If A = begin{bmatrix} 1 & -1 2 & -1 end{bmat | vedclass

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

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$\begin{bmatrix} \frac{1}{2} & 0 \ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$

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(B) First,calculate the sum of matrices $A$ and $B$: $A+B = \begin{bmatrix} 1 & -1 \ 2 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 1 \ 4 & -1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 6 & -2 \end{bmatrix}$ Next,find the determinant of $(A+B)$: $|A+B| = (2)(-2) - (0)(6) = -4 - 0 = -4$ Since $|A+B| 
eq 0$,the inverse exists. For a matrix $M = \begin{bmatrix} a & b \ c & d \end{bmatrix}$,$M^{-1} = \frac{1}{|M|} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}$. Applying this to $A+B$: $(A+B)^{-1} = \frac{1}{-4} \begin{bmatrix} -2 & 0 \ -6 & 2 \end{bmatrix} = \begin{bmatrix} \frac{-2}{-4} & \frac{0}{-4} \ \frac{-6}{-4} & \frac{2}{-4} \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}$

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

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