Question

The matrix $X$ for which $\left[\begin{array}{cc}1& -4\\ 3& -2\end{array}\right]X=\left[\begin{array}{cc}-16& -6\\ 7& 2\end{array}\right],$ is

• A. $\left[\begin{array}{cc}-2& 4\\ -3& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}6& 2\\ \frac{11}{2}& 2\end{array}\right]$
• C. $\left[\begin{array}{cc}-16& -6\\ 7& 2\end{array}\right]$
• D. $\left[\begin{array}{cc}-\frac{1}{5}& \frac{2}{5}\\ -\frac{3}{10}& \frac{1}{5}\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{cc}6& 2\\ \frac{11}{2}& 2\end{array}\right]$

Let, $A=\left[\begin{array}{cc}1& -4\\ 3& -2\end{array}\right]\text{and}B=\left[\begin{array}{cc}-16& -6\\ 7& 2\end{array}\right]$

Then the matrix equation is $AX=B$
$\therefore |A|=\left[\begin{array}{cc}1& -4\\ 3& -2\end{array}\right]=-2+12=10(\ne 0)$
So $A$ is an invertible matrix. Also,
$\text{adj}A=\left[\begin{array}{cc}-2& 4\\ -3& 1\end{array}\right]$
So,
$A^{-1}=\frac{1}{|A|}\text{adj}A=\frac{1}{10}\left[\begin{array}{cc}-2& 4\\ -3& 1\end{array}\right]$

Now, $AX=B$
$\Rightarrow A^{-1}\left(AX\right)=A^{-1}B$
$\left(A^{-1}A\right)X=A^{-1}B$
$\Rightarrow IX=A^{-1}B$
$\Rightarrow X=A^{-1}B$
$\Rightarrow X=\frac{1}{10}\left[\begin{array}{cc}-2& 4\\ -3& 1\end{array}\right]\left[\begin{array}{cc}-16& -6\\ 7& 2\end{array}\right]=\left[\begin{array}{cc}6& 2\\ 11/2& 2\end{array}\right]$