If left{ begin{bmatrix} 3 & 1 & 2 8 & 9 & 5 | vedclass

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(A) Let $A = \begin{bmatrix} 3 & 1 & 2 \ 8 & 9 & 5 \ 1 & 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 & 3 \ 3 & 2 & 7 \ 3 & 7 & 9 \end{bmat

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(A) Let $A = \begin{bmatrix} 3 & 1 & 2 \ 8 & 9 & 5 \ 1 & 1 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 & 3 \ 3 & 2 & 7 \ 3 & 7 & 9 \end{bmatrix}$.Note that $A^T = \begin{bmatrix} 3 & 8 & 1 \ 1 & 9 & 1 \ 2 & 5 & 3 \end{bmatrix}$.The given expression is $X = (ABA^T)^2$.Since $B$ is a symmetric matrix $(B^T = B)$,the matrix $ABA^T$ is also symmetric because $(ABA^T)^T = (A^T)^T B^T A^T = ABA^T$.If a matrix $M$ is symmetric,then $M^2$ is also symmetric.Thus,$X = (ABA^T)^2$ is a symmetric matrix.For a symmetric matrix $X = \begin{bmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{bmatrix}$,we have $a_2 = b_1$,$a_3 = c_1$,and $b_3 = c_2$.Therefore,$|a_2 - b_1| + |a_3 - c_1| + |b_3 - c_2| = |0| + |0| + |0| = 0$.

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