The product of matrices A=[ cos 2 θ cos θ sin θ cos θ sin θ sin 2 θ] and sin B=[ cos 2 φ cos φ sin φ cos φ sin φ sin 2 φ] is a null matrix if θ-φ=

Question

The product of matrices A=[ cos 2 θ cos θ sin θ cos θ sin θ sin 2 θ] and sin B=[ cos 2 φ cos φ sin φ cos φ sin φ sin 2 φ] is a null matrix if θ-φ= (A) 2 n π, n ∈ Z (B) n (π/2), n ∈ Z (C) (2 n+1) (π/2), n ∈ Z (D) n π, n ∈ Z

Tardigrade Admin · Accepted Answer

A B=[ cos 2 θ cos θ sin θ cos θ sin θ sin 2 θ][ cos 2 φ cos φ sin φ cos φ sin φ sin 2 φ] <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/0109570176c51eac21b535b3bc99b183-.png" /> =[ cos θ cos φ( cos (θ-φ)) cos θ sin φ( cos (θ-φ)) sin θ cos φ( cos (θ-φ)) sin θ sin φ( cos (θ-φ))] =( cos (θ-φ))[ cos θ cos φ cos θ sin φ sin θ cos φ sin θ sin φ] Now A B=O ⇒ cos (θ-φ)=0 ⇒ θ-φ=(2 n+1) (π/2), n ∈ Z

Q. The product of matrices $A = [cos^{2} θ cos θ sin θ cos θ sin θ sin^{2} θ]$ and $sin B = [cos^{2} ϕ cos ϕ sin ϕ cos ϕ sin ϕ sin^{2} ϕ]$ is a null matrix if $θ - ϕ =$

$2 nπ, n \in Z$

$n \frac{π}{2}, n \in Z$

$(2 n + 1) \frac{π}{2}, n \in Z$

$nπ, n \in Z$

Q. The product of matrices A=[cos2θcosθsinθ​cosθsinθsin2θ​] and sinB=[cos2ϕcosϕsinϕ​cosϕsinϕsin2ϕ​] is a null matrix if θ−ϕ=

2nπ,n∈Z

n2π​,n∈Z

(2n+1)2π​,n∈Z

nπ,n∈Z

Q. The product of matrices $A = [cos^{2} θ cos θ sin θ cos θ sin θ sin^{2} θ]$ and $sin B = [cos^{2} ϕ cos ϕ sin ϕ cos ϕ sin ϕ sin^{2} ϕ]$ is a null matrix if $θ - ϕ =$

$2 nπ, n \in Z$

$n \frac{π}{2}, n \in Z$

$(2 n + 1) \frac{π}{2}, n \in Z$

$nπ, n \in Z$