((♭emath)/(•••••)) use cayley − hamilton theorem to calculate A^(−1) for A= (((1 2 2)),((1 2 −1)),((−1 1 4)) )


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Question Number 109240 by bemath last updated on 22/Aug/20

$\frac{♭ e m a t h}{∙ ∙ ∙ ∙ ∙}$ $u s e c a y l e y - h a m i l t o n t h e o r e m$ $t o c a l c u l a t e A^{- 1} f o r A = (\begin{matrix} 1 2 2 \\ 1 2 - 1 \\ - 1 1 4 \end{matrix})$
	Answered by bobhans last updated on 22/Aug/20

	$You can't use 'macro parameter character #' in math mode$ $t h e c h a r a c t e r i s t i c e q u a t i o n$ $λ^{3} - (t r A) λ^{2} + (\begin{matrix} m i n o r \\ f r o m d i a g o n a l \end{matrix}) λ - d e t (A) = 0$ $λ^{3} - 9 λ^{2} + (\| \begin{matrix} 2 - 1 \\ 1 4 \end{matrix} \| + \| \begin{matrix} 1 2 \\ - 1 4 \end{matrix} \| + \| \begin{matrix} 1 2 \\ 1 2 \end{matrix} \|) λ - d e t (A) = 0$ $λ^{3} - 9 λ^{2} + 15 λ - 9 = 0$ $B y C a y l e y - H a m i l t o n t h e o r e m$ $A^{3} - 9 A^{2} + 15 A - 9 I = 0$ $m u l t i p l y b o t h s i d e s b y A^{- 1}$ $\Rightarrow A^{2} - 9 A + 15 I - 9 A^{- 1} = 0$ $9 A^{- 1} = A^{2} - 9 A + 15 I$ $9 A^{- 1} = (\begin{matrix} 1 8 8 \\ 4 5 - 4 \\ - 4 4 13 \end{matrix}) - (\begin{matrix} 9 18 18 \\ 9 18 - 9 \\ - 9 9 36 \end{matrix}) + (\begin{matrix} 15 0 0 \\ 0 15 0 \\ 0 0 15 \end{matrix})$ $9 A^{- 1} = (\begin{matrix} 7 - 10 - 10 \\ - 5 2 5 \\ 5 - 5 - 8 \end{matrix})$ $∴ A^{- 1} = \frac{1}{9} (\begin{matrix} 7 - 10 - 10 \\ - 5 2 5 \\ 5 - 5 - 8 \end{matrix})$
	Commented by john santu last updated on 22/Aug/20

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