If $A - 2 B = (\begin{matrix} 1 & - 2 \\ 3 & 0 \end{matrix})$ and 2A–3B= $[\begin{matrix} - 3 3 \\ 1 - 1 \end{matrix}]$ then B =

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$(\begin{matrix} - 5 & 7 \\ 5 & 1 \end{matrix})$

$(\begin{matrix} - 5 & 7 \\ - 5 & - 1 \end{matrix})$

$(\begin{matrix} - 5 & 7 \\ 5 & - 1 \end{matrix})$

$(\begin{matrix} - 5 & - 7 \\ - 5 & - 1 \end{matrix})$

answer is B.

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Detailed Solution

$(2 A - 4 B) - (2 A - 3 B)$ = $[\begin{matrix} 2 & - 4 \\ 6 & 0 \end{matrix}] - [\begin{matrix} - 3 & 3 \\ 1 & - 1 \end{matrix}]$

$\Rightarrow - B = [\begin{matrix} 2 & - 4 \\ 6 & 0 \end{matrix}] - [\begin{matrix} - 3 & 3 \\ 1 & - 1 \end{matrix}] = [\begin{matrix} 5 & - 7 \\ 5 & 1 \end{matrix}]$

$∴ B = [\begin{matrix} - 5 & 7 \\ - 5 & - 1 \end{matrix}]$

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