# 7.7 Solving systems with inverses  (Page 3/8)

 Page 3 / 8

Use the formula to find the inverse of matrix $\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$ Verify your answer by augmenting with the identity matrix.

$A=\left[\begin{array}{cc}1& -1\\ 2& \text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]$

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

## Finding the inverse of the matrix, if it exists

Find the inverse, if it exists, of the given matrix.

$A=\left[\begin{array}{cc}3& 6\\ 1& 2\end{array}\right]$

We will use the method of augmenting with the identity.

$\left[\begin{array}{cc}3& 6\\ 1& 3\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
1. Switch row 1 and row 2.
$\left[\begin{array}{cc}1& 3\\ 3& 6\text{\hspace{0.17em}}\text{​}\end{array}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{​}\text{​}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
2. Multiply row 1 by −3 and add it to row 2.
$\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{cc}1& 0\\ -3& 1\end{array}\right]$
3. 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 $\text{\hspace{0.17em}}2\text{}×\text{}2\text{\hspace{0.17em}}$ matrix to find the inverse of a $\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$ matrix. Instead, we will augment the original matrix with the identity matrix and use row operations    to obtain the inverse.

Given a $\text{\hspace{0.17em}}3\text{}×\text{}3\text{\hspace{0.17em}}$ matrix

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

augment $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ with the identity matrix

To begin, we write the augmented matrix    with the identity on the right and $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}A{A}^{-1}=I\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{A}^{-1}A=I.$

## Finding the inverse of a 3 × 3 matrix

Given the $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$ matrix $\text{\hspace{0.17em}}A,\text{\hspace{0.17em}}$ find the inverse.

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

Augment $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ with the identity matrix, and then begin row operations until the identity matrix replaces $\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$ The matrix on the right will be the inverse of $\text{\hspace{0.17em}}A.\text{\hspace{0.17em}}$

$-{R}_{2}+{R}_{1}={R}_{1}\to \left[\begin{array}{ccc}1& 0& 0\\ 2& 3& 1\\ 2& 4& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]$
$-{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 2& 3& 1\\ 0& 1& 0\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\end{array}\right]$
$-2{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 3& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 3& \hfill -2& \hfill 0\end{array}\right]$
$-3{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{rrr}\hfill -1& \hfill 1& \hfill 0\\ \hfill -1& \hfill 0& \hfill 1\\ \hfill 6& \hfill -2& \hfill -3\end{array}\right]$

Thus,

${A}^{-1}=B=\left[\begin{array}{ccc}-1& \text{\hspace{0.17em}}1& \text{\hspace{0.17em}}0\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}6& -2& -3\end{array}\text{\hspace{0.17em}}\right]$

Find the inverse of the $\text{\hspace{0.17em}}3×3\text{\hspace{0.17em}}$ matrix.

$A=\left[\begin{array}{ccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2& -17& 11\\ -1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}11& -7\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3& -2\end{array}\right]$

${A}^{-1}=\left[\begin{array}{ccc}1& 1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ 2& 4& -3\\ 3& 6& -5\end{array}\right]$

## 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: $\text{\hspace{0.17em}}X\text{\hspace{0.17em}}$ is the matrix representing the variables of the system, and $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ be the coefficient matrix    , let $\text{\hspace{0.17em}}X\text{\hspace{0.17em}}$ be the variable matrix, and let $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ be the constant matrix. Thus, we want to solve a system $\text{\hspace{0.17em}}AX=B.\text{\hspace{0.17em}}$ For example, look at the following system of equations.

$\begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}$

From this system, the coefficient matrix is

$A=\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]$

The variable matrix is

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

And the constant matrix is

$B=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right]$

Then $\text{\hspace{0.17em}}AX=B\text{\hspace{0.17em}}$ looks like

Recall the discussion earlier in this section regarding multiplying a real number by its inverse, $\text{\hspace{0.17em}}\left({2}^{-1}\right)\text{\hspace{0.17em}}2=\left(\frac{1}{2}\right)\text{\hspace{0.17em}}2=1.\text{\hspace{0.17em}}$ To solve a single linear equation $\text{\hspace{0.17em}}ax=b\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ Thus,

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