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प्रश्न
Show that the matrix, \[A = \begin{bmatrix}1 & 0 & - 2 \\ - 2 & - 1 & 2 \\ 3 & 4 & 1\end{bmatrix}\] satisfies the equation, \[A^3 - A^2 - 3A - I_3 = O\] . Hence, find A−1.
उत्तर
\[\text{ We have, }A = \begin{bmatrix} 1 & 0 &- 2\\ - 2 & - 1 & 2\\ 3 & 4 & 1 \end{bmatrix} \]
\[ \Rightarrow \left| A \right| = \begin{vmatrix}| 1 & 0 &- 2\\ - 2 & - 1 & 2\\ 3 & 4 & 1 \end{vmatrix} = 1\left( - 9 \right) + 0 - 2\left( - 8 \right) = - 9 + 16 = 7 \]
\[\text{ Since, }\left| A \right| \neq 0\]
\[\text{ Hence, }A^{- 1}\text{ exists .} \]
Now,
\[ A^2 = \begin{bmatrix} 1 & 0 & - 2\\ - 2 & - 1 & 2\\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 &- 2\\ - 2 & - 1 & 2\\ 3 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 1 + 0 - 6 & 0 + 0 - 8 & - 2 + 0 - 2\\ - 2 + 2 + 6 & 0 + 1 + 8 & 4 - 2 + 2\\ 3 - 8 + 3 & 0 - 4 + 4 & - 6 + 8 + 1 \end{bmatrix} = \begin{bmatrix} - 5 & - 8 & - 4\\ 6 & 9 & 4\\ - 2 & 0 & 3 \end{bmatrix}\]
\[ A^3 = A^2 . A = \begin{bmatrix} - 5 & - 8 & - 4\\ 6 & 9 & 4\\ - 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 & - 2\\ - 2 & - 1 & 2\\ 3 & 4 & 1 \end{bmatrix} = \begin{bmatrix} - 5 + 16 - 12 & 0 + 8 - 16 & 10 - 16 - 4\\ 6 - 18 + 12 & 0 - 9 + 16 & - 12 + 18 + 4\\ - 2 + 0 + 9 & 0 + 0 + 12 & 4 + 0 + 3 \end{bmatrix} = \begin{bmatrix} - 1 & - 8 & - 10\\ 0 & 7 & 10\\ 7 & 12 & 7 \end{bmatrix} \]
\[\text{ Now, }A^3 - A^2 - 3A - I_3 = \begin{bmatrix} - 1 & - 8 & - 10\\ 0 & 7 & 10\\ 7 & 12 & 7 \end{bmatrix} - \begin{bmatrix} - 5 & - 8 & - 4\\ 6 & 9 & 4\\ - 2 & 0 & 3 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 & - 2\\ - 2 & - 1 & 2\\ 3 & 4 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \\ = \begin{bmatrix} - 1 + 5 - 3 - 1 & - 8 + 8 + 0 + 0 & - 10 + 4 + 6 - 0\\0 - 6 + 6 - 0 & 7 - 9 + 3 - 1 & 10 - 4 - 6 - 0\\ 7 + 2 - 9 - 0 & 12 + 0 - 12 - 0 & 7 - 3 - 3 - 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} = O\]
Hence proved .
\[\text{ Now,} A^3 - A^2 - 3A - I_3 = O (\text{ Null matrix })\]
\[ \Rightarrow A^{- 1} \left( A^3 - A^2 - 3A - I_3 \right) = A^{- 1} O (\text{ Pre - multiplying by A}^{- 1} )\]
\[ \Rightarrow A^2 - A^1 - 3 I_3 = A^{- 1} \]
\[ \Rightarrow \begin{bmatrix} - 5 & - 8 & - 4\\ 6 & 9 & 4\\ - 2 & 0 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 0 & - 2\\ - 2 & - 1 & 2\\ 3 & 4 & 1 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} = A^{- 1} \]
\[ \Rightarrow \begin{bmatrix} - 5 - 1 - 3 & - 8 - 0 - 0 & - 4 + 2 + 0\\ 6 + 2 + 0 & 9 + 1 - 3 & 4 - 2\\- 2 - 3 - 0 & 0 - 4 - 0 & 3 - 1 - 3 \end{bmatrix} = \begin{bmatrix} - 9 & - 8 & - 2\\ 8 & 7 &2\\ - 5 & - 4 & - 1 \end{bmatrix} = A^{- 1} \]
\[ \Rightarrow A^{- 1} = \begin{bmatrix} - 9 & - 8 & - 2\\ 8 & 7 & 2\\ - 5 & - 4 & - 1 \end{bmatrix}\]
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