Question

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

• A. $\left(\begin{array}{cc}1& 3\\ -1& 0\\ -2& -4\end{array}\right)$
• B. $\left(\begin{array}{cc}1& 3\\ 1& 0\\ 2& 4\end{array}\right)$
• C. $\left(\begin{array}{cc}1& 3\\ -1& 0\\ 2& 4\end{array}\right)$
• D. $\left(\begin{array}{cc}1& -3\\ 1& 0\\ 2& 4\end{array}\right)$

Answer 1

The correct option is C $\left(\begin{array}{cc}1& 3\\ -1& 0\\ 2& 4\end{array}\right)$

Given,

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

Taking Transpose of above equation

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

$3A^T+4B=\left(\begin{array}{cc}7& 0\\ -10& 6\\ 17& 31\end{array}\right)-\left(1\right)$

Also

$2B-3A^T=\left(\begin{array}{cc}-1& 18\\ 4& -6\\ -5& -7\end{array}\right)-\left(2\right)$

On Adding eqs.(1) and (2) we get,

$6B=\left(\begin{array}{cc}6& 18\\ -6& 0\\ 12& 24\end{array}\right)⟹B=\left(\begin{array}{cc}1& 3\\ -1& 0\\ 2& 4\end{array}\right)$