4 The inverse of a 3 $×$ 3 matrix - determinant method

This method which employs determinants, is of importance from a theoretical perspective. The numerical computations involved are too heavy for matrices of higher order than $3×3$ and in such cases the Gauss elimination approach is prefered.

To obtain ${A}^{-1}$ using the determinant approach the steps in the following keypoint are followed:

Key Point 10

Matrix Inverse $-$ the Determinant Method

Given a square matrix $A$ :

• Find $|A|$ . If $|A|=0$ then ${A}^{-1}$ does not exist. If $|A|\ne 0$ we can proceed to find the inverse matrix, as follows.
• Replace each element of $A$ by its cofactor (see Section 7.3).
• Transpose the result to form the adjoint matrix , denoted by adj $\left(A\right)$
• Then calculate ${A}^{-1}=\frac{1}{\left|A\right|}$ adj $\left(A\right)$ .

Find the inverse of $A=\left[\begin{array}{ccc}\hfill 0& \hfill 1& \hfill 1\\ \hfill 2& \hfill 3& \hfill -1\\ \hfill -1& \hfill 2& \hfill 1\end{array}\right]$ . This will require five stages.

1. First find $\left|A\right|$ :

$\left|A\right|=0×5+1×\left(-1\right)+1×7=6$

2. Now replace each element of $A$ by its minor:

$\left[\begin{array}{ccc}\hfill \left|\begin{array}{cc}\hfill 3& \hfill -1\\ \hfill 2& \hfill 1\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill 2& \hfill -1\\ \hfill -1& \hfill 1\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill 2& \hfill 3\\ \hfill -1& \hfill 2\end{array}\right|\hfill \\ \hfill \hfill \\ \hfill \left|\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill 0& \hfill 1\\ \hfill -1& \hfill 1\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill 0& \hfill 1\\ \hfill -1& \hfill 2\end{array}\right|\hfill \\ \hfill \hfill \\ \hfill \left|\begin{array}{cc}\hfill 1& \hfill 1\\ \hfill 3& \hfill -1\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill 0& \hfill 1\\ \hfill 2& \hfill -1\end{array}\right|\hfill & \hfill \left|\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 3\hfill \end{array}\right|\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 5& \hfill 1& \hfill 7\\ \hfill -1& \hfill 1& \hfill 1\\ \hfill -4& \hfill -2& \hfill -2\end{array}\right]$

3. Now attach the signs from the array

$\begin{array}{ccc}\hfill +\hfill & \hfill -\hfill & \hfill +\hfill \\ \hfill -\hfill & \hfill +\hfill & \hfill -\hfill \\ \hfill +\hfill & \hfill -\hfill & \hfill +\hfill \end{array}$

(so that where a $+$ sign is met no action is taken and where a $-$ sign is met the sign is change

4. to obtain the matrix of cofactors:

$\left[\begin{array}{ccc}\hfill 5& \hfill -1& \hfill 7\\ \hfill 1& \hfill 1& \hfill -1\\ \hfill -4& \hfill 2& \hfill -2\end{array}\right]$

(d) Then transpose the result to obtain the adjoint matrix:

Transposing, adj $\left(A\right)=\left[\begin{array}{ccc}\hfill 5& \hfill 1& \hfill -4\\ \hfill -1& \hfill 1& \hfill 2\\ \hfill 7& \hfill -1& \hfill -2\end{array}\right]$

5. Finally obtain ${A}^{-1}$ :

${A}^{-1}=\frac{1}{\text{det}\left(A\right)}\text{adj}\left(A\right)=\frac{1}{6}\left[\begin{array}{ccc}\hfill 5& \hfill 1& \hfill -4\\ \hfill -1& \hfill 1& \hfill 2\\ \hfill 7& \hfill -1& \hfill -2\end{array}\right]$ as before using Gauss elimination.

Exercises
1. Find the inverses of the following matrices
1. $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$
2. $\left[\begin{array}{cc}\hfill -1& \hfill 0\\ \hfill 0& \hfill 4\end{array}\right]$
3. $\left[\begin{array}{cc}\hfill 1& \hfill 1\\ \hfill -1& \hfill 1\end{array}\right]$
2. Use the determinant method and also the Gauss elimination method to find the inverse of the following matrices
1. $A=\left[\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$
2. $B=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$
1. $-\frac{1}{2}\phantom{\rule{1em}{0ex}}\left[\begin{array}{cc}\hfill 4& \hfill -2\\ \hfill -3& \hfill 1\end{array}\right]$
2. $\left[\begin{array}{cc}\hfill -1& \hfill 0\\ \hfill 0& \hfill \frac{1}{4}\end{array}\right]$
3. $\frac{1}{2}\left[\begin{array}{cc}\hfill 1& \hfill -1\\ \hfill 1& \hfill 1\end{array}\right]$
1. ${A}^{-1}=-\frac{1}{2}{\left[\begin{array}{ccc}\hfill 0& \hfill -2& \hfill 1\\ \hfill -2& \hfill 4& \hfill 2\\ \hfill 0& \hfill 0& \hfill -1\end{array}\right]}^{T}=-\frac{1}{2}\left[\begin{array}{ccc}\hfill 0& \hfill -2& \hfill 0\\ \hfill -2& \hfill 4& \hfill 0\\ \hfill 1& \hfill 2& \hfill -1\end{array}\right]$
2. ${B}^{-1}={\left[\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 0\\ \hfill -1& \hfill 1& \hfill 0\\ \hfill 0& \hfill -1& \hfill 1\end{array}\right]}^{T}=\left[\begin{array}{ccc}\hfill 1& \hfill -1& \hfill 0\\ \hfill 0& \hfill 1& \hfill -1\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]$