If (BA)^{-1} = C where B = begin{bmatrix} 2 & | vedclass

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(B) Given $(BA)^{-1} = C$. Using the property of inverse matrices $(BA)^{-1} = A^{-1}B^{-1}$,we have $A^{-1}B^{-1} = C$. Multiplying both sides by $B$

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

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(B) Given $(BA)^{-1} = C$. Using the property of inverse matrices $(BA)^{-1} = A^{-1}B^{-1}$,we have $A^{-1}B^{-1} = C$. Multiplying both sides by $B$ on the right,we get $A^{-1}B^{-1}B = CB$. Since $B^{-1}B = I$,it follows that $A^{-1} = CB$. Now,calculate the product $CB$: $A^{-1} = \begin{bmatrix} -1 & 0 & 1 \ 1 & 1 & 3 \ 2 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 6 & 4 \ 1 & 0 & 1 \ -1 & 1 & -1 \end{bmatrix}$ $A^{-1} = \begin{bmatrix} (-1)(2) + (0)(1) + (1)(-1) & (-1)(6) + (0)(0) + (1)(1) & (-1)(4) + (0)(1) + (1)(-1) \ (1)(2) + (1)(1) + (3)(-1) & (1)(6) + (1)(0) + (3)(1) & (1)(4) + (1)(1) + (3)(-1) \ (2)(2) + (0)(1) + (2)(-1) & (2)(6) + (0)(0) + (2)(1) & (2)(4) + (0)(1) + (2)(-1) \end{bmatrix}$ $A^{-1} = \begin{bmatrix} -3 & -5 & -5 \ 0 & 9 & 2 \ 2 & 14 & 6 \end{bmatrix}$

If $(BA)^{-1} = C$ where $B = \begin{bmatrix} 2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1 \end{bmatrix}$ and $C = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix}$,then $A^{-1}$ is given by

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