## Problem on using inverses to solve a 3x3 matrix equation

Find $\bfA^{-1}$ and use it to solve $\bfA \bfx = \bfb$, where $$\bfA = \begin{pmatrix}1 & -1& 1 \\ 1 & 1 & 1\\1 & 2 & 4 \end{pmatrix}, \quad \bfb = \begin{pmatrix} 1\\3\\2 \end{pmatrix}$$
• ## Solution

Recall that
For example, the minor of the $1,1$ entry is $\begin{vmatrix}1 & 1 \\ 2 & 4\end{vmatrix}.$

The minor of the $1,2$ entry is $\begin{vmatrix} 1 & 1 \\ 1 & 4 \end{vmatrix}$.
Repeating this process, the minors of all entries are $$\begin{pmatrix} \begin{vmatrix}1 & 1 \\ 2 & 4\end{vmatrix} & \begin{vmatrix}1 & 1 \\ 1 & 4\end{vmatrix} & \begin{vmatrix}1 & 1 \\ 1 & 2\end{vmatrix} \\ \begin{vmatrix}-1 & 1 \\ 2 & 4\end{vmatrix} & \begin{vmatrix}1 & 1 \\ 1 & 4\end{vmatrix} & \begin{vmatrix}1 & -1 \\ 1 & 2\end{vmatrix} \\ \begin{vmatrix}-1 & 1 \\ 1 & 1\end{vmatrix} & \begin{vmatrix}1 & 1 \\ 1 & 1\end{vmatrix} & \begin{vmatrix}1 & -1 \\ 1 & 1\end{vmatrix} \end{pmatrix}$$
Recall that
Computing all the determinants, the matrix of minors is $$\begin{pmatrix} 2 & 3 & 1 \\ -6 & 3 & 3 \\ -2 &0 & 2 \end{pmatrix}.$$
Next we negate every other entry, according to the pattern $$\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & +\end{pmatrix}.$$
The matrix of minors becomes $$\begin{pmatrix} 2 & -3 & 1 \\6 & 3 & -3 \\ -2 & 0 & 2\end{pmatrix}.$$
Transposing, we get $$\begin{pmatrix} 2 & 6 & -2 \\ -3 & 3 & 0 \\ 1 & -3 & 2 \end{pmatrix}.$$
In order to divide by the determinant of $\bfA$, we must first compute it.
Hence, $$\bfA^{-1} = \frac{1}{6}\begin{pmatrix} 2 & 6 & -2 \\ -3 & 3 & 0 \\ 1 & -3 & 2 \end{pmatrix}.$$
To solve the linear system, recall that
Computing, we have $$\bfx = \frac{1}{6}\begin{pmatrix} 2 & 6 & -2 \\ -3 & 3 & 0 \\ 1 & -3 & 2 \end{pmatrix} \begin{pmatrix} 1\\3\\2 \end{pmatrix}$$
My matrix multiplication we get $$\bfx = \begin{pmatrix} 8/3 \\ 1 \\ -2/3\end{pmatrix}.$$