Question

Let $A=\begin{bmatrix}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{bmatrix}$ and $B=\left[b_{i j}\right]_{3 \times 3}$ with
$b_{11}=2, b_{13}=-2, b_{12}=0$ is such that
$A B=\begin{bmatrix}2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12\end{bmatrix}$, then $|B|+$ trace $(B)=$

• A. -2
• B. 10
• C. -8
• D. 6

Answer 1

Answer: (A)

We have, $A=\begin{bmatrix}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{bmatrix}$
and $B=\begin{bmatrix}b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{bmatrix}=\begin{bmatrix}2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{bmatrix}$
$\therefore A B=\begin{bmatrix}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{bmatrix}\begin{bmatrix}2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{bmatrix}$
$\begin{bmatrix}2+4 b_{21}+2 b_{31} & 0+4 b_{22}+2 b_{32} & -2+4 b_{23}+2 b_{33} \\ 4-b_{21}+4 b_{31} & 0+b_{22}+4 b_{32} & -4-b_{23}+4 b_{33} \\ -6+7 b_{21}-6 b_{31} & 0+7 b_{22}-6 b_{32} & 6+7 b_{23}-6 b_{33}\end{bmatrix}$
$=\begin{bmatrix}2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12\end{bmatrix}$
On solving above equal matrices with corresponding elements,
we get $b_{21}=b_{31}=0, b_{22}=3, b_{32}=1, b_{23}=0$ and $b_{33}=-1$
$\therefore B=\begin{bmatrix}2 & 0 & -2 \\ 0 & 3 & 0 \\ 0 & 1 & -1\end{bmatrix}$
$\therefore |B|=2(-3-0)+(-2)(0-0)=-6$
and Trace $(B)=2+3-1=4$
$\therefore |B|+$ trace $(B)=-6+4=-2$