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Inequalities Involving Modulus Function

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Question

Show that the matrix, $A = [\begin{matrix} 1 & 0 & - 2 \\ - 2 & - 1 & 2 \\ 3 & 4 & 1 \end{matrix}]$ satisfies the equation, $A^{3} - A^{2} - 3 A - I_{3} = O$ . Hence, find A⁻¹.

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Solution

$We have, A = [1 0 - 2 - 2 - 1 2 3 4 1] \Rightarrow |A| = |1 0 - 2 - 2 - 1 2 3 4 1| = 1 (- 9) + 0 - 2 (- 8) = - 9 + 16 = 7 Since, |A| \neq 0 Hence, A^{- 1} exists . Now, A^{2} = [1 0 - 2 - 2 - 1 2 3 4 1] [1 0 - 2 - 2 - 1 2 3 4 1] = [1 + 0 - 6 0 + 0 - 8 - 2 + 0 - 2 - 2 + 2 + 6 0 + 1 + 8 4 - 2 + 2 3 - 8 + 3 0 - 4 + 4 - 6 + 8 + 1] = [- 5 - 8 - 4 6 9 4 - 2 0 3] A^{3} = A^{2} . A = [- 5 - 8 - 4 6 9 4 - 2 0 3] [1 0 - 2 - 2 - 1 2 3 4 1] = [- 5 + 16 - 12 0 + 8 - 16 10 - 16 - 4 6 - 18 + 12 0 - 9 + 16 - 12 + 18 + 4 - 2 + 0 + 9 0 + 0 + 12 4 + 0 + 3] = [- 1 - 8 - 10 0 7 10 7 12 7] Now, A^{3} - A^{2} - 3 A - I_{3} = [- 1 - 8 - 10 0 7 10 7 12 7] - [- 5 - 8 - 4 6 9 4 - 2 0 3] - 3 [1 0 - 2 - 2 - 1 2 3 4 1] - [1 0 0 0 1 0 0 0 1] = [- 1 + 5 - 3 - 1 - 8 + 8 + 0 + 0 - 10 + 4 + 6 - 0 0 - 6 + 6 - 0 7 - 9 + 3 - 1 10 - 4 - 6 - 0 7 + 2 - 9 - 0 12 + 0 - 12 - 0 7 - 3 - 3 - 1] = [0 0 0 0 0 0 0 0 0] = O Hence proved . Now, A^{3} - A^{2} - 3 A - I_{3} = O (Null matrix) \Rightarrow A^{- 1} (A^{3} - A^{2} - 3 A - I_{3}) = A^{- 1} O (Pre - multiplying by A^{- 1}) \Rightarrow A^{2} - A^{1} - 3 I_{3} = A^{- 1} \Rightarrow [- 5 - 8 - 4 6 9 4 - 2 0 3] - [1 0 - 2 - 2 - 1 2 3 4 1] - 3 [1 0 0 0 1 0 0 0 1] = A^{- 1} \Rightarrow [- 5 - 1 - 3 - 8 - 0 - 0 - 4 + 2 + 0 6 + 2 + 0 9 + 1 - 3 4 - 2 - 2 - 3 - 0 0 - 4 - 0 3 - 1 - 3] = [- 9 - 8 - 2 8 7 2 - 5 - 4 - 1] = A^{- 1} \Rightarrow A^{- 1} = [- 9 - 8 - 2 8 7 2 - 5 - 4 - 1]$

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Show that $A = [\begin{matrix} 53 - 1 - 2 \end{matrix}]$ satisfies the equation $A^{2}$ - 3A - 7I = 0 and hence find the value of $A^{- 1}$ .

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Inequalities Involving Modulus Function

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