For the Matrix a = ⎡ ⎢ ⎣ 1 1 1 1 2 − 3 2 − 1 3 ⎤ ⎥ ⎦ . Show that a − 3 − 6 a 2 + 5 a + 11 I 3 = O . Hence, Find A−1. - Mathematics

Question

For the matrix \[A = \begin{bmatrix}1 & 1 & 1 \\ 1 & 2 & - 3 \\ 2 & - 1 & 3\end{bmatrix}\] . Show that

\[A^{- 3} - 6 A^2 + 5A + 11 I_3 = O\]. Hence, find A⁻¹.

Solution

\[A = \begin{bmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 &- 1& 3 \end{bmatrix} \]

\[ \Rightarrow \left| A \right| = \begin{vmatrix} 1 & 1 & 1\\1 & 2 & - 3\\2 & - 1 & 3 \end{vmatrix} = \left( 1 \times 3 \right) - \left( 1 \times 9 \right) + \left( 1 \times - 5 \right) = 3 - 9 - 5 = - 11 \]

\[\text{ Since, }\left| A \right| \neq 0\]

\[\text{Hence, }A^{- 1}\text{ exists . }\]

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

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Q 24 Q 23 Q 25

Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [Page 24]

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RD Sharma Mathematics [English] Class 12

Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 24 | Page 24

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