If b and c are non-zero real numbers,A = begi | vedclass

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(A) Given matrices $A = \begin{bmatrix} 1 & b & c \ b & 2 & 3 \ c & 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & b & c \ -b & 0 & 2 \ -c & -2

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$3$

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(A) Given matrices $A = \begin{bmatrix} 1 & b & c \ b & 2 & 3 \ c & 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & b & c \ -b & 0 & 2 \ -c & -2 & 0 \end{bmatrix}$.First,we calculate the sum $A+B$:$A+B = \begin{bmatrix} 1+0 & b+b & c+c \ b-b & 2+0 & 3+2 \ c-c & 3-2 & 4+0 \end{bmatrix} = \begin{bmatrix} 1 & 2b & 2c \ 0 & 2 & 5 \ 0 & 1 & 4 \end{bmatrix}$.Now,we find the determinant of the resulting matrix:$\det(A+B) = \begin{vmatrix} 1 & 2b & 2c \ 0 & 2 & 5 \ 0 & 1 & 4 \end{vmatrix}$.Expanding along the first column:$\det(A+B) = 1 	imes \begin{vmatrix} 2 & 5 \ 1 & 4 \end{vmatrix} - 0 + 0 = 1 	imes (2 	imes 4 - 5 	imes 1) = 1 	imes (8 - 5) = 3$.

If $b$ and $c$ are non-zero real numbers,$A = \begin{bmatrix} 1 & b & c \\ b & 2 & 3 \\ c & 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & b & c \\ -b & 0 & 2 \\ -c & -2 & 0 \end{bmatrix}$,then $\det(A+B) = $

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