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(D) Given $A = \begin{bmatrix} 1 & 2 & 3 \ 3 & 1 & 2 \ 2 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -5 & 7 & 1 \ 1 & -5 & 7 \ 7 & 1 & -5 \end

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$18I_3$

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(D) Given $A = \begin{bmatrix} 1 & 2 & 3 \ 3 & 1 & 2 \ 2 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -5 & 7 & 1 \ 1 & -5 & 7 \ 7 & 1 & -5 \end{bmatrix}$.Calculating the product $AB$:$AB = \begin{bmatrix} 1 & 2 & 3 \ 3 & 1 & 2 \ 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} -5 & 7 & 1 \ 1 & -5 & 7 \ 7 & 1 & -5 \end{bmatrix}$Performing matrix multiplication:Row $1$: $(1)(-5) + (2)(1) + (3)(7) = -5 + 2 + 21 = 18$; $(1)(7) + (2)(-5) + (3)(1) = 7 - 10 + 3 = 0$; $(1)(1) + (2)(7) + (3)(-5) = 1 + 14 - 15 = 0$.Row $2$: $(3)(-5) + (1)(1) + (2)(7) = -15 + 1 + 14 = 0$; $(3)(7) + (1)(-5) + (2)(1) = 21 - 5 + 2 = 18$; $(3)(1) + (1)(7) + (2)(-5) = 3 + 7 - 10 = 0$.Row $3$: $(2)(-5) + (3)(1) + (1)(7) = -10 + 3 + 7 = 0$; $(2)(7) + (3)(-5) + (1)(1) = 14 - 15 + 1 = 0$; $(2)(1) + (3)(7) + (1)(-5) = 2 + 21 - 5 = 18$.Thus,$AB = \begin{bmatrix} 18 & 0 & 0 \ 0 & 18 & 0 \ 0 & 0 & 18 \end{bmatrix} = 18 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = 18I_3$.

If $A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -5 & 7 & 1 \\ 1 & -5 & 7 \\ 7 & 1 & -5 \end{bmatrix}$,then $AB$ is equal to

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