Question

If $\left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]A\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, then $A=$

• A. $\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]$
• D. $\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$

Given:

$\left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]A\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
Let, $P=\left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right],Q=\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]$
$\Rightarrow PAQ=I$
$\Rightarrow PA=Q^{-1}$
$\Rightarrow A=P^{-1}{Q}^{-1}$
Now, since we have $P=\left[\begin{array}{cc}2& 1\\ 3& 2\end{array}\right]$
Also, $|P|=1$
$adjP=C^T={\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]}^T$
$\Rightarrow adjP=\left[\begin{array}{cc}2& -1\\ -3& 2\end{array}\right]$
Hence, $P^{-1}=\left[\begin{array}{cc}2& -1\\ -3& 2\end{array}\right]$
Now, we have $Q=\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]$
Also, $|Q|=-1$
$adjQ=C^T={\left[\begin{array}{cc}-3& -5\\ -2& -3\end{array}\right]}^T$
$\Rightarrow adjQ=\left[\begin{array}{cc}-3& -2\\ -5& -3\end{array}\right]$
Hence, $Q^{-1}=\left[\begin{array}{cc}3& 2\\ 5& 3\end{array}\right]$
Hence, $A=\left[\begin{array}{cc}2& -1\\ -3& 2\end{array}\right]\left[\begin{array}{cc}3& 2\\ 5& 3\end{array}\right]$
$\Rightarrow A=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$

Hence, option A.