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.

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

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