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(D) Given $A = \begin{bmatrix} 3 & 2 \ 1 & 1 \end{bmatrix}$.First,calculate $A^{2} = A 	imes A = \begin{bmatrix} 3 & 2 \ 1 & 1 \end{bmatrix} \begin

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$a = -4, b = 1$

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(D) Given $A = \begin{bmatrix} 3 & 2 \ 1 & 1 \end{bmatrix}$.First,calculate $A^{2} = A 	imes A = \begin{bmatrix} 3 & 2 \ 1 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 9+2 & 6+2 \ 3+1 & 2+1 \end{bmatrix} = \begin{bmatrix} 11 & 8 \ 4 & 3 \end{bmatrix}$.Substitute $A^{2}$,$A$,and $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ into the equation $A^{2} + aA + bI = 0$:$\begin{bmatrix} 11 & 8 \ 4 & 3 \end{bmatrix} + a \begin{bmatrix} 3 & 2 \ 1 & 1 \end{bmatrix} + b \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.This gives the matrix equation:$\begin{bmatrix} 11 + 3a + b & 8 + 2a \ 4 + a & 3 + a + b \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}$.Comparing the elements:$8 + 2a = 0 \Rightarrow 2a = -8 \Rightarrow a = -4$.$4 + a = 0 \Rightarrow 4 + (-4) = 0$ (Consistent).$11 + 3a + b = 0 \Rightarrow 11 + 3(-4) + b = 0 \Rightarrow 11 - 12 + b = 0 \Rightarrow -1 + b = 0 \Rightarrow b = 1$.Thus,$a = -4$ and $b = 1$.

For the matrix $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$,find the values of $a$ and $b$ such that $A^{2} + aA + bI = 0$.

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