ -4&3 \end{array}\right]\left[\begin{array}{*{20}{r}} 5 \\ 7 \end{array}\right] \\ \left[\begin{array}{*{20}{r}} 1&0 \\ 0&1 \end{array}\right]\left[\begin{array}{*{20}{r}} x \\ y \end{array}\right] &= \left[\begin{array}{c} 11(5) + (-8)7 \\ - 4(5) + 3(7)\end{array}\right] \\ \left[\begin{array}{*{20}{r}} x \\ y \end{array}\right] &= \left[\begin{array}{*{20}{r}} -1 \\ 1 \end{array}\right]. \end{align*}

The solution is $$(-1,1)$$.

Note that you cannot solve for $$X$$ by finding the product $$BA^{-1}$$.  Recall that matrix multiplication is not commutative, so $$A^{-1}B \ne B.$$ Consider our steps for solving the matrix equation.

\begin{align*} \left( A^{-1} \right)AX = \left( A^{-1} \right)B \\ \left[\left( A^{-1} \right)A\right]X = \left( A^{-1} \right)B \\ IX = \left( A^{-1} \right)B \\ X = \left( A^{-1} \right)B \end{align*} \nonumber

Notice in the first step we multiplied both sides of the equation by $$A^{-1}$$, but the $$A^{-1}$$ was to the left of $$A$$ on the left side and to the left of $$B$$ on the right side. Because matrix multiplication is not commutative, order matters.

Example $$\PageIndex{7}$$

Solve the following system using the inverse of a matrix.

\begin{align*} 5x + 15y + 56z = 35 \\ - 4x-11y - 41z = -26 \\ - x - 3y-11z = - 7 \end{align*} \nonumber

Solution

Write the equation $$AX = B$$.

$\left[\begin{array}{*{20}{r}} 5&{15}&{56} \\ -4&{-11}&{ - 41} \\ -1&-3&{-11} \end{array}\right]\left[\begin{array}{*{20}{r}} x \\ y \\ z \end{array}\right] = \left[\begin{array}{*{20}{r}} {35} \\ {-26} \\ -7 \end{array}\right] \nonumber$

First, we will find the inverse of $$A$$ by augmenting with the identity.

$\left[\begin{array}{*{20}{r}} 5&{15}&{56} \\ -4&{-11}&{ - 41} \\ -1&-3&{-11} \end{array} \left| { \begin{array}{*{20}{r}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right.\right] \nonumber$

Multiply row 1 by $$\frac{1}{5}$$.

$\left[\begin{array}{*{20}{r}} 1&3&{\frac{56}{5}} \\ -4&{-11}&{ - 41} \\ -1&-3&{-11} \end{array} \left| { \begin{array}{*{20}{r}} {\frac{1}{5}}&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right.\right] \nonumber$

Multiply row 1 by 4 and add to row 2.

$\left[\begin{array}{*{20}{r}} 1&3&{\frac{56}{5}} \\ 0&1&{\frac{19}{5}} \\ -1&-3&{-11} \end{array} \left| { \begin{array}{*{20}{r}} {\frac{1}{5}}&0&0 \\ {\frac{4}{5}}&1&0 \\ 0&0&1 \end{array}} \right.\right] \nonumber$

Add row 1 to row 3.

$\left[\begin{array}{*{20}{r}} 1&3&\frac{56}{5} \\ 0&1&\frac{19}{5}\\ 0&0&\frac{1}{5} \end{array} \left| { \begin{array}{*{20}{r}} \frac{1}{5}&0&0 \\ \frac{4}{5}&1&0 \\ \frac{1}{5}&0&1 \end{array}} \right.\right] \nonumber$

Multiply row 2 by −3 and add to row 1.

$\left[\begin{array}{*{20}{r}} 1&0&-\frac{1}{5} \\ 0&1&\frac{19}{5} \\ 0&0&{\frac{1}{5}} \end{array} \left| { \begin{array}{*{20}{r}} -\frac{11}{5}&-3&0 \\ \frac{4}{5}&1&0 \\ \frac{1}{5}&0&1 \end{array}} \right.\right] \nonumber$

Multiply row 3 by 5.

$\left[\begin{array}{*{20}{r}} 1&0&{ - \frac{1}{5}} \\ 0&1&\frac{19}{5}\\ 0&0&1 \end{array} \left| { \begin{array}{*{20}{r}} -\frac{11}{5}&-3&0 \\ \frac{4}{5}&1&0 \\ 1&0&5 \end{array}} \right.\right] \nonumber$

Multiply row 3 by $$\frac{1}{5}$$ and add to row 1.

$\left[\begin{array}{*{20}{r}} 1&0&0 \\ 0&1&\frac{19}{5} \\ 0&0&1 \end{array} \left| { \begin{array}{*{20}{r}} -2&-3&1 \\ \frac{4}{5}&1&0 \\ 1&0&5 \end{array}} \right.\right] \nonumber$

Multiply row 3 by $$-\frac{19}{5}$$  and add to row 2.

$\left[\begin{array}{*{20}{r}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array} \left| { \begin{array}{*{20}{r}} -2&-3&1 \\ -3&1&{-19} \\ 1&0&5 \end{array}} \right.\right] \nonumber$

So,

$A^{-1}= \left[\begin{array}{*{20}{r}} -2&-3&1 \\ -3&1&{-19} \\ 1&0&5 \end{array}\right]. \nonumber$

Multiply both sides of the equation by $$A^{-1}$$. We want $$A^{-1}AX = B$$:

$\left[\begin{array}{*{20}{r}} -2& -3&1 \\ - 3&1&-19 \\ 1&0&5 \end{array}\right]\left[\begin{array}{*{20}{r}} 5&15&56 \\ -4&{-11}&{ - 41} \\ -1&-3&{-11} \end{array}\right]\left[\begin{array}{*{20}{r}} x \\ y \\ z \end{array}\right] = \left[\begin{array}{*{20}{r}} -2&-3&1 \\ -3&1&{-19} \\ 1&0&5 \end{array}\right]\left[\begin{array}{*{20}{r}} {35} \\ {-26} \\ -7 \end{array}\right]. \nonumber$

Thus,

$A^{-1}B = \left[\begin{array}{*{20}{c}} - 70 + 78 - 7 \\ -105-26 + 133 \\ 35 + 0 - 35 \end{array}\right] = \left[\begin{array}{*{20}{r}} 1 \\ 2 \\ 0 \end{array}\right]. \nonumber$

The solution is $$(1, 2, 0)$$.

Exercise $$\PageIndex{4}$$

Solve the system using the inverse of the coefficient matrix.

\begin{align*} 2x-17y + 11z &= 0 \\ - x + 11y - 7z &= 8 \\ 3y-2z &= -2 \end{align*} \nonumber

$X = \left[\begin{array}{*{20}{r}} 4 \\ 38 \\ 58 \end{array}\right], \quad \text{so} \quad x=4, y= 38, z = 58 \nonumber$

Systems of equations in many variables can get very complex, but solving using matrices can easily be handled by calculators or computers.

Given a system of equations, solve with matrix inverses using a caluclator

1. Save the coefficient matrix and the constant matrix as matrix variables $$\left[ A \right]$$ and $$\left[ B \right].$$
2. Enter the multiplication into the calculator, calling up each matrix variable as needed.
3. If the coefficient matrix is invertible, the calculator will present the solution matrix; if the coefficient matrix is not invertible, the calculator will present an error message.

Important Topics of this Section

Finding the inverse of a matrix:

Using the formula for 2 by 2 matrices

By augmenting with the identity and row reducing

Using the inverse to solve a system of equations

2.4: Solving Systems with Inverses is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.