Given the systemÂ
[tex]4x-2y=-12\ .\ .\ .\ (1) \\ 3x-y=-3\ .\ .\ .\ (2)[/tex]
A matrix equation is of the form AX = B, where A is the matrix of the coefficients of the variables, X is the matrix of the variables, and B is the matrix of the constants.
Thus, rewriting the system into matrix equation we have:
[tex]\left[\begin{array}{cc}4&-2\\3&-1\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right] = \left[\begin{array}{c}-12\\-3\end{array}\right] \\ \\ \Rightarrow\left[\begin{array}{c}x\\y\end{array}\right]= \left[\begin{array}{cc}4&-2\\3&-1\end{array}\right]^{-1}\left[\begin{array}{c}-12\\-3\end{array}\right] \\ \\ = \frac{1}{-4-(-6)} \left[\begin{array}{cc}-1&2\\-3&4\end{array}\right]\left[\begin{array}{c}-12\\-3\end{array}\right]= \frac{1}{-4+6} \left[\begin{array}{c}12-6\\36-12\end{array}\right][/tex]
[tex] \frac{1}{2} \left[\begin{array}{c}6\\24\end{array}\right]=\left[\begin{array}{c}3\\12\end{array}\right][/tex]
Therefore, x = 3 and y = 12.
