Question

If $3A+4B^{\prime }$ =$\left[\begin{array}{ccc}7& -10& 17\\ 0& 6& 31\end{array}\right]$ and $2B-3A^{\prime }$=$\left[\begin{array}{cc}-1& 18\\ 4& 0\\ -5& -7\end{array}\right]$ then $B$=

• A. $\left[\begin{array}{cc}-1& -18\\ 4& -16\\ -5& -7\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 3\\ -1& 1\\ 2& 4\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 3\\ -1& 1\\ 2& -4\end{array}\right]$
• D. $\left[\begin{array}{cc}1& -3\\ -1& 1\\ 2& 4\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{cc}1& 3\\ -1& 1\\ 2& 4\end{array}\right]$

$3A+4B^{\prime }$ =$\left[\begin{array}{ccc}7& -10& 17\\ 0& 6& 31\end{array}\right]$..............(1)

$2B-3A^{\prime }=\left[\begin{array}{cc}-1& 18\\ 4& 0\\ -5& -7\end{array}\right]$
$(2B-3A^{\prime }{)}^{\prime }=\left[\begin{array}{ccc}-1& 4& -5\\ 18& 0& -7\end{array}\right]=2B^{\prime }-3A..........(2)$
By adding (1) and (2) we have
$6B^{\prime }$ =$\left[\begin{array}{ccc}6& -6& 12\\ 18& 6& 24\end{array}\right]$

$(B^{\prime }{)}^{\prime }$ =$\frac{1}{6}{\left[\begin{array}{ccc}6& -6& 12\\ 18& 6& 24\end{array}\right]}^{\prime }=$ $\left[\begin{array}{cc}1& 3\\ -1& 1\\ 2& 4\end{array}\right]$