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प्रश्न
\[A = \begin{bmatrix}2 & 3 \\ - 1 & 0\end{bmatrix}\],show that A2 − 2A + 3I2 = O
उत्तर
\[Given: A = \begin{bmatrix}2 & 3 \\ - 1 & 0\end{bmatrix}\]
\[Now, \]
\[ A^2 = AA\]
\[ \Rightarrow A^2 = \begin{bmatrix}2 & 3 \\ - 1 & 0\end{bmatrix}\begin{bmatrix}2 & 3 \\ - 1 & 0\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}4 - 3 & 6 + 0 \\ - 2 + 0 & - 3 + 0\end{bmatrix}\]
\[ \Rightarrow A^2 = \begin{bmatrix}1 & 6 \\ - 2 & - 3\end{bmatrix}\]
\[\]
\[ A^2 - 2A + 3 I_2 \]
\[ \Rightarrow A^2 - 2A + 3 I_2 = \begin{bmatrix}1 & 6 \\ - 2 & - 3\end{bmatrix} - 2\begin{bmatrix}2 & 3 \\ - 1 & 0\end{bmatrix} + 3\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\]
\[ \Rightarrow A^2 - 2A + 3 I_2 = \begin{bmatrix}1 & 6 \\ - 2 & - 3\end{bmatrix} - \begin{bmatrix}4 & 6 \\ - 2 & 0\end{bmatrix} + \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix}\]
\[ \Rightarrow A^2 - 2A + 3 I_2 = \begin{bmatrix}1 - 4 + 3 & 6 - 6 + 0 \\ - 2 + 2 + 0 & - 3 + 0 + 3\end{bmatrix}\]
\[ \Rightarrow A^2 - 2A + 3 I_2 = \begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix} = 0\]
\[\]
Hence proved .
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