If A = begin{bmatrix} 3 & 2 & 6 1 & 1 & 2 2 | vedclass

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(D) Step $1$: Given $XA = B^T$, multiply by $A^{-1}$ on the right: $X = B^T A^{-1}$.Step $2$: Given $A^{-1}Y = B$, we have $Y = AB$.Step $3$: Calculat

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$[2]$

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(D) Step $1$: Given $XA = B^T$, multiply by $A^{-1}$ on the right: $X = B^T A^{-1}$.Step $2$: Given $A^{-1}Y = B$, we have $Y = AB$.Step $3$: Calculate $XY = (B^T A^{-1})(AB) = B^T (A^{-1}A) B = B^T I B = B^T B$.Step $4$: $B^T = \begin{bmatrix} 1 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}$.Step $5$: $XY = \begin{bmatrix} 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix} = (1 	imes 1) + (0 	imes 0) + (1 	imes 1) = 1 + 0 + 1 = [2]$.

If $A = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$ such that $XA = B^T$ and $A^{-1}Y = B$, then find the value of $XY$.

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