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.
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. $$A^{-1}=\left[ \begin{array}{ccc}\frac{3}{2}&\frac{2}{7}&\frac{1}{7}\\ \frac{4}{7}&-\frac{9}{7}&\frac{6}{7}\\\frac{6}{7}&-\frac{10}{7}&\frac{9}{7}\end{array}\right]$$
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. $$A=\left[ \begin{array}{ccc}-5&-3&2 \\ 5&5&-1 \\ -2&-2 & 1\end{array}\right]$$
Please choose the best answer.
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. $$A=\left[ \begin{array}{ccc}4&0&-4 \\ 1&-1&-1 \\ 2&-3 & 0\end{array}\right]$$
Please choose the best answer.
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
$$A^{-1}=\left[ \begin{array}{ccc}-\frac{3}{10}&0&-\frac{1}{10}\\ -1&-1&0\\-\frac{1}{5}&0&-\frac{2}{5}\end{array}\right]$$
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