If P = begin{bmatrix} 1 & 2 & 3 2 & 3 & 4 3 | vedclass

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(A) Let $A = \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 4 \ 3 & 4 & 5 \end{bmatrix}$,$B = \begin{bmatrix} -1 & -2 \ -2 & 0 \ 0 & -4 \end{bmatrix}$,and $C

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$40$

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(A) Let $A = \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 4 \ 3 & 4 & 5 \end{bmatrix}$,$B = \begin{bmatrix} -1 & -2 \ -2 & 0 \ 0 & -4 \end{bmatrix}$,and $C = \begin{bmatrix} -4 & -5 & -6 \ 0 & 0 & 1 \end{bmatrix}$.First,calculate $D = A 	imes B$:$D = \begin{bmatrix} (1)(-1)+(2)(-2)+(3)(0) & (1)(-2)+(2)(0)+(3)(-4) \ (2)(-1)+(3)(-2)+(4)(0) & (2)(-2)+(3)(0)+(4)(-4) \ (3)(-1)+(4)(-2)+(5)(0) & (3)(-2)+(4)(0)+(5)(-4) \end{bmatrix} = \begin{bmatrix} -5 & -14 \ -8 & -20 \ -11 & -26 \end{bmatrix}$.Now,calculate $P = D 	imes C$:$P = \begin{bmatrix} -5 & -14 \ -8 & -20 \ -11 & -26 \end{bmatrix} \begin{bmatrix} -4 & -5 & -6 \ 0 & 0 & 1 \end{bmatrix}$.The element $P_{22}$ is the product of the second row of $D$ and the second column of $C$:$P_{22} = (-8)(-5) + (-20)(0) = 40 + 0 = 40$.

If $P = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{bmatrix} \begin{bmatrix} -1 & -2 \\ -2 & 0 \\ 0 & -4 \end{bmatrix} \begin{bmatrix} -4 & -5 & -6 \\ 0 & 0 & 1 \end{bmatrix}$,then $P_{22} = $

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