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Let $A=\left[\begin{array}{ccc}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{array}\right]$ and $B=\left[b_{i j}\right]_{3 \times 3}$ with
$b_{11}=2, b_{13}=-2, b_{12}=0$ is such that $A B=\left[\begin{array}{ccc}2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12\end{array}\right]$, then $|B|+\operatorname{trace}(B)=$
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$b_{11}=2, b_{13}=-2, b_{12}=0$ is such that $A B=\left[\begin{array}{ccc}2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12\end{array}\right]$, then $|B|+\operatorname{trace}(B)=$
Solution:
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Verified Answer
The correct answer is:
-2
We have, $A=\left[\begin{array}{ccc}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{array}\right]$
and $B=\left[\begin{array}{lll}b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right]=\left[\begin{array}{ccc}2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right]$
$\therefore A B=\left[\begin{array}{ccc}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{array}\right]\left[\begin{array}{ccc}2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}2+4 b_{21}+2 b_{31} & 0+4 b_{22}+2 b_{32} & -2+4 b_{23}+2 b_{33} \\ 4-b_{21}+4 b_{31} & 0+b_{22}+4 b_{32} & -4-b_{23}+4 b_{33} \\ -6+7 b_{21}-6 b_{31} & 0+7 b_{22}-6 b_{32} & 6+7 b_{23}-6 b_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12\end{array}\right]$
On solving above equal matrices with corresponding elements, we get
$b_{21}=b_{31}=0, b_{22}=3, b_{32}=1, b_{23}=0 \text { and } b_{33}=-1$
$\therefore \quad B=\left[\begin{array}{ccc}2 & 0 & -2 \\ 0 & 3 & 0 \\ 0 & 1 & -1\end{array}\right]$
$\therefore|B|=2(-3-0)+(-2)(0-0)=-6$
and Trace $(B)=2+3-1=4$
$\therefore|B|+\operatorname{trace}(B)=-6+4=-2$
and $B=\left[\begin{array}{lll}b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right]=\left[\begin{array}{ccc}2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right]$
$\therefore A B=\left[\begin{array}{ccc}1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & -6\end{array}\right]\left[\begin{array}{ccc}2 & 0 & -2 \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}2+4 b_{21}+2 b_{31} & 0+4 b_{22}+2 b_{32} & -2+4 b_{23}+2 b_{33} \\ 4-b_{21}+4 b_{31} & 0+b_{22}+4 b_{32} & -4-b_{23}+4 b_{33} \\ -6+7 b_{21}-6 b_{31} & 0+7 b_{22}-6 b_{32} & 6+7 b_{23}-6 b_{33}\end{array}\right]$
$=\left[\begin{array}{ccc}2 & 14 & -4 \\ 4 & 1 & -8 \\ -6 & 15 & 12\end{array}\right]$
On solving above equal matrices with corresponding elements, we get
$b_{21}=b_{31}=0, b_{22}=3, b_{32}=1, b_{23}=0 \text { and } b_{33}=-1$
$\therefore \quad B=\left[\begin{array}{ccc}2 & 0 & -2 \\ 0 & 3 & 0 \\ 0 & 1 & -1\end{array}\right]$
$\therefore|B|=2(-3-0)+(-2)(0-0)=-6$
and Trace $(B)=2+3-1=4$
$\therefore|B|+\operatorname{trace}(B)=-6+4=-2$
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