Question

If A= `[[1           0           2],[0        2           1],[2            0           3]]` , then show that A is a root of the polynomial f (x) = x³ − 6x² + 7x + 2.

 

Answer 1

\[Given: \hspace{0.167em} f\left( x \right) = x^3 - 6 x^2 + 7x + 2\]
\[f\left( A \right) = A^3 - 6 A^2 + 7A + 2 I_3 \]
\[Now, \]
\[ A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix}\begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}1 + 0 + 4 & 0 + 0 + 0 & 2 + 0 + 6 \\ 0 + 0 + 2 & 0 + 4 + 0 & 0 + 2 + 3 \\ 2 + 0 + 6 & 0 + 0 + 0 & 4 + 0 + 9\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{bmatrix}\]
\[\]
\[ A^3 = A^2 A\]
\[ \Rightarrow A^3 = \begin{bmatrix}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{bmatrix}\begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix}\]
\[ \Rightarrow A^3 = \begin{bmatrix}5 + 0 + 16 & 0 + 0 + 0 & 10 + 0 + 24 \\ 2 + 0 + 10 & 0 + 8 + 0 & 4 + 4 + 15 \\ 8 + 0 + 26 & 0 + 0 + 0 & 16 + 0 + 39\end{bmatrix}\]
\[ \Rightarrow A^3 = \begin{bmatrix}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{bmatrix}\]
\[\]
\[ A^3 - 6 A^2 + 7A + 2 I_3 \]
\[ \Rightarrow A^3 - 6 A^2 + 7A + 2 I_3 = \begin{bmatrix}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{bmatrix} - 6\begin{bmatrix}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{bmatrix} + 7\begin{bmatrix}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{bmatrix} + 2\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^3 - 6 A^2 + 7A + 2 I_3 = \begin{bmatrix}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{bmatrix} - \begin{bmatrix}30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78\end{bmatrix} + \begin{bmatrix}7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21\end{bmatrix} + \begin{bmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{bmatrix}\]
\[ \Rightarrow A^3 - 6 A^2 + 7A + 2 I_3 = \begin{bmatrix}21 - 30 + 7 + 2 & 0 - 0 + 0 + 0 & 34 - 48 + 14 + 0 \\ 12 - 12 + 0 + 0 & 8 - 24 + 14 + 2 & 23 - 30 + 7 + 0 \\ 34 - 48 + 14 + 0 & 0 - 0 + 0 + 0 & 55 - 78 + 21 + 2\end{bmatrix}\]
\[ \Rightarrow A^3 - 6 A^2 + 7A + 2 I_3 = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} = 0\]
\[\]
since `f(A)=0,A ` is the root of `f(A)x^3-6x^2+7x+2.`