Question

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

• A. $\frac{1}{10}\left[\begin{array}{cc}2& 2\\ -3& 2\end{array}\right]$
• B. $\frac{1}{10}\left[\begin{array}{cc}2& -2\\ 2& 3\end{array}\right]$
• C. $\frac{1}{10}\left[\begin{array}{cc}2& 2\\ -2& 3\end{array}\right]$
• D. $\frac{1}{10}\left[\begin{array}{cc}-2& 2\\ 2& 3\end{array}\right]$

Answer 1

The correct option is B $\frac{1}{10}\left[\begin{array}{cc}2& -2\\ 2& 3\end{array}\right]$

$(AB{)}^{-1}=B^{-1}{A}^{-1}$

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

$\therefore {\left(AB\right)}^{-1}=\frac{1}{10}\left[\begin{array}{cc}2& -2\\ 2& 3\end{array}\right]$

Hence, $B^{-1}{A}^{-1}=\frac{1}{10}\left[\begin{array}{cc}2& -2\\ 2& 3\end{array}\right]$