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Question
\[A = \begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix} and \text{ I }= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\], then prove that A2 − A + 2I = O.
Solution
\[Given: \hspace{0.167em} A = \begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix}\]
\[Now, \]
\[ A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix}\begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}9 - 8 & - 6 + 4 \\ 12 - 8 & - 8 + 4\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}1 & - 2 \\ 4 & - 4\end{bmatrix}\]
\[ A^2 - A + 2I = \begin{bmatrix}1 & - 2 \\ 4 & - 4\end{bmatrix} - \begin{bmatrix}3 & - 2 \\ 4 & - 2\end{bmatrix} + 2\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^2 - A + 2I = \begin{bmatrix}1 - 3 & - 2 + 2 \\ 4 - 4 & - 4 + 2\end{bmatrix} + \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}\]
\[ \Rightarrow A^2 - A + 2I = \begin{bmatrix}- 2 & 0 \\ 0 & - 2\end{bmatrix} + \begin{bmatrix}2 & 0 \\ 0 & 2\end{bmatrix}\]
\[ \Rightarrow A^2 - A + 2I = \begin{bmatrix}- 2 + 2 & 0 + 0 \\ 0 + 0 & - 2 + 2\end{bmatrix}\]
\[ \Rightarrow A^2 - A + 2I = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\]
\[ \Rightarrow A^2 - A + 2I = 0\]
Hence proved .
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