If A=[\table \cos \alpha,\sin \alpha;-\sin \alpha,\cos \alpha], then for any positive integer n, A^n is

Correct Answer is : [cosnαsinnα−sinnαcosnα]

[\table \sin n\alpha,\cos n\alpha \cos n\alpha,-\sin n\alpha]

[\table \cos n\alpha,\sin n\alpha;\sin n\alpha,\cos n\alpha]

[\table \cos n\alpha,\sin n\alpha;\sin n\alpha,-\cos n\alpha]

[\table \cos n\alpha,\sin n\alpha;-\sin n\alpha,\cos n\alpha]

NIMCET 2020 Question Paper with solutions

Question : 13 of 120

Marks: +1, -0

A=[\table \cos α,\sin α;-\sin α,\cos α]

, then for any positive integer n,

A^n

Solution:

A=[\table \cos α,\sinα;-\sinα,\cosα]

A^2=[\table \cos α,\sinα;-\sinα,\cosα].[\table \cos α,\sinα;-\sinα,\cosα]

A^2=[\table \cos2α,\sin 2α;-\sin2α,\cos2α]

A^3=A^2 .A

A^2=[\table \cos2 α\sin2α;-\sin2α,\cos2α].[\table \cos α,\sinα;-\sinα,\cosα]

=[\table \cos3 α,\sin3α;-\sin3α,\cos3α]

By induction n=1,2,.....

A^n=[\table \cos nα,\sin nα;-\sin nα,\cos nα]

Go to Question:

Prev Question

Next Question