STD 12 - 3 and 4 . determinant and metrices — Mathematics STD 12 Science — Question

$\therefore A^{2}=\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]\left[\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}9+2 & 6+2 \\ 3+1 & 2+1\end{array}\right]=\left[\begin{array}{cc}11 & 8 \\ 4 & 3\end{array}\right]$

Now,

$A^{2}+a A+b I=0$

$\Rightarrow(A A) A^{-1}+a A A^{-1}+b L A^{-1}=0 \quad\left[\text { Post-multiplying by } A^{-1} \text { as }|A| \neq 0\right]$

$\Rightarrow A\left(A A^{-1}\right)+a I+b\left(L A^{-1}\right)=0$

$\Rightarrow A I+a I+b A^{-1}=0$

$\Rightarrow A+a I=-b A^{-1}$

$\Rightarrow A^{-1}=\frac{1}{b}(A+a I)$

Now,

$A^{-1}=\frac{1}{|A|}$ adj. $A=\frac{1}{1}\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right]=\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right]$

We have:

$\left[\begin{array}{cc}1 & -2 \\ -1 & 3\end{array}\right]=-\frac{1}{b}\left(\left[\begin{array}{cc}3 & 2 \\ 1 & 1\end{array}\right]+\left[\begin{array}{cc}a & 0 \\ 0 & a\end{array}\right]\right)=-\frac{1}{b}\left[\begin{array}{cc}3+a & 2 \\ 1 & 1+a\end{array}\right]=\left[\begin{array}{cc}\frac{-3-a}{b} & -\frac{2}{b} \\ -\frac{1}{b} & \frac{-1-a}{b}\end{array}\right]$

Comparing the corresponding elements of the two matrices, we have:

$-\frac{1}{b}=-1 \Rightarrow b=1$

$\frac{-3-a}{b}=1 \Rightarrow-3-a \Rightarrow a=-4$

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