The inverse of a matrix can be calculated as:
[tex]\begin{gathered} \text{When} \\ A=\begin{bmatrix}{a} & {b} & {} \\ {c} & {d} & {} \\ {} & {} & \end{bmatrix} \\ \text{Then A\textasciicircum-1 is:} \\ A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}{d} & -{b} & {} \\ {-c} & {a} & {} \\ {} & {} & \end{bmatrix} \end{gathered}[/tex]Then, let's start by calculating the inverse of the given matrix:
[tex]\begin{gathered} \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}=\frac{1}{4\cdot3-1\cdot(-2)}\begin{bmatrix}{3} & -{1} & {} \\ {-(-2)} & {4} & {} \\ {} & {} & \end{bmatrix} \\ \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}=\frac{1}{14}\begin{bmatrix}{3} & -{1} & {} \\ {2} & {4} & {} \\ {} & {} & \end{bmatrix} \end{gathered}[/tex]The problem says he multiplies the left side of the coefficient matrix by the inverse matrix, thus:
[tex]\begin{gathered} \begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}\begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}\cdot\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{4} & {1} & {} \\ {-2} & {3} & {} \\ {} & {} & \end{bmatrix}^{-1}\begin{bmatrix}{2} & {} & {} \\ {-22} & {} & {} \\ {} & {} & {}\end{bmatrix} \\ \end{gathered}[/tex]*These matrices will be the options to put on the first and second boxes.
Then:
[tex]\begin{gathered} \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\frac{1}{14}\begin{bmatrix}{3} & -{1} & {} \\ {2} & {4} & {} \\ {} & {} & \end{bmatrix}\cdot\begin{bmatrix}{2} & {} & {} \\ {-22} & {} & {} \\ {} & {} & {}\end{bmatrix}\text{ This is for the third box} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\frac{1}{14}\begin{bmatrix}{3\times2+(-1)\times(-22)} & & {} \\ {2\times2+4\times(-22)} & & {} \\ {} & {} & \end{bmatrix}=\frac{1}{14}\begin{bmatrix}{28} & & {} \\ {-84} & & {} \\ {} & {} & \end{bmatrix}\text{ This is the 4th box} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{28/14} & & {} \\ {-84/14} & & {} \\ {} & {} & \end{bmatrix}=\begin{bmatrix}{2} & & {} \\ {-6} & & {} \\ {} & {} & \end{bmatrix}\text{ And finally this is the last box} \end{gathered}[/tex]