3.4 Solving Systems with Inverses

Terri Manthey

3.4 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

The multiplicative inverse of a real number $a$ is $a^{-1}$ , where the product is 1: $aa^{-1}=a^{-1}a=\left(\dfrac{1}{a} \right)a=1$ . For example, $2^{-1}=\dfrac{1}{2}2=1$ . and 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.

${I_2} = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\,\,\,\,\,\,\,\qquad{I_3} = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]$

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:

$\begin{array}{l} A{A^{ - 1}} = I\\ {A^{ - 1}}A = I \end{array}$

A matrix that has a multiplicative inverse is called an invertible matrix. Only a square matrix may have a multiplicative inverse, as the reversibility, $AA^{-1}=A^{-1}A=I$ , 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 technology approach that can be used on any size matrix.

The Identity Matrix and the Multiplicative Inverse

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

$\begin{array}{l} {I_2} = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\,\,\,\,\,\,\,\,\,{I_3} = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\\ \,\,\,\quad \quad 2 \times 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad \quad 3 \times 3 \end{array}$

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

Example of the Multiplicative Inverse in Action

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

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

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

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

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

Showing that one Matrix is the multiplicative inverse of the another

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

Example showing two matrices are multiplicative inverses

Show that the given matrices are multiplicative inverses of each other.
$A = \left[ {\begin{array}{*{20}{c}} 1&5\\ { - 2}&{ - 9} \end{array}} \right],\quad B = \left[ {\begin{array}{*{20}{c}} { - 9}&{ - 5}\\ 2&1 \end{array}} \right]$

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

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

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

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

Try it Now 1

Show that the following two matrices are inverses of each other.
$A = \left[ {\begin{array}{*{20}{c}} {\,\,\,1}&{\,\,4}\\ { - 1}&{ - 3} \end{array}} \right],\quad B = \left[ {\begin{array}{*{20}{c}} { - 3}&{ - 4}\\ {\,\,\,1}&{\,\,\,1} \end{array}} \right]$

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

Augment $A$ with the identity
$\left[ {\begin{array}{*{20}{c}} 2&1\\ 5&3 \end{array}\,\,\,\left| {\,\,\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right.} \right]$

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

Divide row 1 by 2.
$\left[ {\begin{array}{*{20}{c}} 1&\frac{1}{2}\\ 5&3 \end{array}\,\,\,\left| {\,\,\,\begin{array}{*{20}{c}} \frac{1}{2}&0\\ 0&1 \end{array}} \right.} \right]$
Multiply row 1 by –5 and add to row 2.
$\left[ {\begin{array}{*{20}{c}} 1&\frac{1}{2}\\ 0&\frac{1}{2} \end{array}\,\,\,\left| {\,\,\,\begin{array}{*{20}{c}} \frac{1 }{2}&0\\ \frac{-5}{2}&1 \end{array}} \right.} \right]$
Multiply row 2 by 2.
$\left[ {\begin{array}{*{20}{c}} 1&\frac{1}{2}\\ 0&1 \end{array}\,\,\,\left| {\,\,\begin{array}{*{20}{c}} \frac{1 }{2}&0\\ -5&{ 2} \end{array}} \right.} \right]$
Multiply row 2 by $-\frac{1}{2}$ and add to row 1
$\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}\,\,\,\left| {\,\,\,\begin{array}{*{20}{c}} 3&{- 1}\\ -5&{2} \end{array}} \right.} \right]$

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

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

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

When we need to find the multiplicative inverse of a 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 $A = \left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right]$

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

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

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

A good way to remember this is that we switch the diagonals of the matrix and make the off-diagonals opposite.

Example Finding the Inverse of a $\color{white}2 \times 2$ Matrix with the Formula

Use the formula to find the multiplicative inverse of
$A = \left[ {\begin{array}{*{20}{c}} 2&1\\ 5&3 \end{array}} \right]$

Using the formula, we have
$\begin{array}{c} {A^{ - 1}} = \frac{1}{{\left( 2 \right)\left( 3 \right) - \left( 5 \right)\left( 1 \right)}}\left[ {\begin{array}{*{20}{c}} 3&{ - 1}\\ { - 5}&2 \end{array}} \right]\\ = \frac{1}{1}\left[ {\begin{array}{*{20}{c}} 3&{ - 1}\\ { - 5}&2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 1}\\ { - 5}&2 \end{array}} \right] \end{array}$

Notice this matches the inverse we found above by augmenting with the identity matrix but was a lot quicker and easier.

Try it Now 2

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

Example of a non-invertible matrix

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

We will use the method of augmenting with the identity.
$\left[ {\begin{array}{*{20}{c}} 3&6\\ 1&3 \end{array}\,\,\,\left| {\,\,\,\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right.} \right]$

Switch row 1 and row 2.
$\left[ {\begin{array}{*{20}{c}} 1&3\\ 3&{6\,} \end{array}\,\,\left| {\,\,\,\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right.} \right]$

Multiply row 1 by and add it to row 2.
$\left[ {\begin{array}{*{20}{c}} 1&2\\ 0&0 \end{array}\,\,\,\left| {\,\,\,\begin{array}{*{20}{c}} 1&0\\ { - 3}&1 \end{array}} \right.} \right]$

Now that we have a row of zeroes on the left, there is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.

