Question

If $\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]A\left[\begin{array}{cc}-3& 5\\ 2& -4\end{array}\right]=\left[\begin{array}{cc}1& -5\\ 11& -27\end{array}\right]$, then the matrix $A$ is equal to

• A. $\frac{1}{4}\left[\begin{array}{cc}-4& 11\\ 8& 4\end{array}\right]$
• B. $\left[\begin{array}{cc}-1& 3\\ 1& 2\end{array}\right]$
• C. $\frac{1}{4}\left[\begin{array}{cc}-4& 11\\ 12& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right]$

Answer 1

The correct option is D $\left[\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right]$

$\text{Let}B=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],C=\left[\begin{array}{cc}-3& 5\\ 2& -4\end{array}\right]\text{and}D=\left[\begin{array}{cc}1& -5\\ 11& -27\end{array}\right]$

$BAC=D\Rightarrow AC=B^{-1}D$
$A=B^{-1}D{C}^{-1}$
$B^{-1}=-\frac{1}{2}\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right],C^{-1}=-\frac{1}{2}\left[\begin{array}{cc}4& 5\\ 2& 3\end{array}\right]$
$A=\frac{1}{4}\left[\begin{array}{cc}4& -2\\ -3& 1\end{array}\right]\left[\begin{array}{cc}1& -5\\ 11& -27\end{array}\right]\left[\begin{array}{cc}4& 5\\ 2& 3\end{array}\right]$
$=\frac{1}{4}\left[\begin{array}{cc}-18& 34\\ 8& -12\end{array}\right]\left[\begin{array}{cc}4& 5\\ 2& 3\end{array}\right]$
$=\frac{1}{4}\left[\begin{array}{cc}-4& 12\\ 8& 4\end{array}\right]=\left[\begin{array}{cc}-1& 3\\ 2& 1\end{array}\right]$