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(B) Given $A = \begin{bmatrix} 1 & -1 \ 2 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} a & 1 \ b & -1 \end{bmatrix}$.First,calculate $A + B = \begin

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$a = 1, b = 4$

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(B) Given $A = \begin{bmatrix} 1 & -1 \ 2 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} a & 1 \ b & -1 \end{bmatrix}$.First,calculate $A + B = \begin{bmatrix} 1+a & 0 \ 2+b & -2 \end{bmatrix}$.Next,calculate $A^2 = \begin{bmatrix} 1 & -1 \ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \ 2 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$.Calculate $B^2 = \begin{bmatrix} a & 1 \ b & -1 \end{bmatrix} \begin{bmatrix} a & 1 \ b & -1 \end{bmatrix} = \begin{bmatrix} a^2+b & a-1 \ ab-b & b+1 \end{bmatrix}$.Thus,$A^2 + B^2 = \begin{bmatrix} a^2+b-1 & a-1 \ ab-b & b \end{bmatrix}$.Now,calculate $(A+B)^2 = \begin{bmatrix} 1+a & 0 \ 2+b & -2 \end{bmatrix} \begin{bmatrix} 1+a & 0 \ 2+b & -2 \end{bmatrix} = \begin{bmatrix} (1+a)^2 & 0 \ (2+b)(1+a) - 2(2+b) & 4 \end{bmatrix} = \begin{bmatrix} (1+a)^2 & 0 \ (2+b)(a-1) & 4 \end{bmatrix}$.Equating $(A+B)^2 = A^2 + B^2$:$\begin{bmatrix} (1+a)^2 & 0 \ (2+b)(a-1) & 4 \end{bmatrix} = \begin{bmatrix} a^2+b-1 & a-1 \ ab-b & b \end{bmatrix}$.Comparing the elements:$1$) $a-1 = 0 \Rightarrow a = 1$.$2$) $b = 4$.Therefore,$a = 1$ and $b = 4$.

If $A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$,$B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}$ and $(A + B)^2 = A^2 + B^2$,then the values of $a$ and $b$ are:

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