Let A=begin{bmatrix} a & 3 & 5 5 & -1 & 3 2 | vedclass

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(C) Given: $A=\begin{bmatrix} a & 3 & 5 \ 5 & -1 & 3 \ 2 & 3 & -4 \end{bmatrix}$ and $B=\begin{bmatrix} b & 1 & 4 \ 4 & c & 1 \ -3 & 1 & d \end{bm

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$3$

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(C) Given: $A=\begin{bmatrix} a & 3 & 5 \ 5 & -1 & 3 \ 2 & 3 & -4 \end{bmatrix}$ and $B=\begin{bmatrix} b & 1 & 4 \ 4 & c & 1 \ -3 & 1 & d \end{bmatrix}$.Since the trace of $A$ is $-4$, we have $a - 1 - 4 = -4$, which implies $a = 1$.Now, compute the product $AB$:$AB = \begin{bmatrix} 1 & 3 & 5 \ 5 & -1 & 3 \ 2 & 3 & -4 \end{bmatrix} \begin{bmatrix} b & 1 & 4 \ 4 & c & 1 \ -3 & 1 & d \end{bmatrix} = \begin{bmatrix} b+12-15 & 1+3c+5 & 4+3+5d \ 5b-4-9 & 5-c+3 & 20-1+3d \ 2b+12+12 & 2+3c-4 & 8+3-4d \end{bmatrix} = \begin{bmatrix} b-3 & 3c+6 & 5d+7 \ 5b-13 & 8-c & 3d+19 \ 2b+24 & 3c-2 & 11-4d \end{bmatrix}$.Comparing this with the given matrix $\begin{bmatrix} -1 & 0 & 17 \ -3 & 10 & 25 \ 28 & -8 & 3 \end{bmatrix}$:From $b-3 = -1$, we get $b = 2$.From $3c+6 = 0$, we get $c = -2$.From $5d+7 = 17$, we get $5d = 10$, so $d = 2$.Thus, $a+b+c+d = 1 + 2 - 2 + 2 = 3$.

Let $A=\begin{bmatrix} a & 3 & 5 \\ 5 & -1 & 3 \\ 2 & 3 & -4 \end{bmatrix}$ and $B=\begin{bmatrix} b & 1 & 4 \\ 4 & c & 1 \\ -3 & 1 & d \end{bmatrix}$. If the trace of $A$ is $-4$ and $AB=\begin{bmatrix} -1 & 0 & 17 \\ -3 & 10 & 25 \\ 28 & -8 & 3 \end{bmatrix}$, then $a+b+c+d=$

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