Question

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

Answer 1

Given $A=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]$,
$A^2=AA=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]=\left[\begin{array}{cc}9+2& 6+2\\ 3+1& 2+1\end{array}\right]=\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]$
Given $A^2+aA+bI=0$
On putting the values of $A^2,A$ and I, we get
$\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]+a\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]+b\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0\Rightarrow \left[\begin{array}{cc}11+2a+b& 8+2a+0\\ 4+a+0& 3+a+b\end{array}\right]=0$
$\Rightarrow \left[\begin{array}{cc}11+3a+b& 8+2a\\ 4+a& 3+a+b\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
If two matrices are equal, then their corresponding elements are equal.
$\Rightarrow 11+3a+b=0\dots \left(i\right)8+2a=0\dots \left(ii\right)4+a=0\dots \left(iii\right)and3+a+b=0\dots \left(iv\right)$
Solving Eqs. (iii) and (iv), we get
$4+a=0\Rightarrow a=-4$
and $3+a+b=0\Rightarrow 3-4+b=0\Rightarrow b=1$
Thus, $a=-4$ and $b=1$