Q. If $3A + 4B' = $$\begin{bmatrix}7&-10&17\\ 0&6&31\end{bmatrix} $ and $2B - 3A'$ $ \begin{bmatrix}-1&18\\ 4&0\\ -5&-7\end{bmatrix}$ then $B =$

Solution:

$3A+4B' =\left[\begin{matrix}7&-10&17\\ 0&6&31\end{matrix}\right] \quad \dots\left(1\right)$
$\left(2B-3A'\right)'=\left(2B\right)'-\left(3A'\right)'=2B'-3A$
$\Rightarrow\, 2B' -3A=\left[\begin{matrix}-1&4&-5\\ 18&0&-7\end{matrix}\right]\quad\dots\left(2\right)$
Adding 1 and 2, we get $6B'=\left[\begin{matrix}6&-6&12\\ 18&6&24\end{matrix}\right]$
$\Rightarrow\, B' =\left[\begin{matrix}1&-1&2\\ 3&1&4\end{matrix}\right] \, \therefore\, B=\left[\begin{matrix}1&3\\ -1&1\\ 2&4\end{matrix}\right]$

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