Simultaneous equations can be solved using inverse matrix operation.
The complete steps of Jacob's solution are:
[tex]\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right][/tex]
[tex]\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right][/tex]
[tex]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right][/tex]
[tex]\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right][/tex]
We have:
[tex]4x + y = 2[/tex]
[tex]-2x + 4y = -22[/tex]
Calculate the determinant of [tex]\left[\begin{array}{cc}4&1\\-2&3\end{array}\right][/tex]
[tex]|A| = 4 \times 3 -1 \times -2[/tex]
[tex]|A| = 12 +2[/tex]
[tex]|A| = 14[/tex]
So, the inverse matrix becomes
[tex]A = \frac{1}{14}\left[\begin{array}{cc}4&1\\-2&3\end{array}\right][/tex]
Replace the first column with [tex]\left[\begin{array}{c}2&-22\end{array}\right][/tex] to calculate the value of x
[tex]x = \frac{1}{14}\left[\begin{array}{cc}2&1\\-22&3\end{array}\right][/tex]
So, we have:
[tex]x = \frac{1}{14}(2 \times 3 - 1 \times -22)[/tex]
[tex]x = \frac{1}{14}(6 +22)[/tex]
[tex]x = \frac{1}{14}(28)[/tex]
[tex]x = 2[/tex]
Replace the second column with [tex]\left[\begin{array}{c}2&-22\end{array}\right][/tex] to calculate the value of y
[tex]y = \frac{1}{14}\left[\begin{array}{cc}4&2\\-2&-22\end{array}\right][/tex]
So, we have:
[tex]y = \frac{1}{14}(4 \times -22 - 2 \times -2)[/tex]
[tex]y = \frac{1}{14}(-88 +4)[/tex]
[tex]y = \frac{1}{14}(-84)[/tex]
[tex]y = -6[/tex]
Hence, the complete process is:
[tex]\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right][/tex]
[tex]\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right][/tex]
[tex]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right][/tex]
[tex]\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right][/tex]
Read more about matrices at:
https://brainly.com/question/11367104
