# 7.7 Solving systems with inverses  (Page 2/8)

 Page 2 / 8

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

$A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right],B=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]$
$\begin{array}{l}AB=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]=\left[\begin{array}{rrr}\hfill 1\left(-3\right)+4\left(1\right)& \hfill & \hfill 1\left(-4\right)+4\left(1\right)\\ \hfill -1\left(-3\right)+-3\left(1\right)& \hfill & \hfill -1\left(-4\right)+-3\left(1\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \\ BA=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill -4\\ \hfill 1& \hfill & \hfill 1\end{array}\right]\begin{array}{r}\hfill \end{array}\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 4\\ \hfill -1& \hfill & \hfill -3\end{array}\right]=\left[\begin{array}{rrr}\hfill -3\left(1\right)+-4\left(-1\right)& \hfill & \hfill -3\left(4\right)+-4\left(-3\right)\\ \hfill 1\left(1\right)+1\left(-1\right)& \hfill & \hfill 1\left(4\right)+1\left(-3\right)\end{array}\right]=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill 0\\ \hfill 0& \hfill & \hfill 1\end{array}\right]\hfill \end{array}$

## 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 .

## Finding the multiplicative inverse using matrix multiplication

Use matrix multiplication to find the inverse of the given matrix.

$A=\left[\begin{array}{rrr}\hfill 1& \hfill & \hfill -2\\ \hfill 2& \hfill & \hfill -3\end{array}\right]$

For this method, we multiply $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by a matrix containing unknown constants and set it equal to the identity.

Find the product of the two matrices on the left side of the equal sign.

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.

Using row operations, multiply and add as follows: $\text{\hspace{0.17em}}\left(-2\right){R}_{1}+{R}_{2}\to {R}_{2}.\text{\hspace{0.17em}}$ Add the equations, and solve for $\text{\hspace{0.17em}}c.$

$\begin{array}{r}\hfill 1a-2c=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill 0+1c=-2\\ \hfill c=-2\end{array}$

Back-substitute to solve for $\text{\hspace{0.17em}}a.$

$\begin{array}{r}\hfill a-2\left(-2\right)=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill a+4=1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \hfill a=-3\end{array}$

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.

$\begin{array}{rr}\hfill 1b-2d=0& \hfill {R}_{1}\\ \hfill 2b-3d=1& \hfill {R}_{2}\end{array}$

Using row operations, multiply and add as follows: $\text{\hspace{0.17em}}\left(-2\right){R}_{1}+{R}_{2}={R}_{2}.\text{\hspace{0.17em}}$ Add the two equations and solve for $\text{\hspace{0.17em}}d.$

$\begin{array}{r}\hfill 1b-2d=0\\ \hfill \frac{0+1d=1}{\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}d=1}\\ \hfill \end{array}$

Once more, back-substitute and solve for $\text{\hspace{0.17em}}b.$

$\begin{array}{r}\hfill b-2\left(1\right)=0\\ \hfill b-2=0\\ \hfill b=2\end{array}$
${A}^{-1}=\left[\begin{array}{rrr}\hfill -3& \hfill & \hfill 2\\ \hfill -2& \hfill & \hfill 1\end{array}\right]$

## Finding the multiplicative inverse by augmenting with the identity

Another way to find the multiplicative inverse is by augmenting with the identity. When matrix $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is transformed into $\text{\hspace{0.17em}}I,\text{\hspace{0.17em}}$ the augmented matrix $\text{\hspace{0.17em}}I\text{\hspace{0.17em}}$ transforms into $\text{\hspace{0.17em}}{A}^{-1}.$

For example, given

$A=\left[\begin{array}{rrr}\hfill 2& \hfill & \hfill 1\\ \hfill 5& \hfill & \hfill 3\end{array}\right]$

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

Perform row operations    with the goal of turning $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ into the identity.

1. Switch row 1 and row 2.
2. Multiply row 2 by $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and add to row 1.
3. Multiply row 1 by $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and add to row 2.
4. Add row 2 to row 1.
5. Multiply row 2 by $\text{\hspace{0.17em}}-1.$

The matrix we have found is $\text{\hspace{0.17em}}{A}^{-1}.$

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

## Finding the multiplicative inverse of 2×2 matrices using a formula

When we need to find the multiplicative inverse of a $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.

If $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is a $\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$ matrix, such as

$A=\left[\begin{array}{rrr}\hfill a& \hfill & \hfill b\\ \hfill c& \hfill & \hfill d\end{array}\right]$

the multiplicative inverse of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is given by the formula

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

where $\text{\hspace{0.17em}}ad-bc\ne 0.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}ad-bc=0,\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ has no inverse.

## Using the formula to find the multiplicative inverse of matrix A

Use the formula to find the multiplicative inverse of

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

Using the formula, we have

$\begin{array}{l}{A}^{-1}=\frac{1}{\left(1\right)\left(-3\right)-\left(-2\right)\left(2\right)}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{-3+4}\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right]\hfill \end{array}$

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