If A=left[begin{array}{ll}2 & 3 1 & 2end{arr | vedclass

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(D) We know that for any invertible matrices $A$ and $B$,the property $(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1}$ holds true. Since $(A^{-1})

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$\left[\begin{array}{ll}1 & 0 \ 0 & 1\end{array}ight]$

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(D) We know that for any invertible matrices $A$ and $B$,the property $(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1}$ holds true. Since $(A^{-1})^{-1} = A$ and $(B^{-1})^{-1} = B$,the expression simplifies to $AB$. Now,we calculate the product $AB$: $AB = \left[\begin{array}{ll}2 & 3 \ 1 & 2\end{array}ight] \left[\begin{array}{cc}2 & -3 \ -1 & 2\end{array}ight]$ $AB = \left[\begin{array}{cc} (2)(2) + (3)(-1) & (2)(-3) + (3)(2) \ (1)(2) + (2)(-1) & (1)(-3) + (2)(2) \end{array}ight]$ $AB = \left[\begin{array}{cc} 4 - 3 & -6 + 6 \ 2 - 2 & -3 + 4 \end{array}ight]$ $AB = \left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}ight]$ Thus,the correct option is $D$.

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 $\left(B^{-1} A^{-1}\right)^{-1}=$

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