Let B=begin{bmatrix} 2 & 6 & 4 1 & 0 & 1 -1 | vedclass

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(D) Given $BAC = I$. Taking the inverse of both sides,we get $(BAC)^{-1} = I^{-1} = I$. Using the property $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$,we have $

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$\begin{bmatrix} -3 & -5 & -5 \ 0 & 9 & 2 \ 2 & 14 & 6 \end{bmatrix}$

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(D) Given $BAC = I$. Taking the inverse of both sides,we get $(BAC)^{-1} = I^{-1} = I$. Using the property $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$,we have $C^{-1}A^{-1}B^{-1} = I$. Multiplying by $C$ on the left and $B$ on the right,we get $A^{-1} = CB$. Now,calculate the product $CB$: $A^{-1} = \begin{bmatrix} -1 & 0 & 1 \ 1 & 1 & 3 \ 2 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 6 & 4 \ 1 & 0 & 1 \ -1 & 1 & -1 \end{bmatrix}$ $A^{-1} = \begin{bmatrix} (-1)(2)+(0)(1)+(1)(-1) & (-1)(6)+(0)(0)+(1)(1) & (-1)(4)+(0)(1)+(1)(-1) \ (1)(2)+(1)(1)+(3)(-1) & (1)(6)+(1)(0)+(3)(1) & (1)(4)+(1)(1)+(3)(-1) \ (2)(2)+(0)(1)+(2)(-1) & (2)(6)+(0)(0)+(2)(1) & (2)(4)+(0)(1)+(2)(-1) \end{bmatrix}$ $A^{-1} = \begin{bmatrix} -3 & -5 & -5 \ 0 & 9 & 2 \ 2 & 14 & 6 \end{bmatrix}$.

Let $B=\begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$ and $C=\begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix}$. If a matrix $A$ is such that $BAC=I$,then $A^{-1}=$

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