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प्रश्न
For the matrix \[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3\end{bmatrix}\] . Show that
उत्तर
\[A = \begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 &- 1& 3 \end{bmatrix} \]
\[ \Rightarrow \left| A \right| = \begin{vmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 & - 1 & 3 \end{vmatrix} = \left( 1 \times 3 \right) - \left( 1 \times 9 \right) + \left( 1 \times - 5 \right) = 3 - 9 - 5 = - 11 \]
\[\text{ Since, }\left| A \right| \neq 0\]
\[\text{Hence, }A^{- 1}\text{ exists . }\]
Now,
\[ A^2 = \begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 & - 1 & 3 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 &- 1 & 3 \end{bmatrix} = \begin{bmatrix} 1 + 1 + 2 & 1 + 2 - 1 & 1 - 3 + 3\\1 + 2 - 6 & 1 + 4 + 3 & 1 - 6 - 9\\2 - 1 + 6 & 2 - 2 - 3 & 2 + 3 + 9 \end{bmatrix} = \begin{bmatrix} 4 & 2 & 1\\ - 3 & 8 & - 14\\ 7 & - 3 & 14 \end{bmatrix}\]
\[\text{ and }A^3 = A^2 . A = \begin{bmatrix} 4 & 2 & 1\\ - 3 & 8 & - 14\\7 & - 3 & 14 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 & - 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 + 2 + 2 & 4 + 4 - 1 & 4 - 6 + 3\\ - 3 + 8 - 28 & - 3 + 16 + 14 & - 3 - 24 - 42\\ 7 - 3 + 28 & 7 - 6 - 14 & 7 + 9 + 42 \end{bmatrix} = \begin{bmatrix} 8 & 7 & 1\\ - 23 & 27 & - 69\\ 32 & - 13 & 58 \end{bmatrix}\]
\[\text{ Now, }A^3 - 6 A^2 + 5A + 11 I_3 = \begin{bmatrix} 8 & 7 & 1 \\ - 23 & 27 & - 69 \\ 32 & - 13 & 58 \end{bmatrix} - 6 \begin{bmatrix} 4 & 2 & 1 \\ - 3 & 8 & - 14 \\ 7 & - 3 & 14 \end{bmatrix} + 5 \begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 & - 1 & 3 \end{bmatrix} + 11 \begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}\]
\[ = \begin{bmatrix} 8 - 24 + 5 + 11 & 7 - 12 + 5 + 0 & 1 - 6 + 5 + 0\\ - 23 + 18 + 5 + 0 & 27 - 48 + 10 + 11 & - 69 + 84 - 15 + 0\\ 32 - 42 + 10 + 0 & - 13 + 18 - 5 + 0 & 58 - 84 + 15 + 11 \end{bmatrix} \]
\[ = \begin{bmatrix} 0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0 \end{bmatrix} = O (\text{ Null matrix})\]
\[\text{ Again, }A^3 - 6 A^2 + 5A + 11 I_3 = O\]
\[ \Rightarrow A^{- 1} \times \left( A^3 - 6 A^2 + 5A + 11 I_3 \right) = A^{- 1} \times O (\text{ Pre - multiplying both sides because }A^{- 1} exists) \]
\[ \Rightarrow \left( A^2 - 6A + 5 I_3 + 11 A^{- 1} \right) = 0\]
\[ \Rightarrow \begin{bmatrix} 4 & 2 & 1\\ - 3 & 8 - 14\\ 7 & - 3 & 14 \end{bmatrix} - 6\begin{bmatrix} 1 & 1 & 1\\1 & 2 &- 3\\2 & - 1 & 3 \end{bmatrix} + 5\begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix} = - 11 A^{- 1} \]
\[ \Rightarrow \begin{bmatrix} 4 - 6 + 5 & 2 - 6 + 0 & 1 - 6 + 0\\ - 3 - 6 + 0 & 8 - 12 + 5 & - 14 + 18 + 0\\ 7 - 12 + 0 & - 3 + 6 + 0 & 14 - 18 + 5 \end{bmatrix} = - 11 A^{- 1} \]
\[ \Rightarrow A^{- 1} = - \frac{1}{11}\begin{bmatrix} 3 & - 4 & - 5\\- 9 & 1 & 4\\ - 5 & 3 & 1 \end{bmatrix} \]
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