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

(C)
\( 3 A+4 B^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right] \ldots .(1) \) \( \left(2 B-3 A^{\prime}\right)^{\prime}=(2 B)^{\prime}-\left(3 A^{\prime}\right)^{\prime}=2 B^{\prime}-3 A \) \( \Rightarrow 2 B^{\prime}-3 A=\left[\begin{array}{ccc}-1 & 4 & -5 \\ 18 & 0 & -7\end{array}\right] \ldots .(2) \)
\((1)\)
(2)
Adding (1) and (2) we get, \( 6 \mathrm{~B}^{\prime}=\left[\begin{array}{ccc}6 & -6 & 12 \\ 18 & 6 & 24\end{array}\right] \)
\[
\begin{array}{l}
\Rightarrow B^{\prime}=\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 1 & 4
\end{array}\right] \\
\therefore B=\left[\begin{array}{ccc}
1 & 3 & \\
-1 & 1 & \\
2 & 4 &
\end{array}\right]
\end{array}
\]