The given system of equations is,
[tex]\begin{gathered} x-9y=-4 \\ -x+7y=5 \end{gathered}[/tex]A system of equations can be defined as,
[tex]AX=B\text{ ------(1)}[/tex]Here, A is the ceofficient matrix, X is the variable matrix and B is the constant matrix.
Coefficient matrix gives the coefficients of the variables in the system of equations. Hence,
[tex]A=\begin{bmatrix}{1} & {-9} & {} \\ {-1} & {7} & {} \\ {} & {} & {}\end{bmatrix}[/tex]Hence, equation (1) can be written as,
[tex]\begin{bmatrix}{1} & {-9} & {} \\ {-1} & {7} & {} \\ {} & {} & {}\end{bmatrix}\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{-4} & & {} \\ {5} & & {} \\ {} & {} & {}\end{bmatrix}\text{ -------(2)}[/tex]If a 2x2 matrix A is given by,
[tex]A=\begin{bmatrix}{a} & {b} & {} \\ {c} & {d} & \\ {} & {} & {}\end{bmatrix}[/tex]Then, adj A is, Hence, the adjoint of the given matrix A is,
[tex]adjA=\begin{bmatrix}{d} & {-b} & {} \\ {-c} & {a} & \\ {} & {} & {}\end{bmatrix}[/tex]The determinat of A is,
[tex]|A|=ad-bc[/tex]Hence, the inverse of the given matrix A is,
[tex]\begin{gathered} A^{-1}=\frac{1}{|A|}adj\text{ A} \\ =\frac{1}{7\times1-(-9\times(-1))}\begin{bmatrix}{7} & {9} & {} \\ {1} & {1} & {} \\ {} & {} & {}\end{bmatrix} \\ =\frac{1}{7-9}\begin{bmatrix}{7} & {9} & {} \\ {1} & {1} & {} \\ {} & {} & {}\end{bmatrix} \\ =\frac{1}{-2}\begin{bmatrix}{7} & {9} & {} \\ {1} & {1} & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}[/tex]
Now, multiply equation (1) by inverse of A to solve for the variable matrix.
[tex]\begin{gathered} A^{-1}AX=A^{-1}B \\ IX=A^{-1}B \\ \begin{bmatrix}{1} & {0} & {} \\ {0} & {1} & {} \\ {} & {} & {}\end{bmatrix}\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\frac{1}{-2}\begin{bmatrix}{7} & {9} & {} \\ {1} & {1} & {} \\ {} & {} & {}\end{bmatrix}\begin{bmatrix}{-4} & & {} \\ {5} & & {} \\ {} & {} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{\frac{-7}{2}} & {\frac{-9}{2}} & {} \\ {\frac{-1}{2}} & {\frac{-1}{2}} & {} \\ {} & {} & {}\end{bmatrix}\begin{bmatrix}{-4} & & {} \\ {5} & & {} \\ {} & {} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{\frac{-7}{2}\times(-4)+(\frac{-9}{2}\times5)} & & {} \\ {\frac{-1}{2}\times(-4)+(\frac{-1}{2}\times5} & {} & {} \\ {} & {} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & & {} \\ {y} & & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{14-\frac{45}{2}} & & {} \\ {2-\frac{5}{2}} & & {} \\ {} & {} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{\frac{-17}{2}} & & {} \\ {\frac{-1}{2}} & & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}[/tex]Here, I is identity matrix.
Therefore, the inverse of matrix A is,
[tex]A^{-1}=\frac{-1}{2}\begin{bmatrix}{7} & {9} & {} \\ {1} & {1} & {} \\ {} & {} & {}\end{bmatrix}[/tex]x=-17/2 and y=-1/2.