We could have also used the $ad-bc$ formula to see this as well. In this case, it would be $3(2)-6(1)=0$ so it is not invertible.

Finding the Multiplicative Inverse of 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:

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

Augment $A$ with the identity matrix

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

To begin, we write the augmented matrix with the identity on the right and 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.

Finding the Inverse of a $\color{white}3 \times 3$ Matrix

Write the original matrix augmented with the identity matrix on the right.
Use elementary row operations so that the identity appears on the left.
What is obtained on the right is the inverse of the original matrix.
Use matrix multiplication to show that $AA^{-1}$ and $A^{-1}A=I$ .

Example Finding the Inverse of a $\color{white} 3 \times 3$ Matrix

Given the $3 \times 3$ matrix $A$ , find the inverse.
$A = \left[ {\begin{array}{*{20}{c}} 2&3&1\\ 3&3&1\\ 2&4&1 \end{array}} \right]$

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 $A$ .

$Switch R_1 and R_2 \rightarrow \left[ {\begin{array}{*{20}{c}} 2&3&1\\ 3&3&1\\ 2&4&1 \end{array}\,\,\,\left| {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ {\,\,0}&0&1 \end{array}} \right.} \right] \longrightarrow \left[ {\begin{array}{*{20}{c}} 3&3&1\\ 2&3&1\\ 2&4&1 \end{array}\,\,\,\left| {\begin{array}{*{20}{c}} 0&1&0\\ 1&0&0\\ {\,\,\,0}&0&1 \end{array}} \right.} \right]$

$- {R_2} + {R_1} = {R_1} \to \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 2&3&1\\ 2&4&1 \end{array}\,\,\,\left| {\,\,\begin{array}{*{20}{c}} { - 1}&1&0\\ 1&0&0\\ 0&0&1 \end{array}} \right.} \right]$

$- {R_2} + {R_3} = {R_3} \to \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 2&3&1\\ 0&1&0 \end{array}\left| {\,\,\begin{array}{*{20}{c}} { - 1}&1&0\\ {\,\,1}&0&0\\ { - 1}&0&1 \end{array}} \right.} \right]$

${R_3}\, \leftrightarrow {\rm{ }}{R_2} \to \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 2&3&1 \end{array}\,\,\left| {\,\,\begin{array}{*{20}{c}} { - 1}&1&0\\ { - 1}&0&1\\ {\,\,\,1}&0&0 \end{array}} \right.} \right]$

$- 2{R_1} + {R_3} = {R_3} \to \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&3&1 \end{array}\,\,\left| {\,\begin{array}{*{20}{c}} { - 1}&1&0\\ { - 1}&0&1\\ 3&{ - 2}&0 \end{array}} \right.} \right]$

$- 3{R_2} + {R_3} = {R_3} \to \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&{\,1\,\,} \end{array}\left| {\,\,\begin{array}{*{20}{c}} { - 1}&{\,\,1}&{\,\,0}\\ { - 1}&{\,\,0}&{\,\,\,1}\\ {\,\,6}&{ - 2}&{ - 3} \end{array}} \right.} \right]$

Thus,

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

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

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

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

This is of course, a difficult and tedious process. On a TI calculator, once your matrix is in the calculator, you can simply call up matrix [A] by 2nd $x^{-1}$ enter. Then you can use the $x^{-1}$ enter key to invert it. Casio has similar functions and Excel has a =MINVERSE command to do this as well.

Try it Now 3

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

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

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:

$AX=B$ .

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.

$\begin{array}{ccccc} {a_1}x + {b_1}y = {c_1}\\ {a_2}x + {b_2}y = {c_2} \end{array}$

From this system, the coefficient matrix is

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

The variable matrix is

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

And the constant matrix is

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

Then looks like

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

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.

First consider the matrix equation: $AX=B$

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

$\begin{array}{c} \left( {{A^{ - 1}}} \right)AX = \left( {{A^{ - 1}}} \right)B\\ \left[ {\left( {{A^{ - 1}}} \right)A} \right]X = \left( {{A^{ - 1}}} \right)B\\ IX = \left( {{A^{ - 1}}} \right)B\\ X = \left( {{A^{ - 1}}} \right)B \end{array}$

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

$AX=B$

If matrix $A$ is invertible, then the solution to this system is:

$X=A^{-1}B$

Warning Sign 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. At this point, we need to use the row-reduction method to determine which it is and if dependent, give the general solution.

