0

\(A=\left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right]\)
\(\therefore A^{2}=\left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right]\left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right]=\left[\begin{array}{ll} 9+2 & 6+2 \\ 3+1 & 2+1 \end{array}\right]=\left[\begin{array}{ll} 11 & 8 \\ 4 & 3 \end{array}\right]\)
\(\text { Now, }\)
\(A^{2}+a A+b I=O\)
\(\Rightarrow(A A) A^{-1}+a A A^{-1}+b I A^{-1}=O \quad\left[\text { Post-multiplying by } A^{-1} \text { as }|A| \neq 0\right]\)
\(\Rightarrow A\left(A A^{-1}\right)+a I+b\left(I A^{-1}\right)=O\)
\(\Rightarrow A I+a I+b A^{-1}=O\)
\(\Rightarrow A+a I=-b A^{-1} \)
\(\Rightarrow A^{-1}=-\frac{1}{b}(A+a I)\)
\(\text {Now, }A^{-1}=\frac{1}{|A|} a d j A=\frac{1}{1}\left[\begin{array}{cc} 1 & -2 \\ -1 & 3 \end{array}\right]=\left[\begin{array}{cc} 1 & -2 \\ -1 & 3 \end{array}\right]\)
\(\text {We have: }\left[\begin{array}{cc} 1 & -2 \\ -1 & 3 \end{array}\right]=-\frac{1}{b}\left(\left[\begin{array}{ll} 3 & 2 \\ 1 & 1 \end{array}\right]+\left[\begin{array}{cc} a & 0 \\ 0 & a \end{array}\right]\right)=-\frac{1}{b}\left[\begin{array}{cc} 3+a & 2 \\ 1 & 1+a \end{array}\right]=\left[\begin{array}{cc} \frac{-3-a}{b} & -\frac{2}{b} \\ -\frac{1}{b} & \frac{-1-a}{b} \end{array}\right]\)
Comparing the corresponding elements of the two matrices, we have:
\(-\frac{1}{b}=-1 \Rightarrow b=1 \)
\(\frac{-3-a}{b}=1 \Rightarrow-3-a=1 \Rightarrow a=-4 \)

 Hence, −4 and 1 are the required values of a and b respectively.

x = \(-\frac { a }{ 3 } \) or x = \(-\frac { a }{ 2 } \)

Inconsistent