Invertible Matrices

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New answer posted

3 weeks ago

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R
Raj Pandey

Contributor-Level 9

l i m x 1 2 s i n ( c o s 1 x ) x 1 t a n ( c o s 1 x )

Let c o s 1 x = π 4 + θ

= l i m θ 2 s i n θ 2 t a n θ ( 1 t a n θ ) = 1 2

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

A = [, [-1, 4]. |A| = 2 - 1 = 1.
Characteristic equation: λ² - tr (A)λ + |A| = 0 ⇒ λ² - 3λ + 1 = 0.
By Cayley-Hamilton, A² - 3A + I = 0. A? ¹ (A² - 3A + I) = A - 3I + A? ¹ = 0.
A? ¹ = 3I - A.
Comparing with A? ¹ = αI + βA, we get α=3, β=-1.
4 (α - β) = 4 (3 - (-1) = 16.

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