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(A) Given $A = \begin{bmatrix} b & a & 0 \ c & 0 & b \ a & a & b \end{bmatrix}$ and $B = \begin{bmatrix} 0 & a & b \ b & 0 & c \ b & a & a \end{bm

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$14$

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(A) Given $A = \begin{bmatrix} b & a & 0 \ c & 0 & b \ a & a & b \end{bmatrix}$ and $B = \begin{bmatrix} 0 & a & b \ b & 0 & c \ b & a & a \end{bmatrix}$.Performing matrix multiplication $AB$:$AB = \begin{bmatrix} b(0)+a(b)+0(b) & b(a)+a(0)+0(a) & b(b)+a(c)+0(a) \ c(0)+0(b)+b(b) & c(a)+0(0)+b(a) & c(b)+0(c)+b(a) \ a(0)+a(b)+b(b) & a(a)+a(0)+b(a) & a(b)+a(c)+b(a) \end{bmatrix} = \begin{bmatrix} ab & ab & b^2+ac \ b^2 & ac+ab & bc+ab \ ab+b^2 & a^2+ab & 2ab+ac \end{bmatrix}$.Equating this to the given matrix $\begin{bmatrix} 2 & 2 & 7 \ 1 & 8 & 5 \ 3 & 6 & 10 \end{bmatrix}$:From the element at $(2,1)$, $b^2 = 1$.From the element at $(1,1)$, $ab = 2$.From the element at $(1,3)$, $b^2+ac = 7 \implies 1+ac = 7 \implies ac = 6$.Since $ab=2$ and $b^2=1$, we have $a = 2/b$. If $b=1$, $a=2$. If $b=-1$, $a=-2$.Case $1$: $b=1, a=2$. Then $ac=6 \implies 2c=6 \implies c=3$.Case $2$: $b=-1, a=-2$. Then $ac=6 \implies -2c=6 \implies c=-3$.In both cases, $a^2+b^2+c^2 = (\pm 2)^2 + (\pm 1)^2 + (\pm 3)^2 = 4+1+9 = 14$.

If $A=\begin{bmatrix} b & a & 0 \\ c & 0 & b \\ a & a & b \end{bmatrix}$ and $B=\begin{bmatrix} 0 & a & b \\ b & 0 & c \\ b & a & a \end{bmatrix}$ are two matrices such that $AB=\begin{bmatrix} 2 & 2 & 7 \\ 1 & 8 & 5 \\ 3 & 6 & 10 \end{bmatrix}$, then $a^2+b^2+c^2=$

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