If a = ⎡ ⎢ ⎣ 2 − 1 1 − 1 2 − 1 1 − 1 2 ⎤ ⎥ ⎦ . Verify that a 3 − 6 a 2 + 9 a − 4 I = O and Hence Find A−1. - Mathematics

प्रश्न

If \[A = \begin{bmatrix}2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2\end{bmatrix}\].
Verify that \[A^3 - 6 A^2 + 9A - 4I = O\] and hence find A⁻¹.

उत्तर

\[A = \begin{bmatrix}2 & - 1 & 1 \\ - 1 & 2 & - 1 \\ 1 & - 1 & 2\end{bmatrix}\].
\[ \Rightarrow \left| A \right| = \begin{vmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{vmatrix} = 2 \times \left( 4 - 1 \right) + 1\left( - 2 + 1 \right) + 1\left( 1 - 2 \right) = 6 - 1 - 1 = 4 \]
\[\text{ Since, }\left| A \right| \neq 0 \]
\[\text{ Hence, }A^{- 1}\text{ exists }. \]
Now,
\[ A^2 = \begin{bmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{bmatrix} = \begin{bmatrix} 4 + 1 + 1 & - 2 - 2 - 1 & 2 + 1 + 2\\ - 2 - 2 - 1 & 1 + 4 + 1 & - 1 - 2 - 2\\ 2 + 1 + 2 & - 1 - 2 - 2 & 1 + 1 + 4 \end{bmatrix} = \begin{bmatrix} 6 & - 5 & 5\\ - 5 & 6 & - 5\\ 5 & - 5 & 6 \end{bmatrix} \]
\[ A^3 = A^2 . A = \begin{bmatrix} 6 & - 5 & 5\\ - 5 & 6 & - 5\\ 5 & - 5 & 6 \end{bmatrix}\begin{bmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{bmatrix} = \begin{bmatrix} 12 + 5 + 5 & - 6 - 10 - 5 & 6 + 5 + 10\\ - 10 - 6 - 5 & 5 + 12 + 5 & - 5 - 6 - 10\\ 10 + 5 + 6 & - 5 - 10 - 6 & 5 + 5 + 12 \end{bmatrix} = \begin{bmatrix} 22 & - 21 & 21\\ - 21 & 22 & - 21\\ 21 &- 21 & 22 \end{bmatrix} \]
\[\text{ Now, }A^3 - 6 A^2 + 9A - 4I = \begin{bmatrix} 22 & - 21 & 21\\ - 21 & 22 & - 21\\ 21 & - 21 & 22 \end{bmatrix} - 6\begin{bmatrix} 6 & - 5 & 5\\ - 5 & 6 & - 5\\ 5 & - 5 & 6 \end{bmatrix} + 9\begin{bmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{bmatrix} - 4\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} 22 - 36 + 18 - 4 & - 21 + 30 - 9 - 0 & 21 - 30 + 9 - 0\\ - 21 + 30 - 9 - 0 & 22 - 36 + 18 - 4 & - 21 + 30 - 9 - 0\\ 21 - 30 + 9 - 0 & - 21 + 30 - 9 - 0 & 22 - 36 + 18 - 4 \end{bmatrix} \]
\[ = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} = O [\text{ Null matrix }]\]
Hence proved .
\[\text{ Now,} A^3 - 6 A^2 + 9A - 4I = O\]
\[ \Rightarrow A^{- 1} \times \left( A^3 - 6 A^2 + 9A - 4I \right) = A^{- 1} \times O \]
\[ \Rightarrow A^2 - 6A + 9I - 4 A^{- 1} = O\]
\[ \Rightarrow 4 A^{- 1} = A^2 - 6A + 9I\]
\[ \Rightarrow 4 A^{- 1} = \begin{bmatrix} 6 & - 5 & 5\\ - 5 & 6 & - 5\\ 5 & - 5 & 6 \end{bmatrix} - 6\begin{bmatrix} 2 & - 1 & 1\\ - 1 & 2 & - 1\\ 1 & - 1 & 2 \end{bmatrix} + 9\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\]
\[ \Rightarrow 4 A^{- 1} = \begin{bmatrix} 6 - 12 + 9 & - 5 + 6 + 0 & 5 - 6 + 0\\ - 5 + 6 + 0 & 6 - 12 + 9 & - 5 + 6 + 0\\ 5 - 6 + 0 & - 5 + 6 + 0 & 6 - 12 + 9 \end{bmatrix} \]
\[ \Rightarrow 4 A^{- 1} = \begin{bmatrix} 3 & 1 & - 1\\ 1 & 3 & 1\\ - 1 & 1 & 3 \end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{4}\begin{bmatrix} 3 & 1 & - 1\\ 1 & 3 & 1\\ - 1 & 1 & 3 \end{bmatrix}\]

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Inverse of Matrix - Inverse of a Square Matrix by the Adjoint Method

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Q 26 Q 25 Q 27

पाठ 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [पृष्ठ २४]

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आरडी शर्मा Mathematics [English] Class 12

पाठ 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 26 | पृष्ठ २४

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