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. $-\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$

Answer 1

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

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