The given system of equations is,
![\begin{gathered} x-9y=-4 \\ -x+7y=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/arevv86vi2uy3tovk6k0k3wkt7jzpx8gg1.png)
A system of equations can be defined as,
![AX=B\text{ ------(1)}](https://img.qammunity.org/2023/formulas/mathematics/college/ulvrurhzgc4thnc2bhift9u1gi43wb4d55.png)
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,
![A=\begin{bmatrix}{1} & {-9} & {} \\ {-1} & {7} & {} \\ {} & {} & {}\end{bmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/e20yj4vtbc6z172muvp5q1re52n4qhs4bd.png)
Hence, equation (1) can be written as,
![\begin{bmatrix}{1} & {-9} & {} \\ {-1} & {7} & {} \\ {} & {} & {}\end{bmatrix}\begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{-4} & & {} \\ {5} & & {} \\ {} & {} & {}\end{bmatrix}\text{ -------(2)}](https://img.qammunity.org/2023/formulas/mathematics/college/7fv7fonz459ukyhqn7lctai8tspviz2mek.png)
If a 2x2 matrix A is given by,
![A=\begin{bmatrix}{a} & {b} & {} \\ {c} & {d} & \\ {} & {} & {}\end{bmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/ub2h2g8ai6zjswirqjn3z38d4r6ug9f9ax.png)
Then, adj A is, Hence, the adjoint of the given matrix A is,
![adjA=\begin{bmatrix}{d} & {-b} & {} \\ {-c} & {a} & \\ {} & {} & {}\end{bmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/8wjiqn13e2pu5sqgut084prbqh18ss5uaw.png)
The determinat of A is,
![|A|=ad-bc](https://img.qammunity.org/2023/formulas/mathematics/college/23dbqe7wno5hg0pr401mahcgn7uzl4zloz.png)
Hence, the inverse of the given matrix A is,
![\begin{gathered} A^(-1)=(1)/(|A|)adj\text{ A} \\ =(1)/(7*1-(-9*(-1)))\begin{bmatrix}{7} & {9} & {} \\ {1} & {1} & {} \\ {} & {} & {}\end{bmatrix} \\ =(1)/(7-9)\begin{bmatrix}{7} & {9} & {} \\ {1} & {1} & {} \\ {} & {} & {}\end{bmatrix} \\ =(1)/(-2)\begin{bmatrix}{7} & {9} & {} \\ {1} & {1} & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4dgqq8mz5m3h6f0yyz310zuaw386vbdaeu.png)
Now, multiply equation (1) by inverse of A to solve for the variable matrix.
![\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}=(1)/(-2)\begin{bmatrix}{7} & {9} & {} \\ {1} & {1} & {} \\ {} & {} & {}\end{bmatrix}\begin{bmatrix}{-4} & & {} \\ {5} & & {} \\ {} & {} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{(-7)/(2)} & {(-9)/(2)} & {} \\ {(-1)/(2)} & {(-1)/(2)} & {} \\ {} & {} & {}\end{bmatrix}\begin{bmatrix}{-4} & & {} \\ {5} & & {} \\ {} & {} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{(-7)/(2)*(-4)+((-9)/(2)*5)} & & {} \\ {(-1)/(2)*(-4)+((-1)/(2)*5} & {} & {} \\ {} & {} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & & {} \\ {y} & & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{14-(45)/(2)} & & {} \\ {2-(5)/(2)} & & {} \\ {} & {} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & {} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{(-17)/(2)} & & {} \\ {(-1)/(2)} & & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}]()
Here, I is identity matrix.
Therefore, the inverse of matrix A is,
![A^(-1)=(-1)/(2)\begin{bmatrix}{7} & {9} & {} \\ {1} & {1} & {} \\ {} & {} & {}\end{bmatrix}](https://img.qammunity.org/2023/formulas/mathematics/college/o7xjsjntqtrjuvsity0lhy8xa6alaxefol.png)
x=-17/2 and y=-1/2.