If A = begin{bmatrix} 1 & 2 & 3 4 & 5 & 6 7 | vedclass

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(D) Given $A = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix}$.First,find the transpose $A'$ by interchanging rows and columns:$A'

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$\begin{bmatrix} 14 & 32 & 50 \ 32 & 77 & 122 \ 50 & 122 & 194 \end{bmatrix}$

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(D) Given $A = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix}$.First,find the transpose $A'$ by interchanging rows and columns:$A' = \begin{bmatrix} 1 & 4 & 7 \ 2 & 5 & 8 \ 3 & 6 & 9 \end{bmatrix}$.Now,calculate the product $AA'$:$AA' = \begin{bmatrix} 1 & 2 & 3 \ 4 & 5 & 6 \ 7 & 8 & 9 \end{bmatrix} \begin{bmatrix} 1 & 4 & 7 \ 2 & 5 & 8 \ 3 & 6 & 9 \end{bmatrix} = \begin{bmatrix} (1+4+9) & (4+10+18) & (7+16+27) \ (4+10+18) & (16+25+36) & (28+40+54) \ (7+16+27) & (28+40+54) & (49+64+81) \end{bmatrix} = \begin{bmatrix} 14 & 32 & 50 \ 32 & 77 & 122 \ 50 & 122 & 194 \end{bmatrix}$.We know that for any matrix $B$,$(B')' = B$. Since $AA'$ is a symmetric matrix (as $(AA')' = (A')'A' = AA'$),we have $(AA')' = AA'$.Therefore,$(AA')' = \begin{bmatrix} 14 & 32 & 50 \ 32 & 77 & 122 \ 50 & 122 & 194 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$,then $(AA')' = $

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