Wanna go for a 'Det.'?

Algebra Level 5

A and B are two square matrices such that A^2×B = BA and if (AB)^10 = A^k×B^10 . Then find the value of (K-1020)/e . ( e is the base of normal logarithm)


The answer is 1.1036.

Arka Dutta by Arka Dutta , 19, India

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2 solutions

Mark Hennings
Apr 3, 2019

Two simple inductions show that B A n = A 2 n B ( A B ) n = A 2 n − 1 B n BA^n \; = \; A^{2n}B \hspace{2cm} (AB)^n \; = \; A^{2^n-1}B^n for all n ∈ N n \in \mathbb{N} . Thus k = 2 10 − 1 k = 2^{10}-1 , and hence the answer is 3 e \boxed{\tfrac{3}{e}} .

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Yeah I have done the same way

Arka Dutta - 2 years, 2 months ago
Arka Dutta
Apr 3, 2019

I have added a solution here. There may be some minor mistakes in calculation.

[Hint to Matrix Problem] (http://imgur.com/gallery/VHs7niy)

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