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(B) Given $X^T B^T = A^T$. Taking the transpose on both sides,we get $(X^T B^T)^T = (A^T)^T$,which implies $BX = A$. First,calculate the determinant o

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$4$

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(B) Given $X^T B^T = A^T$. Taking the transpose on both sides,we get $(X^T B^T)^T = (A^T)^T$,which implies $BX = A$. First,calculate the determinant of $B$: $|B| = 3(0 - (-8)) - 5(0 - 48) - 7(0 - (-6)) = 3(8) - 5(-48) - 7(6) = 24 + 240 - 42 = 222$. The adjoint of $B$ is $	ext{adj}(B) = \begin{bmatrix} 8 & 7 & 33 \ 48 & 42 & -24 \ 6 & 33 & -3 \end{bmatrix}$. Thus,$X = B^{-1}A = \frac{1}{222} \begin{bmatrix} 8 & 7 & 33 \ 48 & 42 & -24 \ 6 & 33 & -3 \end{bmatrix} \begin{bmatrix} 0 \ -6 \ 8 \end{bmatrix}$. $X = \frac{1}{222} \begin{bmatrix} 0 - 42 + 264 \ 0 - 252 - 192 \ 0 - 198 - 24 \end{bmatrix} = \frac{1}{222} \begin{bmatrix} 222 \ -444 \ -222 \end{bmatrix} = \begin{bmatrix} 1 \ -2 \ -1 \end{bmatrix}$. So,$D = \begin{bmatrix} 1 \ -2 \ -1 \end{bmatrix}$. Finally,$D^T A = \begin{bmatrix} 1 & -2 & -1 \end{bmatrix} \begin{bmatrix} 0 \ -6 \ 8 \end{bmatrix} = (1)(0) + (-2)(-6) + (-1)(8) = 0 + 12 - 8 = 4$.

Let $A=\begin{bmatrix} 0 \\ -6 \\ 8 \end{bmatrix}$,$B=\begin{bmatrix} 3 & 5 & -7 \\ 0 & -1 & 8 \\ 6 & -1 & 0 \end{bmatrix}$ and $X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$. If $D=[\alpha, \beta, \gamma]^{T}$ is the solution of $X^{T} B^{T}=A^{T}$,then $D^{T} A=$

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