Let A=begin{bmatrix} 1 & 4 & 2 2 & -1 & 4 - | vedclass

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(A) We have $A=\begin{bmatrix} 1 & 4 & 2 \ 2 & -1 & 4 \ -3 & 7 & -6 \end{bmatrix}$ and $B=\begin{bmatrix} 2 & 0 & -2 \ b_{21} & b_{22} & b_{23} \

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(A) We have $A=\begin{bmatrix} 1 & 4 & 2 \ 2 & -1 & 4 \ -3 & 7 & -6 \end{bmatrix}$ and $B=\begin{bmatrix} 2 & 0 & -2 \ b_{21} & b_{22} & b_{23} \ b_{31} & b_{32} & b_{33} \end{bmatrix}$.Given $AB=\begin{bmatrix} 2 & 14 & -4 \ 4 & 1 & -8 \ -6 & 15 & 12 \end{bmatrix}$.Performing matrix multiplication:$AB = \begin{bmatrix} 2+4b_{21}+2b_{31} & 4b_{22}+2b_{32} & -2+4b_{23}+2b_{33} \ 4-b_{21}+4b_{31} & -b_{22}+4b_{32} & -4-b_{23}+4b_{33} \ -6+7b_{21}-6b_{31} & 7b_{22}-6b_{32} & 6+7b_{23}-6b_{33} \end{bmatrix}$.Equating corresponding elements:From column $1$: $2+4b_{21}+2b_{31}=2 \implies 2b_{21}+b_{31}=0$; $4-b_{21}+4b_{31}=4 \implies -b_{21}+4b_{31}=0$; $-6+7b_{21}-6b_{31}=-6 \implies 7b_{21}-6b_{31}=0$. Solving these gives $b_{21}=0, b_{31}=0$.From column $2$: $4b_{22}+2b_{32}=14$; $-b_{22}+4b_{32}=1$; $7b_{22}-6b_{32}=15$. Solving these gives $b_{22}=3, b_{32}=1$.From column $3$: $-2+4b_{23}+2b_{33}=-4$; $-4-b_{23}+4b_{33}=-8$; $6+7b_{23}-6b_{33}=12$. Solving these gives $b_{23}=0, b_{33}=-1$.Thus, $B=\begin{bmatrix} 2 & 0 & -2 \ 0 & 3 & 0 \ 0 & 1 & -1 \end{bmatrix}$.$|B| = 2(-3-0) - 0 + (-2)(0-0) = -6$.$\operatorname{trace}(B) = 2+3-1 = 4$.Therefore, $|B|+\operatorname{trace}(B) = -6+4 = -2$.

List-$I$	List-$II$
$(A)$ $\lambda=8, \mu \neq 15$	$1$. Infinitely many solutions
$(B)$ $\lambda \neq 8, \mu \in R$	$2$. No solution
$(C)$ $\lambda=8, \mu=15$	$3$. Unique solution

Let $A=\begin{bmatrix} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6 \end{bmatrix}$ and $B=[b_{ij}]_{3 \times 3}$ with $b_{11}=2, b_{13}=-2, b_{12}=0$ such that $AB=\begin{bmatrix} 2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12 \end{bmatrix}$. Then $|B|+\operatorname{trace}(B)=$

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