Let A be3×3 matrix with eigen values 1,−1,0. Then determinant of I+A^(100 ) =??


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Question Number 59295 by pooja24 last updated on 07/May/19

$L e t A b e 3 \times 3 m a t r i x w i t h e i g e n v a l u e s$ $1, - 1, 0 . T h e n d e t e r m i n a n t o f I + A^{100} = ? ?$
	Answered by alex041103 last updated on 10/May/19

	$b a s i c a l l y$ $A = {P D P}^{- 1}$ $w h e r e D i s a d i a g o n a l m a t r i x$ $\Rightarrow A^{100} = {P D}^{100} P^{- 1}$ $a l s o I = {P I P}^{- 1}$ $\Rightarrow M = I + A^{100} = P (I + D^{100}) P^{- 1}$ $T h e n d e t (M) = d e t (P) d e t (I + D^{100}) d e t (P^{- 1}) =$ $= [d e t (P) d e t (P^{- 1})] d e t (I + D^{100}) =$ $= d e t ({P P}^{- 1}) d e t (I + D^{100}) =$ $= d e t (I) d e t (I + D^{100}) =$ $= d e t (I + D^{100}) = d e t (M)$ $W e d e f i n e P a s t h e m a t r i x w h i c h g e t s$ $u s f r o m t h e n o r m a l s y s t e m o f c o o r d i n a t e s t o t h e$ $s y s t e m o f t h e c o o r d i n a t e s w h i c h u s e s$ $t h e e i g e n v e c t o r s a s b a s i s v e c t o r s . T h e n$ $D i s a d i a g o n a l m a t r i x w i t h t h e e i g e n$ $v a l u e s i n i t .$ $\Rightarrow D = [\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{matrix}]$ $\Rightarrow D^{100} = [\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ $\Rightarrow d e t (M) = \| \begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix} \| = 4$ $\Rightarrow d e t (I + A^{100}) = 4$
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