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(C) First,calculate the sum of matrices $A$ and $B$: $A+B = \begin{bmatrix} 1 & 1 \ 1 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 1 \ 3 & 1 \end{bmatrix

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$\frac{1}{7}\begin{bmatrix} 3 & -2 \ -4 & 5 \end{bmatrix}$

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(C) First,calculate the sum of matrices $A$ and $B$: $A+B = \begin{bmatrix} 1 & 1 \ 1 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 1 \ 3 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 2 \ 4 & 3 \end{bmatrix}$ Next,find the determinant of $(A+B)$: $|A+B| = (5 	imes 3) - (2 	imes 4) = 15 - 8 = 7$ Now,find the adjoint of $(A+B)$ by swapping the diagonal elements and changing the signs of the off-diagonal elements: $	ext{adj}(A+B) = \begin{bmatrix} 3 & -2 \ -4 & 5 \end{bmatrix}$ Finally,use the formula $(A+B)^{-1} = \frac{1}{|A+B|} 	ext{adj}(A+B)$: $(A+B)^{-1} = \frac{1}{7} \begin{bmatrix} 3 & -2 \ -4 & 5 \end{bmatrix}$

If $A=\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$ and $B=\begin{bmatrix} 4 & 1 \\ 3 & 1 \end{bmatrix}$,then $(A+B)^{-1} = $

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