Question

If $A=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$ and $A^2+aA+bI=0,$ then find a & b.

Answer 1

Given:$A=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$

$A^2+aA+bI=0$

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

$\Rightarrow \left[\begin{array}{cc}9+2& 6+2\\ 3+1& 2+1\end{array}\right]+\left[\begin{array}{cc}3a+b& 2a\\ a& a+b\end{array}\right]=0$

$\Rightarrow \left[\begin{array}{cc}11+3a+b& 8+2a\\ 4+a& 3+a+b\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Since the orders are same, we can equate the corresponding rows and columns,

$\Rightarrow 2a+8=0\Rightarrow 2a=-8$ or $a=-4$

Put $a=-4$ in $a+b+3=0$

$\Rightarrow b=-a-3=-(-4)-3=4-3=1$

$\therefore a=-4,b=1$