First, let's write the system in the matrix form AX = B:
[tex]\begin{bmatrix}{1} & {2} \\ {-4} & {3}\end{bmatrix}\begin{bmatrix}{x} & \\ {y} & \end{bmatrix}=\begin{bmatrix}{2} & {} \\ {25} & {}\end{bmatrix}[/tex]Now, to solve the system, let's first find the inverse of the matrix A, using the formula below for the inverse of a 2x2 matrix:
[tex]\begin{gathered} A=\begin{bmatrix}{a} & {b} \\ c & {d}\end{bmatrix}\\ \\ A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}{d} & -b \\ {-c} & {a}\end{bmatrix} \end{gathered}[/tex]So we have:
[tex]\begin{gathered} A=\begin{bmatrix}{1} & {2} \\ {-4} & {3}\end{bmatrix}\\ \\ A^{-1}=\frac{1}{3-(-8)}\begin{bmatrix}{3} & -2{} \\ {4} & {1}\end{bmatrix}\\ \\ A^{-1}=\frac{1}{11}\begin{bmatrix}{3} & -2{} \\ {4} & {1}\end{bmatrix}\\ \\ A^{-1}=\begin{bmatrix}{\frac{3}{11}} & -\frac{2}{11}{} \\ {\frac{4}{11}} & {\frac{1}{11}}\end{bmatrix}\\ \end{gathered}[/tex]Now, to solve the system, we can do the following:
[tex]\begin{gathered} AX=B\\ \\ A^{-1}AX=A^{-1}B\\ \\ IX=A^{-1}B\\ \\ X=A^{-1}B \end{gathered}[/tex]Multiplying the inverse matrix and matrix B, we have:
[tex]\begin{gathered} \begin{bmatrix}{\frac{3}{11}} & {-\frac{2}{11}} \\ {\frac{4}{11}} & {\frac{1}{11}}\end{bmatrix}\cdot\begin{bmatrix}{2} & {} \\ 25 & {}\end{bmatrix}=\begin{bmatrix}{\frac{3}{11}}\cdot2+(-\frac{2}{11})\cdot25 & \\ {\frac{4}{11}}\cdot2+\frac{1}{11}\cdot25 & {}\end{bmatrix}\\ \\ =\begin{bmatrix}{\frac{6}{11}}-\frac{50}{11} & \\ {\frac{8}{11}}+\frac{25}{11} & {}\end{bmatrix}=\begin{bmatrix}-\frac{44}{11} & \\ \frac{33}{11} & {}\end{bmatrix}=\begin{bmatrix}-4 & \\ 3 & {}\end{bmatrix} \end{gathered}[/tex]Therefore the solution is x = -4 and y = 3.
