Respuesta :
SOLUTION
We want to solve the simultaneous equation
[tex]\begin{gathered} x+5=-14 \\ -2x-5y=23 \end{gathered}[/tex]We will rearrage this to become
[tex]\begin{gathered} x+0y=-19 \\ -2x-5y=23 \end{gathered}[/tex]Writing this as a matrix equation, we get
[tex]\begin{bmatrix}{1} & {0} & {} \\ {-2} & {-5} & \end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & \end{bmatrix}=\begin{bmatrix}{-19} & {} \\ {23} & \end{bmatrix}[/tex]From the above matrix equation we will take
[tex]\begin{gathered} A=\begin{bmatrix}{1} & {0} & {} \\ {-2} & {-5} & \end{bmatrix} \\ B=\begin{bmatrix}{x} & {} \\ {y} & \end{bmatrix} \\ C=\begin{bmatrix}{-19} & {} \\ {23} & \end{bmatrix} \end{gathered}[/tex]So, this means that
[tex]\begin{gathered} A\times B=C \\ AB=C \\ B=\frac{C}{A} \\ B=C\times A^{-1} \\ B=CA^{-1} \end{gathered}[/tex]Now, we have to find the inverse matrix, which is
[tex]A^{-1}[/tex]This is given as
[tex]\begin{gathered} A^{-1}=\frac{1}{|A|}\times Adj\text{ A} \\ A^{-1}=\frac{1}{ad-bc}\times\begin{bmatrix}{d} & {-b} & {} \\ {-c} & {a} & \end{bmatrix} \end{gathered}[/tex]So, if we have
[tex]\begin{bmatrix}{a} & {b} & {} \\ {c} & {d} & \end{bmatrix}=\begin{bmatrix}{1} & {0} & {} \\ {-2} & {-5} & \end{bmatrix}[/tex]Then
[tex]\begin{gathered} A^{-1}=\frac{1}{ad-bc}\times\begin{bmatrix}{d} & {-b} & {} \\ {-c} & {a} & \end{bmatrix} \\ A^{-1}=\frac{1}{1(-5)-0(-2)}\times\begin{bmatrix}{-5} & {0} & {} \\ {2} & {1} & \end{bmatrix} \\ A^{-1}=\frac{1}{-5}\begin{bmatrix}{-5} & {0} & {} \\ {2} & {1} & \end{bmatrix} \end{gathered}[/tex]Hence, the inverse matrix is
[tex]A^{-1}=\frac{1}{-5}\begin{bmatrix}{-5} & {0} & {} \\ {2} & {1} & \end{bmatrix}[/tex]Now, from
[tex]B=CA^{-1}[/tex]We will have
[tex]\begin{gathered} \begin{bmatrix}{x} & {} \\ {y} & \end{bmatrix}=\frac{1}{-5}\begin{bmatrix}{-5} & {0} & {} \\ {2} & {1} & \end{bmatrix}\begin{bmatrix}{-19} & {} \\ {23} & \end{bmatrix} \\ \begin{bmatrix}{x} & {} \\ {y} & \end{bmatrix}=\frac{1}{-5}\begin{bmatrix}{95} & {} \\ {-38+23} & \end{bmatrix} \\ \begin{bmatrix}{x} & {} \\ {y} & \end{bmatrix}=\frac{1}{-5}\begin{bmatrix}{95} & {} \\ {-15} & \end{bmatrix} \\ \begin{bmatrix}{x} & {} \\ {y} & \end{bmatrix}=\begin{bmatrix}{-19} & {} \\ {3} & \end{bmatrix} \end{gathered}[/tex]Hence from the matrix,
x = -19 and y = 3
