If A=left[begin{array}{lll}1 & 2 & 3 4 & 3 & | vedclass

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(A) Given $A=\left[\begin{array}{lll}1 & 2 & 3 \ 4 & 3 & 2 \ 3 & 4 & 5\end{array}ight]$.First,find the transpose $A^T$:$A^T=\left[\begin{array}{ll

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$4\left[\begin{array}{lll}3 & 2 & -3 \ 3 & 0 & -3 \ 3 & 2 & -3\end{array}ight]$

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(A) Given $A=\left[\begin{array}{lll}1 & 2 & 3 \ 4 & 3 & 2 \ 3 & 4 & 5\end{array}ight]$.First,find the transpose $A^T$:$A^T=\left[\begin{array}{lll}1 & 4 & 3 \ 2 & 3 & 4 \ 3 & 2 & 5\end{array}ight]$.Now,calculate $(A+A^T)$:$A+A^T=\left[\begin{array}{lll}1+1 & 2+4 & 3+3 \ 4+2 & 3+3 & 2+4 \ 3+3 & 4+2 & 5+5\end{array}ight]=\left[\begin{array}{lll}2 & 6 & 6 \ 6 & 6 & 6 \ 6 & 6 & 10\end{array}ight]$.Next,calculate $(A-A^T)$:$A-A^T=\left[\begin{array}{lll}1-1 & 2-4 & 3-3 \ 4-2 & 3-3 & 2-4 \ 3-3 & 4-2 & 5-5\end{array}ight]=\left[\begin{array}{lll}0 & -2 & 0 \ 2 & 0 & -2 \ 0 & 2 & 0\end{array}ight]$.Finally,multiply the two matrices:$(A+A^T)(A-A^T)=\left[\begin{array}{lll}2 & 6 & 6 \ 6 & 6 & 6 \ 6 & 6 & 10\end{array}ight]\left[\begin{array}{lll}0 & -2 & 0 \ 2 & 0 & -2 \ 0 & 2 & 0\end{array}ight]$$= \left[\begin{array}{lll}0+12+0 & -4+0+12 & 0-12+0 \ 0+12+0 & -12+0+12 & 0-12+0 \ 0+12+0 & -12+0+20 & 0-12+0\end{array}ight]$$= \left[\begin{array}{lll}12 & 8 & -12 \ 12 & 0 & -12 \ 12 & 8 & -12\end{array}ight] = 4\left[\begin{array}{lll}3 & 2 & -3 \ 3 & 0 & -3 \ 3 & 2 & -3\end{array}ight]$.Thus,the correct option is $A$.

If $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 3 & 2 \\ 3 & 4 & 5\end{array}\right]$,then $(A+A^T)(A-A^T)=$

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