Answer
$A^{-1} B=\left[\begin{array}{cc}10 & -1 \\ 9 & 10 \\ -5 & -3\end{array}\right]$
Work Step by Step
$A X=B$ in augmented matrix. Note that solution $X$ is $3 \times 2$ matrix.
$\left[\begin{array}{ccc|cc}1 & 3 & 8 & -3 & 5 \\ 2 & 4 & 11 & 1 & 5 \\ 1 & 2 & 5 & 3 & 4\end{array}\right],$ multiply first row with -2,-1 and add to second, third
$\sim\left[\begin{array}{ccc|cc}1 & 3 & 8 & -3 & 5 \\ 0 & -2 & -5 & 7 & -5 \\ 0 & -1 & -3 & 6 & -1\end{array}\right],$ multiply third row with - 2 and add to second
$\sim\left[\begin{array}{ccc|cc}1 & 3 & 8 & -3 & 5 \\ 0 & 0 & 1 & -5 & -3 \\ 0 & -1 & -3 & 6 & -1\end{array}\right]$
From second row:
$x_{31}=-5$
$x_{32}=-3$
From third row:
$-x_{21}-3 x_{31}=6$
$-x_{21}+15=6$
$x_{21}=9$
$-x_{22}-3 x_{32}=-1$
$-x_{22}+9=-1$
$x_{22}=10$
2
From first row:
$x_{11}+27-40=-3$
$x_{11}=10$
$x_{12}+30-24=5$
$x_{12}=-1$
$A^{-1} B=\left[\begin{array}{cc}10 & -1 \\ 9 & 10 \\ -5 & -3\end{array}\right]$
