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Question
`A=[[3,2, 0],[1,4,0],[0,0,5]]` show that A2 − 7A + 10I3 = 0
Solution
\[Given: \hspace{0.167em} A = \begin{bmatrix}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5\end{bmatrix}\]
\[Here, \]
\[ A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5\end{bmatrix}\begin{bmatrix}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}9 + 2 + 0 & 6 + 8 + 0 & 0 + 0 + 0 \\ 3 + 4 + 0 & 2 + 16 + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 25\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25\end{bmatrix}\]
\[\]
\[Now, \]
\[ A^2 - 7A + 10 I_3 \]
\[ \Rightarrow A^2 - 7A + 10 I_3 = \begin{bmatrix}11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25\end{bmatrix} - 7\begin{bmatrix}3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5\end{bmatrix} + 10\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^2 - 7A + 10 I_3 = \begin{bmatrix}11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25\end{bmatrix} - \begin{bmatrix}21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35\end{bmatrix} + \begin{bmatrix}10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10\end{bmatrix}\]
\[ \Rightarrow A^2 - 7A + 10 I_3 = \begin{bmatrix}11 - 21 + 10 & 14 - 14 + 0 & 0 - 0 + 0 \\ 7 - 7 + 0 & 18 - 28 + 10 & 0 - 0 + 0 \\ 0 - 0 + 0 & 0 - 0 + 0 & 25 - 35 + 10\end{bmatrix}\]
\[ \Rightarrow A^2 - 7A + 10 I_3 = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} = 0\]
\[\]
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