Question

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

• A. -2
• B. 10
• C. -8
• D. 6

Answer 1

Answer: (A)

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