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

Answer 1

Answer: (D)

$\left[\begin{array}{cc}6& 2\\ \frac{11}{2}& 2\end{array}\right]$

Given, $\left[\begin{array}{cc}1& -4\\ 3& -2\end{array}\right]X=\left[\begin{array}{cc}-16& -6\\ 7& 2\end{array}\right]$
Let $A=\left[\begin{array}{cc}1& -4\\ 3& -2\end{array}\right]$ , $B=\left[\begin{array}{cc}-16& -6\\ 7& 2\end{array}\right]$
So, $AX=B$
$\Rightarrow X=A^{-1}B$
Here, $|A|=10$
Now, $adjA=C^T={\left[\begin{array}{cc}-2& -3\\ 4& 1\end{array}\right]}^T$
$\Rightarrow adjA=\left[\begin{array}{cc}-2& 4\\ -3& 1\end{array}\right]$
$A^{-1}=\frac{1}{10}\left[\begin{array}{cc}-2& 4\\ -3& 1\end{array}\right]$
$\Rightarrow A^{-1}B=\frac{1}{10}\left[\begin{array}{cc}60& 20\\ 55& 20\end{array}\right]$
Hence, $X=\left[\begin{array}{cc}6& 2\\ \frac{11}{2}& 2\end{array}\right]$