If A + B = begin{bmatrix} 1 & 0 1 & 1 end{bm | vedclass

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(C) Given equations are:$(1)$ $A + B = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}$$(2)$ $A - 2B = \begin{bmatrix} -1 & 1 \ 0 & -1 \end{bmatrix}$To

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$\begin{bmatrix} 1/3 & 1/3 \ 2/3 & 1/3 \end{bmatrix}$

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(C) Given equations are:$(1)$ $A + B = \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix}$$(2)$ $A - 2B = \begin{bmatrix} -1 & 1 \ 0 & -1 \end{bmatrix}$To eliminate $B$,multiply equation $(1)$ by $2$:$2A + 2B = 2 \begin{bmatrix} 1 & 0 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 2 & 2 \end{bmatrix}$Now,add this to equation $(2)$:$(2A + 2B) + (A - 2B) = \begin{bmatrix} 2 & 0 \ 2 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 1 \ 0 & -1 \end{bmatrix}$$3A = \begin{bmatrix} 2-1 & 0+1 \ 2+0 & 2-1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 2 & 1 \end{bmatrix}$Dividing by $3$,we get:$A = \frac{1}{3} \begin{bmatrix} 1 & 1 \ 2 & 1 \end{bmatrix} = \begin{bmatrix} 1/3 & 1/3 \ 2/3 & 1/3 \end{bmatrix}$

If $A + B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ and $A - 2B = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$,then $A=$

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