Answer :
We are given the following system of equations:
[tex]\begin{gathered} 5x+4y=-9 \\ -3x-2y=-4 \end{gathered}[/tex]We can rewrite the system of equations using matrix notations. Matrix A is the matrix of coefficients on the left side, therefore:
[tex]A=\begin{bmatrix}{5} & {4} \\ {-3} & -{2}\end{bmatrix}[/tex]We get a 2x2 matrix of the form:
[tex]A=\begin{bmatrix}{a} & {b} \\ {c} & {d}\end{bmatrix}[/tex]The inverse of this type of matrix is:
[tex]A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}{d} & {-b} \\ {-c} & {a}\end{bmatrix}[/tex]Applying the formula we get:
[tex]A^{-1}=\frac{1}{(5)(-2)-(4)(-3)}\begin{bmatrix}{-2} & {-4} \\ {3} & {5}\end{bmatrix}[/tex]Solving the operations:
[tex]A^{-1}=\frac{1}{2}\begin{bmatrix}{-2} & {-4} \\ {3} & {5}\end{bmatrix}[/tex]And thus we get the inverse of the matrix A.
The system of equations in matrix form is:
[tex]Ax=b[/tex]Where:
[tex]\begin{gathered} x=\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix} \\ \\ b=\begin{bmatrix}{-9} & {} \\ {-4} & {}\end{bmatrix} \end{gathered}[/tex]The solution of the system is determined by left-multiplying both sides by the inverse of A
[tex]x=A^{-1}b[/tex]Substituting the matrices we get:
[tex]\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\frac{1}{2}\begin{bmatrix}{-2} & {-4} \\ {3} & {5}\end{bmatrix}\begin{bmatrix}{-9} & {} \\ {-4} & {}\end{bmatrix}[/tex]Now, we multiply the matrices. We multiply each row element of the inverse of "A" by its corresponding column element of the matrix "b".
[tex]\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\frac{1}{2}\begin{bmatrix}{(-2)(-9)+(-4)(-4)} & {} \\ {(3)(-9)+(5)(-4)} & {}\end{bmatrix}[/tex]Solving the operations we get:
[tex]\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\frac{1}{2}\begin{bmatrix}{34} & {} \\ {-47} & {}\end{bmatrix}[/tex]Therefore, the value of "x" is:
[tex]x=\frac{34}{2}=17[/tex]The value of "y" is:
[tex]y=-\frac{47}{2}[/tex]