If A=[ cos 2 α cos α sin α cos α sin α sin 2 α] and B=[ cos 2 β cos β sin β cos β sin β sin 2 β] are two matrices such that A B=0, then α-β is equal to

Question

If A=[ cos 2 α cos α sin α cos α sin α sin 2 α] and B=[ cos 2 β cos β sin β cos β sin β sin 2 β] are two matrices such that A B=0, then α-β is equal to (A) 0 (B) π (C) (π/2) (D) None of these

Tardigrade Admin · Accepted Answer

The given matrices are A=[ cos 2 α cos α sin α cos α sin α sin 2 α] and B=[ cos 2 β cos β sin β cos β sin β sin 2 β] It is given that A B=0 ⇒[ cos 2 α cos α sin α cos α sin α sin 2 α] [ cos 2 β cos β sin β cos β sin β sin 2 β]=[0 0 0 0] ⇒[ cos 2 α cos 2 β+ cos α sin α cos β sin β cos 2 α cos β sin β+ cos α sin α sin 2 β cos α sin α cos 2 β+ sin 2 α cos β sin β cos α sin α cos β sin β+ sin 2 α sin 2 β] =[0 0 0 0] ⇒[ cos α cos β ⋅ cos (α-β) cos α sin β cos (α-β) cos β sin α cos (α-β) sin α sin β cos (α-β)] =[0 0 0 0] On equating the corresponding parts, we have cos α ⋅ cos β cos (α-β)=0 cos β sin α cos (α-β)=0 cos α sin β cos (α-β)=0 sin α sin β cos (α-β)=0 ⇒ cos (α-β)=0 ∴ α-β=(π/2)

Q. If $A = [cos^{2} α cos α sin α cos α sin α sin^{2} α]$ and $B = [cos^{2} β cos β sin β cos β sin β sin^{2} β]$ are two matrices such that $A B = 0$ , then $α - β$ is equal to

0

$π$

$\frac{π}{2}$

None of these

Q. If A=[cos2αcosαsinα​cosαsinαsin2α​] and B=[cos2βcosβsinβ​cosβsinβsin2β​] are two matrices such that AB=0, then α−β is equal to

0

π

2π​

None of these

Q. If $A = [cos^{2} α cos α sin α cos α sin α sin^{2} α]$ and $B = [cos^{2} β cos β sin β cos β sin β sin^{2} β]$ are two matrices such that $A B = 0$ , then $α - β$ is equal to

$π$

$\frac{π}{2}$