If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & -3 \\ -1 &…
If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$, then $\left(B^{-1} A^{-1}\right)^{-1}=$
- $\left[\begin{array}{cc}2 & 3 \\ 1 & -2\end{array}\right]$
- $\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
- $\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
- $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Solution
$\left(\mathrm{B}^{-1} \mathrm{~A}^{-1}\right)^{-1}=\left(\mathrm{A}^{-1}\right)^{-1}\left(\mathrm{~B}^{-1}\right)^{-1}=\mathrm{AB}$
$\therefore \quad \mathrm{AB}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]=\left[\begin{array}{cc}4-3 & -6+6 \\ 2-2 & -3+4\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Asked in: MHT CET 2020 (13 Oct Shift 1)
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