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Mathematics Question on Matrices

If 3A+4B′=3A + 4B' = [7−1017 0631]\begin{bmatrix}7&-10&17\\\ 0&6&31\end{bmatrix} and 2B−3A′2B - 3A' [−118 40 −5−7] \begin{bmatrix}-1&18\\\ 4&0\\\ -5&-7\end{bmatrix} then B=B =

A

[−1−18 4−16 −5−7]\begin{bmatrix}-1&-18\\\ 4&-16\\\ -5&-7\end{bmatrix}

B

[13 −11 24]\begin{bmatrix}1&3\\\ -1&1\\\ 2&4\end{bmatrix}

C

[13 −11 2−4]\begin{bmatrix}1&3\\\ -1&1\\\ 2&-4\end{bmatrix}

D

[1−3 −11 24]\begin{bmatrix}1&-3\\\ -1&1\\\ 2&4\end{bmatrix}

Answer

[13 −11 24]\begin{bmatrix}1&3\\\ -1&1\\\ 2&4\end{bmatrix}

Explanation

Solution

3A+4B′=[7−1017 0631]…(1)3A+4B' =\left[\begin{matrix}7&-10&17\\\ 0&6&31\end{matrix}\right] \quad \dots\left(1\right)
(2B−3A′)′=(2B)′−(3A′)′=2B′−3A\left(2B-3A'\right)'=\left(2B\right)'-\left(3A'\right)'=2B'-3A
⇒ 2B′−3A=[−14−5 180−7]…(2)\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′=[6−612 18624]6B'=\left[\begin{matrix}6&-6&12\\\ 18&6&24\end{matrix}\right]
⇒ B′=[1−12 314] ∴ B=[13 −11 24]\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]