If A = begin{bmatrix} 3 & 4 1 & -6 end{bmatr | vedclass

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(D) Given the equation $A + 2X = B$. Subtracting $A$ from both sides,we get $2X = B - A$. First,calculate $B - A$: $B - A = \begin{bmatrix} -2 & 5 \

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None of these

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(D) Given the equation $A + 2X = B$. Subtracting $A$ from both sides,we get $2X = B - A$. First,calculate $B - A$: $B - A = \begin{bmatrix} -2 & 5 \ 6 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 4 \ 1 & -6 \end{bmatrix} = \begin{bmatrix} -2 - 3 & 5 - 4 \ 6 - 1 & 1 - (-6) \end{bmatrix} = \begin{bmatrix} -5 & 1 \ 5 & 7 \end{bmatrix}$. Now,$X = \frac{1}{2} (B - A) = \frac{1}{2} \begin{bmatrix} -5 & 1 \ 5 & 7 \end{bmatrix} = \begin{bmatrix} -2.5 & 0.5 \ 2.5 & 3.5 \end{bmatrix}$. Comparing this with the given options,the correct answer is none of these.

If $A = \begin{bmatrix} 3 & 4 \\ 1 & -6 \end{bmatrix}$ and $B = \begin{bmatrix} -2 & 5 \\ 6 & 1 \end{bmatrix}$,then find $X$ such that $A + 2X = B$.

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