Question

Let A be a square matrix defined by $A=\begin{pmatrix}4&-2&1\\ 2&0&1\\ 2&-2&3\end{pmatrix}\hspace{2mm}$Find the minimal polynomial of A. Then express $A^4$ and $A^{-1}$ in terms of A and I.

Answer 1

Solution

we have $A=\begin{pmatrix}4&-2&1\\ 2&0&1\\ 2&-2&3\end{pmatrix}$  The characteristics polynomial is :

$P(\lambda)=-\bigg(\lambda^3-S_1\lambda^2+S_2\lambda-s_3\bigg)$

$\bullet S_1=tr(A)=4+0+3=7$

$\bullet S_2=\begin{vmatrix}4&-2\\ 2&0\end{vmatrix}+\begin{vmatrix}0&1\\ -2&3\end{vmatrix}+\begin{vmatrix}4&1\\ 2&3\end{vmatrix}=4+2+10=16$

$\bullet S_3=\begin{vmatrix}4&-2&1\\ \:2&0&1\\ \:2&-2&3\end{vmatrix}=12$

$\implies P(\lambda)=-\lambda^3+7\lambda^2-16\lambda+12=-\bigg(\lambda-2\bigg)^2\bigg(\lambda-3\bigg)$

since.  $P(\lambda)=0\implies -\bigg(\lambda-2\bigg)^2\bigg(\lambda-3\bigg)=0 \iff \lambda_1=2,\lambda_2=3$

For $\lambda_1=2$ then $\bigg(A-2I\bigg)=0$

$\implies \begin{pmatrix}2&-2&1\\ 2&-2&1\\ 2&-2&1\end{pmatrix}\sim \begin{pmatrix}2&-2&1\\ 0&0&0\\ 0&0&0\end{pmatrix}\hspace{2mm}$

Let $x_1=t,x_2=S\hspace{2mm},t,S\in \mathbb{R}$

$\implies 2x_1-2x_2+x_3=0\implies x_3=-2t+2S$

so, $\hspace{2mm}x\begin{pmatrix}t\\ S\\ -2t+2S\end{pmatrix}=t\begin{pmatrix}1\\ 0\\ -2\end{pmatrix}+S\begin{pmatrix}0\\ 1\\ 2\end{pmatrix}$

$\implies E_{\lambda_1}=span\left\{\begin{pmatrix}1\\ \:0\\ \:-2\end{pmatrix},\begin{pmatrix}0\\ \:1\\ \:2\end{pmatrix}\right\}$

$\implies gm(\lambda_1)=am(\lambda_1)=2$  and  $Index(\lambda_1)=1$

The minimal polynomial is :

$m(\lambda)=\bigg(\lambda-2\bigg)\bigg(\lambda-3\bigg)$

$\bullet$ Express  $A^4$ and $A^{-1}$ interm of of A and I.

we have $m(\lambda)=\bigg(\lambda-2\bigg)\bigg(\lambda-3\bigg)=\lambda^2-5\lambda+6$

Let $x^4=\bigg(x^2-5x+6\bigg)4x+ax+b$

$\implies A^4=\bigg(A^2-5A+6I\bigg)4A+aA+bI$

$\iff A^4=aA+bI$

$\iff \begin{cases} x=3\implies 3^4=3a+b & \quad \\ x=2\implies 2^4=2a+b & \quad \end{cases}\hspace{2mm}\implies a=65,b=-114$

Thus, $A^4=65A-114I$

$\bullet$Find $A^{-1}$

$\implies A^2-5A+6I=0\iff -A^2+5A=6I$

$\iff -\frac{1}{6}A\bigg(A-5I\bigg)=I$

Therefore. $A^{-1}=-\frac{1}{6}A\bigg(A-5I\bigg)$

Answer

Therefore.

$m(\lambda)=\bigg(\lambda-2\bigg)\bigg(\lambda-3\bigg)$

$A^{-1}=-\frac{1}{6}A\bigg(A-5I\bigg)$