Example Solving a $\color{white} 2 \times 2$ System using Inverse

Solve the given system of equations using the inverse of a matrix.
$\begin{array}{c} 3x + 8y = 5\\ 4x + 11y = 7 \end{array}$

Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.
$A = \left[ {\begin{array}{*{20}{c}} 3&8\\ 4&{11} \end{array}} \right],\quad X = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right],\quad B = \left[ {\begin{array}{*{20}{c}} 5\\ 7 \end{array}} \right]$

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

Using the formula to calculate $A^{-1}$ , we have:
$\begin{array}{l} {A^{ - 1}} = \frac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}} d&{ - b}\\ { - c}&a \end{array}} \right]\\ \;\;\;\;\; = \frac{1}{{3(11) - 8(4)}}\left[ {\begin{array}{*{20}{c}} {11}&{ - 8}\\ { - 4}&3 \end{array}} \right]\\ \;\;\;\;\; = \frac{1}{1}\left[ {\begin{array}{*{20}{c}} {11}&{ - 8}\\ { - 4}&3 \end{array}} \right] \end{array}$

${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {11}&{ - 8}\\ { - 4}&{\,\,3} \end{array}} \right]$

Now we are ready to solve. Multiply both sides of the equation by $A^{-1}$ .
$\begin{array}{c} X = \left( {{A^{ - 1}}} \right)B\\ \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {11}&{ - 8}\\ { - 4}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 5\\ 7 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = \left[ \begin{array}{c} 11\left( 5 \right) + \left( { - 8} \right)7\\ - 4\left( 5 \right) + 3\left( 7 \right) \end{array} \right]\\ \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}\\ {\,\,1} \end{array}} \right] \end{array}$

The solution is (-1,1).

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

$\begin{array}{c} \left( {{A^{ - 1}}} \right)AX = \left( {{A^{ - 1}}} \right)B\\ \left[ {\left( {{A^{ - 1}}} \right)A} \right]X = \left( {{A^{ - 1}}} \right)B\\ IX = \left( {{A^{ - 1}}} \right)B\\ X = \left( {{A^{ - 1}}} \right)B \end{array}$

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

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

Solve Systems of Equations using Matrix Inversion in a Calculator

Save the coefficient matrix and the constant matrix as matrix variables $[A]$ and $[B]$ .
Enter the multiplication into the calculator, calling up each matrix variable as needed. $[A]\rightarrow x^{-1} \rightarrow [B]$
If the coefficient matrix, $[A]$ is invertible, the calculator will present the solution matrix; if $[A]$ is not invertible, the calculator will present an error message.

See these videos for this on a TI-83/84 and CASIO.

This can also be accomplished in Excel using =MINVERSE and =MMULT.

Example Solving a System Using Inverses and Technology

Solve the following system using the inverse of a matrix.

$\begin{array}{c} 5x + 15y + 56z = 35\\ - 4x - 11y - 41z = - 26\\ - x - 3y - 11z = - 7 \end{array}$

Write the equation $AX=B$

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

Step 1: Enter the coefficient matrix into [A]:

2nd $x^{-1} \; \rightarrow\, \rightarrow \, EDIT \;Enter$
3 rows and 3 columns and enter in all the values hitting Enter in between each.
2nd Mode to leave the screen.

Step 2: Enter the constant matrix into [B]:

2nd $x^{-1} \; \rightarrow\, \rightarrow \, EDIT \; \downarrow \;Enter$
3 rows and 1 columns and enter in all the values hitting Enter in between each.
2nd Mode to leave the screen.

Step 3: Type in the equation for X

2nd $x^{-1}$ Enter to bring up [A]
$x^{-1}$ to make A inverse
2nd $x^{-1} \downarrow$ to bring up [B] Enter.

The calculator now displays:

$\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ 0 \end{array}} \right]$

So we know the solution is (1,2,0).

Try it Now 4

Solve the system using the inverse of the coefficient matrix.

$\begin{array}{c} 2x - 17y + 11z = 0\\ - x + 11y - 7z = 8\\ 3y - 2z = - 2 \end{array}$

Try it Now Answers

$\begin{array}{l} AB = \left[ {\begin{array}{*{20}{c}} {\,\,1}&{\,4}\\ { - 1}&{ - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 3}&{ - 4}\\ {\,\,\,1}&{\,\,\,1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {1\left( { - 3} \right) + 4\left( 1 \right)}&{1\left( { - 4} \right) + 4\left( 1 \right)}\\ { - 1\left( { - 3} \right) + - 3\left( 1 \right)}&{ - 1\left( { - 4} \right) + - 3\left( 1 \right)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\\ BA = \left[ {\begin{array}{*{20}{c}} { - 3}&{ - 4}\\ {\,\,\,1}&{\,\,\,1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\,\,1}&{\,4}\\ { - 1}&{ - 3} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 3(1) + - 4( - 1)}&{ - 3(4) + - 4( - 3)}\\ {1(1) + 1( - 1)}&{1(4) + 1( - 3)} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] \end{array}$
${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {\frac{3}{5}}&{\frac{1}{5}}\\ { - \frac{2}{5}}&{\frac{1}{5}} \end{array}} \right]$
${A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 1&1&{\,\,2}\\ 2&4&{ - 3}\\ 3&6&{ - 5} \end{array}} \right]$
$X = \left[ {\begin{array}{*{20}{c}} 4\\ {38}\\ {58} \end{array}} \right]$