((♥JS♥)/(°js°)) Given a matrix A= ((( 3 2)),((−5 −4)) ) and A^2 +♭A−2I=0 where ♭ is a constant , I= (((1 0)),((0 1)) ). If B = (((−3♭ 2)),(( 5♭ −1)) ) , then A^(−1) B =


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Question Number 108059 by john santu last updated on 14/Aug/20

$\frac{♡ J S ♡}{° j s °}$ $G i v e n a m a t r i x A = (\begin{matrix} 3 2 \\ - 5 - 4 \end{matrix})$ $a n d A^{2} + ♭ A - 2 I = 0 w h e r e ♭ i s a$ $c o n s t a n t, I = (\begin{matrix} 1 0 \\ 0 1 \end{matrix}) . I f B =$ $(\begin{matrix} - 3 ♭ 2 \\ 5 ♭ - 1 \end{matrix}), t h e n A^{- 1} B =$
	Answered by bemath last updated on 14/Aug/20

	$\frac{⋋ B e M a t h ⋌}{⋎}$ $s i n c e ∣ A ∣ = - 12 + 10 = - 2 = - 2 ∣ I ∣$ $t h e n ♭ = - t r a c e (A) = 1, s o t h e e q u a t i o n b e c o m e s$ $A^{2} + A = 2 I \to 2 I = A (A + I)$ $b y C a y l e y - H a m i l t o n t h e o r e m$ $2 A^{- 1} = A + I; A^{- 1} = \frac{1}{2} (A + I)$ $A^{- 1} = \frac{1}{2} (\begin{matrix} 4 2 \\ - 5 - 3 \end{matrix}) a n d B = (\begin{matrix} - 3 2 \\ 5 - 1 \end{matrix})$ $t h e r e f o r e A^{- 1} B = \frac{1}{2} (\begin{matrix} - 2 6 \\ 0 - 7 \end{matrix})$ $= (\begin{matrix} - 1 3 \\ 0 - \frac{7}{2} \end{matrix})$
	Commented by bemath last updated on 14/Aug/20

	$s a n t u y y$
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