# Find the matrix X for which

Question:

Find the matrix X for which

Solution:

Let $A=\left[\begin{array}{ll}3 & 2\end{array}\right.$

$\left.\begin{array}{ll}7 & 5\end{array}\right], B=\left[\begin{array}{ll}-1 & 1\end{array}\right.$

$-2 \quad 1]$ and $C=\left[\begin{array}{ll}2 & -1\end{array}\right]$

$\left.\begin{array}{ll}0 & 4\end{array}\right]$

Now,

$|A|=\mid \begin{array}{ll}3 & 2\end{array}$

$7 \quad 5 \mid=15-14=1$

$|B|=\mid-1 \quad 1$

$\begin{array}{ll}-2 & 1\end{array}=-1+2=1$

Since, $|A| \neq 0$ and $|B| \neq 0$

Hence, $A \& B$ are invertible, so $A^{-1}$ and $B^{-1}$ exist.

Cofactors of matrix $A$ are

$\begin{array}{llll}\mathrm{A}_{11}=5 & \mathrm{~A}_{12}=-7 & \mathrm{~A}_{21}=-2 & \mathrm{~A}_{22}=3\end{array}$

Now,

$\operatorname{adj} A=\left[\begin{array}{ll}5 & -7\end{array}\right.$

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

$\left.\begin{array}{ll}-7 & 3\end{array}\right]$

$A^{-1}=\frac{1}{|A|}$ adj $A=\left[\begin{array}{ll}5 & -2\end{array}\right.$

$\left.\begin{array}{ll}-7 & 3\end{array}\right]$

Cofactors of matrix $B$ are

$\begin{array}{llll}B_{11}=1 & B_{12}=2 & B_{21}=-1 & B_{22}=-1\end{array}$

Now,

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

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

$2-1]$

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

$2 \quad-1]$

The given equation becomes $A X B=C$

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

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

$\Rightarrow X=\left[\begin{array}{ll}5 & -2\end{array}\right.$

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

$0 \quad 4]\left[\begin{array}{ll}1 & -1\end{array}\right.$

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

$\Rightarrow X=\left[\begin{array}{ll}5 & -2\end{array}\right.$

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

$0+8 \quad 0-4]$

$\Rightarrow \mathrm{X}=\left[\begin{array}{ll}5 & -2\end{array}\right.$

$\left.\begin{array}{ll}-7 & 3\end{array}\right]\left[\begin{array}{ll}0 & -1\end{array}\right.$

$\left.\begin{array}{ll}8 & -4\end{array}\right]$

$\Rightarrow X=\left[\begin{array}{ll}-16 & 3\end{array}\right.$

$24-5]$