Question

If $2A+3B =\begin{bmatrix} {2}&{-1} &{4}\\ {3}&{2}& {5} \\ \end{bmatrix} $ and $A+2B \begin{bmatrix} {5}&{0} &{3}\\ {1}&{6}& {2} \\ \end{bmatrix} $then $B =$

• A. $\begin{bmatrix} {8}&{1} &{2}\\ {1}&{10}& {1} \\ \end{bmatrix} $
• B. $\begin{bmatrix} {8}&{1} &{-2}\\ {-1}&{10}& {-1} \\ \end{bmatrix} $
• C. $\begin{bmatrix} {8}&{1} &{2}\\ {-1}&{10}& {-1} \\ \end{bmatrix} $
• D. $\begin{bmatrix} {8}&{-1} &{2}\\ {-1}&{10}& {-1} \\ \end{bmatrix} $

Answer 1

Answer: (C)

We have
$2A + 3B = \begin{bmatrix}2&-1&4\\ 3&2&5\end{bmatrix}\,\,\,\,\, \dots(i)$
and $ A + 2B = \begin{bmatrix}5&0&3\\ 1&6&2\end{bmatrix} \,\,\,\,\,\dots(ii)$
Multiply Eq. (ii) by 2 and subtracting Eq.(i) from (ii), we get
$ B = 2 \begin{bmatrix}5&0&3\\ 1&6&2\end{bmatrix} -\begin{bmatrix}2&-1&4\\ 3&2&5\end{bmatrix} $
$= \begin{bmatrix}8&1&2\\ -1&10&-1\end{bmatrix}$