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Mathematics Question on Matrices

If 2A+3B=[2−14 325 ]2A+3B =\begin{bmatrix} {2}&{-1} &{4}\\\ {3}&{2}& {5} \\\ \end{bmatrix} and A+2B[503 162 ]A+2B \begin{bmatrix} {5}&{0} &{3}\\\ {1}&{6}& {2} \\\ \end{bmatrix} then B=B =

A

[812 1101 ]\begin{bmatrix} {8}&{1} &{2}\\\ {1}&{10}& {1} \\\ \end{bmatrix}

B

[81−2 −110−1 ]\begin{bmatrix} {8}&{1} &{-2}\\\ {-1}&{10}& {-1} \\\ \end{bmatrix}

C

[812 −110−1 ]\begin{bmatrix} {8}&{1} &{2}\\\ {-1}&{10}& {-1} \\\ \end{bmatrix}

D

[8−12 −110−1 ]\begin{bmatrix} {8}&{-1} &{2}\\\ {-1}&{10}& {-1} \\\ \end{bmatrix}

Answer

[812 −110−1 ]\begin{bmatrix} {8}&{1} &{2}\\\ {-1}&{10}& {-1} \\\ \end{bmatrix}

Explanation

Solution

We have
2A+3B=[2−14 325]     …(i)2A + 3B = \begin{bmatrix}2&-1&4\\\ 3&2&5\end{bmatrix}\,\,\,\,\, \dots(i)
and A+2B=[503 162]     …(ii) A + 2B = \begin{bmatrix}5&0&3\\\ 1&6&2\end{bmatrix} \,\,\,\,\,\dots(ii)
Multiply E (ii) by 2 and subtracting E(i) from (ii), we get
B=2[503 162]−[2−14 325]B = 2 \begin{bmatrix}5&0&3\\\ 1&6&2\end{bmatrix} -\begin{bmatrix}2&-1&4\\\ 3&2&5\end{bmatrix}
=[812 −110−1]= \begin{bmatrix}8&1&2\\\ -1&10&-1\end{bmatrix}