Q73 / 180 · KCET 2019 · Mathematics
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 \( B= \)
- A\( \left[\begin{array}{cc}1 & -3 \\ -1 & 1 \\ 2 & 4\end{array}\right] \)
- B\( \left[\begin{array}{cc}1 & 3 \\ -1 & 1 \\ 2 & -4\end{array}\right] \)
- Correct answerC\( \left[\begin{array}{cc}1 & 3 \\ -1 & 1 \\ 2 & 4\end{array}\right] \)
- D\( \left[\begin{array}{cc}-1 & -18 \\ 4 & -16 \\ -5 & -7\end{array}\right] \)
Solution
(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, \( 6 \mathrm{~B}^{\prime}=\left[\begin{array}{ccc}6 & -6 & 12 \\ 18 & 6 & 24\end{array}\right] \)
\[ \begin{array}{l} \Rightarrow B^{\prime}=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 1 & 4 \end{array}\right] \\ \therefore B=\left[\begin{array}{ccc} 1 & 3 & \\ -1 & 1 & \\ 2 & 4 & \end{array}\right] \end{array} \]
\( 3 A+4 B^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right] \ldots .(1) \) \( \left(2 B-3 A^{\prime}\right)^{\prime}=(2 B)^{\prime}-\left(3 A^{\prime}\right)^{\prime}=2 B^{\prime}-3 A \) \( \Rightarrow 2 B^{\prime}-3 A=\left[\begin{array}{ccc}-1 & 4 & -5 \\ 18 & 0 & -7\end{array}\right] \ldots .(2) \)
$(1)$
(2)
Adding (1) and (2) we get, \( 6 \mathrm{~B}^{\prime}=\left[\begin{array}{ccc}6 & -6 & 12 \\ 18 & 6 & 24\end{array}\right] \)
\[ \begin{array}{l} \Rightarrow B^{\prime}=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 1 & 4 \end{array}\right] \\ \therefore B=\left[\begin{array}{ccc} 1 & 3 & \\ -1 & 1 & \\ 2 & 4 & \end{array}\right] \end{array} \]
Hence, the correct answer is option: 'C'.