Question

For the matrix $A=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$, find the numbers $a$ and $b$ such that $A^2+aA+bI=0$

Answer 1

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

Also, $A^2+aA+bI=O$ ....(1)

Now, $A^2=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$

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

$\Rightarrow A^2=\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]$

Substituting the values in (1), we get

$\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]+a\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]+b\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

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

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

On comparing corresponding elements of both matrix, we get

$a+4=0$

$\Rightarrow a=-4$

Also, $3+a+b=0$

$\Rightarrow 3-4+b=0$

$\Rightarrow b=1$

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