Find the matrix X satisfying the matrix equation

Solution:

Let:

$A=\left[\begin{array}{ll}5 & 3\end{array}\right.$

$-1 \quad-2]$

$\Rightarrow|\mathrm{A}|=\mid 5 \quad 3$

$-1-2 \mid=-10+3=-7 \neq 0$

Hence, $A$ is invertible.

If $C_{i j}$ is a cofactor of $a_{i j}$ in $A$, then $\mathrm{C}_{11}=-2, \mathrm{C}_{12}=1, \mathrm{C}_{21}=-3$ and $\mathrm{C}_{22}=5$.

Now,

adj $A=\left[\begin{array}{ll}-2 & 1\end{array}\right.$

$\left.\begin{array}{ll}-3 & 5\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ll}-2 & -3\end{array}\right.$

$\left.\begin{array}{ll}1 & 5\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A=\frac{-1}{7}\left[\begin{array}{ll}-2 & -3\end{array}\right.$

$\left.\begin{array}{ll}1 & 5\end{array}\right]$

Let:

$B=\left[\begin{array}{ll}14 & 7\end{array}\right.$

$\left.\begin{array}{ll}7 & 7\end{array}\right]$

$\Rightarrow|B|=\mid \begin{array}{ll}14 & 7\end{array}$

$7 \quad 7 \mid=98-49=49 \neq 0$

Hence, $B$ is invertible.

The given matrix equation becomes $X A=B$.

$\Rightarrow(X A) A^{-1}=B A^{-1}$

$\Rightarrow X\left(A A^{-1}\right)=\left[\begin{array}{ll}14 & 7\end{array}\right.$

$7 \quad 7 \quad] \times \frac{-1}{7} \times\left[\begin{array}{ll}-2 & -3\end{array}\right.$

$-14+7 \quad-21+35]$

$\Rightarrow X=\frac{-1}{7}\left[\begin{array}{ll}-21 & -7\end{array}\right.$

$\left.\begin{array}{ll}-7 & 14\end{array}\right]$

$\Rightarrow X=\left[\begin{array}{ll}3 & 1\end{array}\right.$

$\left.\begin{array}{ll}1 & -2\end{array}\right]$