A= [cos θ&-sin θ sin θ&cos θ] and AB= BA =I , then B is equal to

Question

A= [cos θ&-sin θ sin θ&cos θ] and AB= BA =I , then B is equal to (A)  [-cos θ&sin θ sin θ&cos θ]  (B)  [cos θ&sin θ -sin θ&cos θ]  (C)  [-sin θ&cos θ cos θ&sin θ]  (D)  [sin θ&-cos θ -cos θ&sin θ]

Tardigrade Admin · Accepted Answer

Given, A=[cos θ&-sin θ sin θ&cos θ] and AB = BA = I ⇒ B = A-1 I = A-1 =(1/cos2 θ +sin 2θ )[cos θ&sin θ -sin θ&cos θ] ⇒ B=[cos θ&sin θ -sin θ&cos θ]

Q. $A = [cos θ s in θ - s in θ cos θ]$ and $A B = B A = I$ , then $B$ is equal to

$[- cos θ s in θ s in θ cos θ]$

$[cos θ - s in θ s in θ cos θ]$

$[- s in θ cos θ cos θ s in θ]$

$[s in θ - cos θ - cos θ s in θ]$

Given, $A = [cos θ s in θ - s in θ cos θ]$
and $A B = B A = I$
$\Rightarrow B = A^{- 1} I = A^{- 1}$
$= \frac{1}{co s ^{2} θ + s i n ^{2} θ} [cos θ - s in θ s in θ cos θ]$
$\Rightarrow B = [cos θ - s in θ s in θ cos θ]$

Q. A=[cosθsinθ​−sinθcosθ​] and AB=BA=I , then B is equal to

[−cosθsinθ​sinθcosθ​]

[cosθ−sinθ​sinθcosθ​]

[−sinθcosθ​cosθsinθ​]

[sinθ−cosθ​−cosθsinθ​]