Question
If $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right]$ and $C=\left[\begin{array}{ll}24 & 7 \\ 31 & 9\end{array}\right]$ then find matrix $X$ such that $A X B=C$

Answer

$\text{AXB}=C \\ \therefore\left(\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right)(\mathrm{XB})=\left[\begin{array}{ll}24 & 7 \\ 31 & 9\end{array}\right]$
First we perform the row transformations.
By $R_2-R_1$, we get,
$\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right](X B)=\left[\begin{array}{rr}24 & 7 \\ 7 & 2\end{array}\right]$
By $R_1-R_2$, we get,
$\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right](X B)=\left[\begin{array}{rr}17 & 5 \\ 7 & 2\end{array}\right]$
$\therefore X B=\left[\begin{array}{rr}17 & 5 \\ 7 & 2\end{array}\right] $
$ \therefore X\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right]=\left[\begin{array}{rr}17 & 5 \\ 7 & 2\end{array}\right]$
Now, we perform the column transformations.
By $C_1 \leftrightarrow C_3$, we get,
$X\left[\begin{array}{ll}1 & 4 \\ 1 & 3\end{array}\right]=\left[\begin{array}{rr}5 & 17 \\ 2 & 7\end{array}\right]$
By $C_2-4 C_1$, we get,
$X\left[\begin{array}{rr}1 & 0 \\ 1 & -1\end{array}\right]=\left[\begin{array}{ll}5 & -3 \\ 2 & -1\end{array}\right]$
By $(-1) C_2$, we get,
$X\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}5 & 3 \\ 2 & 1\end{array}\right]$
By $C_1-C_2$, we get,
$X\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}2 & 3 \\ 1 & 1\end{array}\right]$
$\therefore X=\left[\begin{array}{ll}2 & 3 \\ 1 & 1\end{array}\right]$

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