Given
[tex]\begin{gathered} |A_x|=det\begin{bmatrix}{20} & {-3} \\ {-192} & {8}\end{bmatrix} \\ |A_y|=det\begin{bmatrix}{2} & {20} \\ {12} & {-192}\end{bmatrix} \end{gathered}[/tex]
To find:
The system of equation.
Explanation:
It is given that,
[tex]\begin{gathered} |A_x|=det\begin{bmatrix}{20} & {-3} \\ {-192} & {8}\end{bmatrix} \\ |A_y|=det\begin{bmatrix}{2} & {20} \\ {12} & {-192}\end{bmatrix} \end{gathered}[/tex]
That implies,
Since,
[tex]\begin{gathered} |A_x|=det\begin{bmatrix}{20} & {-3} \\ {-192} & {8}\end{bmatrix} \\ |A_y|=det\begin{bmatrix}{2} & {20} \\ {12} & {-192}\end{bmatrix} \end{gathered}[/tex]
Then,
[tex]AX=B\Rightarrow\begin{bmatrix}{2} & {-3} \\ {12} & {8}\end{bmatrix}\begin{bmatrix}{x} & {} \\ {y} & {}\end{bmatrix}=\begin{bmatrix}{20} & {} \\ {-192} & \end{bmatrix}[/tex]
Therefore,
The system of equation is,
[tex]\begin{gathered} 2x-3y=20 \\ 12x+8y=-192 \end{gathered}[/tex]
Hence, the answer is option A).
