Question

if $\left[\begin{array}{cc}2& 1\\ 7& 4\end{array}\right]$A$\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, then matrix A equals

• A. $\left[\begin{array}{cc}7& 5\\ -11& -8\end{array}\right]$
• B. $\left[\begin{array}{cc}2& 1\\ 5& 3\end{array}\right]$
• C. $\left[\begin{array}{cc}7& 34\\ 1& 5\end{array}\right]$
• D. $\left[\begin{array}{cc}5& 13\\ 3& 8\end{array}\right]$

Answer 1

The correct option is A $\left[\begin{array}{cc}7& 5\\ -11& -8\end{array}\right]$

$\left[\begin{array}{cc}2& 1\\ 7& 4\end{array}\right]$A$\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$P=\left[\begin{array}{cc}2& 1\\ 7& 4\end{array}\right],Q=\left[\begin{array}{cc}-3& 2\\ 5& -3\end{array}\right],R=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$PAQ=R\Rightarrow A=P^{-1}R{Q}^{-1}$

$\Rightarrow A=P^{-1}{Q}^{-1}=(QP{)}^{-1}$

$QP=\left[\begin{array}{cc}8& 5\\ -11& -7\end{array}\right]$

$\therefore A=(QP{)}^{-1}=\left[\begin{array}{cc}7& 5\\ -11& -8\end{array}\right]$

Hence, option A.