Lesson Objectives
• Learn how to find the inverse using the adjoint and determinant

## How to Find the Inverse Using the Adjoint and the Determinant

In this lesson, we will learn how to find the inverse of a matrix using the determinant and its adjoint. At this point, we are ready to talk about an alternative approach that can be used to find the inverse of a nonsingular square matrix. So far, we have already learned the shortcut for finding the inverse of a 2 x 2 matrix, but with a 3 x 3 or higher, we are left to use row operations.

### Inverse using the Adjoint and the Determinant Method

$$A^{-1}=\frac{1}{det(A)}\cdot adj(A)$$ To find the inverse for a given matrix A, we can multiply 1/det(A) by the adjoint of A. Let's look at an example.
Example #1: Find the inverse of A.  $$A=\left[ \begin{array}{ccc}3&4&-3\\ 0&-3&2\\-2&-6&5\end{array}\right]$$ First, we will find our determinant. $$|A|=-7$$ Next, we will find our adjoint. $$adj(A)=\left[ \begin{array}{ccc}-3&-2&-1\\ -4&9&-6\\-6&10&-9\end{array}\right]$$ Finally, we will multiply -1/7 by each entry of our adjoint. This will give us the inverse of matrix A. $$A^{-1}=\left[ \begin{array}{ccc}\frac{3}{2}&\frac{2}{7}&\frac{1}{7}\\ \frac{4}{7}&-\frac{9}{7}&\frac{6}{7}\\\frac{6}{7}&-\frac{10}{7}&\frac{9}{7}\end{array}\right]$$

#### Skills Check:

Example #1

Find the inverse of A. $$A=\left[ \begin{array}{ccc}-5&-3&2 \\ 5&5&-1 \\ -2&-2 & 1\end{array}\right]$$

A
$$A^{-1}=\left[ \begin{array}{ccc}\frac{5}{2}&\frac{3}{7}&\frac{1}{9}\\ \frac{14}{7}&-\frac{2}{7}&0\\4&-\frac{3}{7}&\frac{9}{7}\end{array}\right]$$
B
$$A^{-1}=\left[ \begin{array}{ccc}5&6&-2\\ 1&-\frac{4}{7}&\frac{16}{7}\\\frac{8}{5}&-\frac{10}{3}&\frac{9}{17}\end{array}\right]$$
C
$$A^{-1}=\left[ \begin{array}{ccc}-\frac{1}{2}&\frac{1}{6}&\frac{7}{6}\\ \frac{1}{2}&\frac{1}{6}&-\frac{5}{6}\\ 0& \frac{2}{3}&\frac{5}{3}\end{array}\right]$$
D
$$A^{-1}=\left[ \begin{array}{ccc}\frac{11}{2}&\frac{62}{7}&\frac{61}{7}\\ -2&-3&8\\-1&-8&\frac{19}{7}\end{array}\right]$$
E
$$A^{-1}=\left[ \begin{array}{ccc}\frac{13}{2}&\frac{12}{7}&\frac{11}{7}\\ \frac{4}{17}&-\frac{9}{7}&\frac{6}{7}\\7&3&8\end{array}\right]$$

Example #2

Find the inverse of A. $$A=\left[ \begin{array}{ccc}4&0&-4 \\ 1&-1&-1 \\ 2&-3 & 0\end{array}\right]$$

A
$$A^{-1}=\left[ \begin{array}{ccc}1&-1&5\\ 3&-7&-1\\ \frac{1}{3}&-\frac{4}{3}&\frac{6}{11}\end{array}\right]$$
B
$$A^{-1}=\left[ \begin{array}{ccc}\frac{3}{8}&-\frac{3}{2}&\frac{1}{2}\\ \frac{1}{4}&-1&0\\\frac{1}{8}&-\frac{3}{2}&\frac{1}{2}\end{array}\right]$$
C
$$A^{-1}=\left[ \begin{array}{ccc}-\frac{3}{8}&-\frac{1}{4}&-\frac{5}{7}\\ -\frac{3}{2}&-1&-\frac{3}{2}\\\frac{1}{2}&0&\frac{1}{2}\end{array}\right]$$
D
$$A^{-1}=\left[ \begin{array}{ccc}\frac{3}{2}&\frac{3}{2}&-\frac{1}{2}\\ \frac{1}{2}&-\frac{1}{2}&\frac{5}{8}\\\frac{3}{8}&-\frac{1}{8}&-2\end{array}\right]$$
E
$$A^{-1}=\left[ \begin{array}{ccc}-\frac{3}{10}&0&-\frac{1}{10}\\ -1&-1&0\\-\frac{1}{5}&0&-\frac{2}{5}\end{array}\right]$$