if

Question:

If $A=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$, then show that $\mathrm{A}$ is a root of the polynomial $f(x)=x^{3}-6 x^{2}+7 x+2$

Solution:

Given : $f(x)=x^{3}-6 x^{2}+7 x+2$

$f(A)=A^{3}-6 A^{2}+7 A+2 I_{3}$

Now,

$A^{2}=A A$

$\Rightarrow A^{2}=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{lll}1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{llc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]$

$A^{3}=A^{2} A$

$\Rightarrow A^{3}=\left[\begin{array}{lcc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]$

$\Rightarrow A^{3}=\left[\begin{array}{lll}5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39\end{array}\right]$

$\Rightarrow A^{3}=\left[\begin{array}{lll}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]$

$A^{3}-6 A^{2}+7 A+2 I_{3}$

$\Rightarrow A^{3}-6 A^{2}+7 A+2 I_{3}=\left[\begin{array}{ccc}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]-6\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]+7\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]+2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\Rightarrow A^{3}-6 A^{2}+7 A+2 I_{3}=\left[\begin{array}{ccc}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]-\left[\begin{array}{ccc}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{array}\right]+\left[\begin{array}{ccc}7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21\end{array}\right]+\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$

$\Rightarrow A^{3}-6 A^{2}+7 A+2 I_{3}=\left[\begin{array}{ccc}21-30+7+2 & 0-0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0-0+0+0 & 55-78+21+2\end{array}\right]$

$\Rightarrow A^{3}-6 A^{2}+7 A+2 I_{3}=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=0$

Since $f(A)=0, A$ is the root of $f(x)=x^{3}-6 x^{2}+7 x+2$.

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