If A-B=begin{bmatrix} 2 & 5 9 & 0 end{bmatri | vedclass

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(A) Given the equations: $A - B = \begin{bmatrix} 2 & 5 \ 9 & 0 \end{bmatrix}$  (Equation $1$) $A + B = \begin{bmatrix} 6 & 3 \ -1 & 0 \end{bmatrix}

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$\begin{bmatrix} 4 & 4 \ 4 & 0 \end{bmatrix}$

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(A) Given the equations: $A - B = \begin{bmatrix} 2 & 5 \ 9 & 0 \end{bmatrix}$  (Equation $1$) $A + B = \begin{bmatrix} 6 & 3 \ -1 & 0 \end{bmatrix}$  (Equation $2$) Adding Equation $1$ and Equation $2$: $(A - B) + (A + B) = \begin{bmatrix} 2 & 5 \ 9 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 3 \ -1 & 0 \end{bmatrix}$ $2A = \begin{bmatrix} 2+6 & 5+3 \ 9+(-1) & 0+0 \end{bmatrix}$ $2A = \begin{bmatrix} 8 & 8 \ 8 & 0 \end{bmatrix}$ Dividing by $2$: $A = \frac{1}{2} \begin{bmatrix} 8 & 8 \ 8 & 0 \end{bmatrix} = \begin{bmatrix} 4 & 4 \ 4 & 0 \end{bmatrix}$ Thus,the correct option is $A$.

If $A-B=\begin{bmatrix} 2 & 5 \\ 9 & 0 \end{bmatrix}$ and $A+B=\begin{bmatrix} 6 & 3 \\ -1 & 0 \end{bmatrix}$,then matrix $A =$ . . . . . .

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