# Solve the matrix equation

Question:

Solve the matrix equation $\left[\begin{array}{ll}5 & 4 \\ 1 & 1\end{array}\right] X=\left[\begin{array}{cc}1 & -2 \\ 1 & 3\end{array}\right]$, where $X$ is a $2 \times 2$ matrix.

Solution:

Let:

$A=\left[\begin{array}{ll}5 & 4\end{array}\right.$

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

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

$\begin{array}{ll}1 & 3]\end{array}$

Now,

$|\mathrm{A}|=\mid \begin{array}{ll}5 & 4\end{array}$

$1 \quad 1 \mid=5-4=1 \neq 0$

Hence, $A$ is invertible.

If $C_{i j}$ is cofactor of $a_{i j}$ in $A$, then $C_{11}=1, C_{12}=-1, C_{21}=-4$ and $C_{22}=5$.

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

$-4-5]^{T}=\left[\begin{array}{ll}1 & -4\end{array}\right.$

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

$\therefore A^{-1}=\frac{1}{|A|}$ adj $A=\left[\begin{array}{ll}1 & -4\end{array}\right.$

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

Now, the given equation becomes $A X=B$.

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

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

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

$\Rightarrow X=\left[\begin{array}{ll}1 & -4\end{array}\right.$

$\left.\begin{array}{ll}-1 & 5\end{array}\right] \times\left[\begin{array}{ll}1 & -2\end{array}\right.$

$\begin{array}{ll}1 & 3]\end{array}$

$\Rightarrow X=\left[\begin{array}{cc}1-4 & -2-12\end{array}\right.$

$\left.\begin{array}{ll}-1+5 & 2+15\end{array}\right]$

$\Rightarrow X=\left[\begin{array}{ll}-3 & -14\end{array}\right.$

$\left.\begin{array}{ll}4 & 17\end{array}\right]$

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