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प्रश्न
If
उत्तर
\[Given: A = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\]
\[Now, \]
\[ A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}9 - 1 & 3 + 2 \\ - 3 - 2 & - 1 + 4\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}8 & 5 \\ - 5 & 3\end{bmatrix}\]
\[ A^2 - 5A + 7 I_2 \]
\[ \Rightarrow A^2 - 5A + 7 I_2 = \begin{bmatrix}8 & 5 \\ - 5 & 3\end{bmatrix} - 5\begin{bmatrix}3 & 1 \\ - 1 & 2\end{bmatrix} + 7\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^2 - 5A + 7 I_2 = \begin{bmatrix}8 & 5 \\ - 5 & 3\end{bmatrix} - \begin{bmatrix}15 & 5 \\ - 5 & 10\end{bmatrix} + \begin{bmatrix}7 & 0 \\ 0 & 7\end{bmatrix}\]
\[ \Rightarrow A^2 - 5A + 7 I_2 = \begin{bmatrix}8 - 15 + 7 & 5 - 5 + 0 \\ - 5 + 5 + 0 & 3 - 10 + 7\end{bmatrix}\]
\[ \Rightarrow A^2 - 5A + 7 I_2 = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}\]
\[ \Rightarrow A^2 - 5A + 7 I_2 = 0\]
Hence proved .
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