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 $A$ is equal to

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

Answer 1

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

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

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