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Question
If f (x) = x3 + 4x2 − x, find f (A), where\[A = \begin{bmatrix}0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0\end{bmatrix}\]
Solution
\[Given: f\left( x \right) = x^3 + 4 x^2 - x\]
\[f\left( A \right) = A^3 + 4 A^2 - A\]
\[Now, \]
\[ A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0\end{bmatrix}\begin{bmatrix}0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}0 + 2 + 2 & 0 - 3 - 2 & 0 + 0 + 0 \\ 0 - 6 + 0 & 2 + 9 - 0 & 4 - 0 + 0 \\ 0 - 2 + 0 & 1 + 3 - 0 & 2 - 0 + 0\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}4 & - 5 & 0 \\ - 6 & 11 & 4 \\ - 2 & 4 & 2\end{bmatrix}\]
\[\]
\[ A^3 = A^2 A\]
\[ \Rightarrow A^3 = \begin{bmatrix}4 & - 5 & 0 \\ - 6 & 11 & 4 \\ - 2 & 4 & 2\end{bmatrix}\begin{bmatrix}0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0\end{bmatrix}\]
\[ \Rightarrow A^3 = \begin{bmatrix}0 - 10 + 0 & 4 + 15 - 0 & 8 - 0 + 0 \\ 0 + 22 + 4 & - 6 - 33 - 4 & - 12 + 0 + 0 \\ 0 + 8 + 2 & - 2 - 12 - 2 & - 4 + 0 + 0\end{bmatrix}\]
\[ \Rightarrow A^3 = \begin{bmatrix}- 10 & 19 & 8 \\ 26 & - 43 & - 12 \\ 10 & - 16 & - 4\end{bmatrix}\]
\[\]
\[f\left( A \right) = A^3 + 4 A^2 - A\]
\[ \Rightarrow f\left( A \right) = \begin{bmatrix}- 10 & 19 & 8 \\ 26 & - 43 & - 12 \\ 10 & - 16 & - 4\end{bmatrix} + 4\begin{bmatrix}4 & - 5 & 0 \\ - 6 & 11 & 4 \\ - 2 & 4 & 2\end{bmatrix} - \begin{bmatrix}0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0\end{bmatrix}\]
\[ \Rightarrow f\left( A \right) = \begin{bmatrix}- 10 & 19 & 8 \\ 26 & - 43 & - 12 \\ 10 & - 16 & - 4\end{bmatrix} + \begin{bmatrix}16 & - 20 & 0 \\ - 24 & 44 & 16 \\ - 8 & 16 & 8\end{bmatrix} - \begin{bmatrix}0 & 1 & 2 \\ 2 & - 3 & 0 \\ 1 & - 1 & 0\end{bmatrix}\]
\[ \Rightarrow f\left( A \right) = \begin{bmatrix}- 10 + 16 - 0 & 19 - 20 - 1 & 8 + 0 - 2 \\ 26 - 24 - 2 & - 43 + 44 + 3 & - 12 + 16 - 0 \\ 10 - 8 - 1 & - 16 + 16 + 1 & - 4 + 8 + 0\end{bmatrix}\]
\[ \Rightarrow f\left( A \right) = \begin{bmatrix}6 & - 2 & 6 \\ 0 & 4 & 4 \\ 1 & 1 & 4\end{bmatrix}\]
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