Question

For the matrix A=$\begin{bmatrix}3&2\\\1&1\end{bmatrix}$,find the numbers a and b such that A2+ aA+bI=O.

Answer 1

Let A=$\begin{bmatrix}3&2\\\1&1\end{bmatrix}$

A2=\begin{bmatrix}3&2\\\1&1\end{bmatrix}$$\begin{bmatrix}3&2\\\1&1\end{bmatrix}=$\begin{bmatrix}9+2&6+2\\\3+1&2+1\end{bmatrix}$

=$\begin{bmatrix}11&8\\\4&3\end{bmatrix}$
Now, A2+aA+bI=0 [post multiplying by A-1 as IAI≠0]
$\Rightarrow$ (AA)A-1+aAA-1+bAA-1=0
$\Rightarrow$ A(AA-1)+aI+b(IA-1)=0
AI+aI+bA-1=0
$\Rightarrow$ A+aI=-bA-1
$\Rightarrow$ A-1=-$\frac{1}{b}$(A+aI)
Now, A-1=$\frac{1}{\mid A\mid}$adj A=\frac{1}{1}$$\begin{bmatrix}1&-2\\\\-1&3\end{bmatrix}
we have:
$\begin{bmatrix}1&-2\\\\-1&3\end{bmatrix}$=-\frac{1}{b}$$\bigg($$\begin{bmatrix}3&2\\\1&1\end{bmatrix}+\begin{bmatrix}a&0\\\0&a\end{bmatrix}$$\bigg)=-\frac{1}{b}$$\begin{bmatrix}3+a&2\\\1&1+a\end{bmatrix}=$\begin{bmatrix}\frac{-3-a}{b}&\frac{-2][b}2\\\\\frac{-1}{b}&\frac{-1-a}{b}\end{bmatrix}$
Comparing the corresponding elements of the two matrices have:
-$\frac{1}{b}=-1\geq b=1$
$\frac{-3-a}{b}=1\geq -3-a=1$
$\Rightarrow$ a=-4

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