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

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(B) Given $A = \begin{bmatrix} 1 & -1 & 2 \ -2 & 3 & -3 \ 4 & -4 & 5 \end{bmatrix}$.Then $A^T = \begin{bmatrix} 1 & -2 & 4 \ -1 & 3 & -4 \ 2 & -3

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$\begin{bmatrix} 4 & -8 & 12 \ -8 & 16 & -28 \ 12 & -28 & 47 \end{bmatrix}$

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(B) Given $A = \begin{bmatrix} 1 & -1 & 2 \ -2 & 3 & -3 \ 4 & -4 & 5 \end{bmatrix}$.Then $A^T = \begin{bmatrix} 1 & -2 & 4 \ -1 & 3 & -4 \ 2 & -3 & 5 \end{bmatrix}$.First,calculate $AA^T$:$AA^T = \begin{bmatrix} 1 & -1 & 2 \ -2 & 3 & -3 \ 4 & -4 & 5 \end{bmatrix} \begin{bmatrix} 1 & -2 & 4 \ -1 & 3 & -4 \ 2 & -3 & 5 \end{bmatrix} = \begin{bmatrix} 6 & -11 & 18 \ -11 & 22 & -35 \ 18 & -35 & 57 \end{bmatrix}$.Next,calculate $A + A^T$:$A + A^T = \begin{bmatrix} 1+1 & -1-2 & 2+4 \ -2-1 & 3+3 & -3-4 \ 4+2 & -4-3 & 5+5 \end{bmatrix} = \begin{bmatrix} 2 & -3 & 6 \ -3 & 6 & -7 \ 6 & -7 & 10 \end{bmatrix}$.Finally,calculate $AA^T - (A + A^T)$:$\begin{bmatrix} 6 & -11 & 18 \ -11 & 22 & -35 \ 18 & -35 & 57 \end{bmatrix} - \begin{bmatrix} 2 & -3 & 6 \ -3 & 6 & -7 \ 6 & -7 & 10 \end{bmatrix} = \begin{bmatrix} 4 & -8 & 12 \ -8 & 16 & -28 \ 12 & -28 & 47 \end{bmatrix}$.Thus,the correct option is $B$.

If $A = \begin{bmatrix} 1 & -1 & 2 \\ -2 & 3 & -3 \\ 4 & -4 & 5 \end{bmatrix}$ is the given matrix and $A^T$ represents the transpose of $A$,then $AA^T - A - A^T =$

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