The matrix equation AX=B, where A and B are numerical matrices and X is unknown matrix has a solution [tex]X= A^{-1} B[/tex], where [tex] A^{-1} [/tex] is inverse matrix of X.
1. We rewrite given system as matrix equation [tex] \left[\begin{array}{cc}4&-2\\3&-1\end{array}\right] X=\left[\begin{array}{c}- 12\\- 3\end{array}\right][/tex];
2. Find[tex] A^{-1} = \left[\begin{array}{cc}4&-2\\3&-1\end{array}\right] ^{-1} [/tex] by the rule [tex] A^{-1}= \frac{1}{det \ A} [ A_{ij} ] ^{T}[/tex]. So, [tex]det\ A=4[/tex]×[tex](-1)-3[/tex]×[tex](-2)=-4+6=2[/tex] and algebraic supplements are
[tex] A_{11} =-1 \\ A_{12} =-3 \\ A_{21}=2 \\ A_{22} =4 [/tex]. Then [tex] A^{-1} = \frac{1}{2} \left[\begin{array}{cc}-1&-3\\2&4\end{array}\right] ^{T}= \left[\begin{array}{cc}- \frac{1}{2} &1\\- \frac{3}{2} &2\end{array}\right] [/tex];
3. Calculate [tex]X= \left[\begin{array}{cc}- \frac{1}{2} &1\\- \frac{3}{2} &2\end{array}\right] \left[\begin{array}{c} -12\\-3\end{array}\right]=[/tex] [tex] \left[\begin{array}{c}3\\12\end{array}\right] [/tex];
4. We obtain [tex] X= \left[\begin{array}{c}3\\12\end{array}\right] [/tex], from where x=3 and y=12.
