Let A=[ beginmatrix cos 2θ sin θ cos θ cos θ sin θ sin 2θ endmatrix ] and B=[ beginmatrix cos 2φ sin φ cos φ cos φ sin φ sin 2φ endmatrix ] then AB=0 if:

Question

Let A=[ beginmatrix cos 2θ sin θ cos θ cos θ sin θ sin 2θ endmatrix ] and B=[ beginmatrix cos 2φ sin φ cos φ cos φ sin φ sin 2φ endmatrix ] then AB=0 if: (A)  θ =nφ ,n=0,1,2,....  (B)  θ +φ =nπ ,n=0,1,2,....  (C)  θ =φ +(2n+1)(π /2),n=0,1,2,....  (D)  θ =φ +n(π /2),n=0,1,2,....

Tardigrade Admin · Accepted Answer

∴ AB=[ beginmatrix cos 2θ sin θ cos θ cos θ sin θ sin 2θ endmatrix ] [ beginmatrix cos 2φ sin φ cos φ cos φ sin φ sin 2φ endmatrix ] =[ beginmatrix cos 2θ cos 2φ + sin θ cos θ cos φ sin φ cos 2φ cos θ sin θ + sin 2θ sin φ cos φ endmatrix . . beginmatrix cos 2θ sin φ cos φ + sin 2φ sin θ cos θ cos θ sin θ sin φ cos φ + sin 2θ sin 2φ endmatrix ] =[ beginmatrix cos θ cos φ cos (θ -φ ) sin θ cos φ cos (θ -φ ) endmatrix . . beginmatrix sin φ cos θ cos (θ -φ ) sin θ sin φ cos (θ -φ ) endmatrix ] ∵ AB=0 ⇒ cos (θ -φ )=0 ⇒ cos (θ -φ )= cos (2n+1)(π /2) ⇒ θ =(2n+1)(π /2)+φ where, n=0,1, text 2,...

Q. Let $A = [cos^{2} θ cos θ sin θ sin θ cos θ sin^{2} θ]$ and $B = [cos^{2} ϕ cos ϕ sin ϕ sin ϕ cos ϕ sin^{2} ϕ]$ then $A B = 0$ if:

$θ = n ϕ, n = 0, 1, 2, ....$

$θ + ϕ = nπ, n = 0, 1, 2, ....$

$θ = ϕ + (2 n + 1) \frac{π}{2}, n = 0, 1, 2, ....$

$θ = ϕ + n \frac{π}{2}, n = 0, 1, 2, ....$

$θ = ϕ + 3 n \frac{π}{2}, n = 0, 1, 2, ....$

Q. Let A=[cos2θcosθsinθ​sinθcosθsin2θ​] and B=[cos2ϕcosϕsinϕ​sinϕcosϕsin2ϕ​] then AB=0 if:

θ=nϕ,n=0,1,2,....

θ+ϕ=nπ,n=0,1,2,....

θ=ϕ+(2n+1)2π​,n=0,1,2,....

θ=ϕ+n2π​,n=0,1,2,....