If `A=[[cos alpha, -sin alpha] , [sin alpha, cos alpha]]` and `B=[[cos beta, -sin beta] , [sin beta, cos beta]]`then show that `AB=BA`

If A = [(cos alpha, sin alpha),(sin alpha, cos alpha)] and B = [(cos beta, sin beta),(sin beta, cos beta)] show that AB = BA

If A,=[[cos alpha,-sin alphasin alpha,cos alpha]] and B=,[[cos beta,cos alpha]] and B=[[cos beta,-sin betasin beta,cos betasin beta,cos beta]] then show that AB=BA

A = [[0, sin alpha, sin alpha sin beta-sin alpha, 0, cos alpha cos beta-sin alpha sin beta, -cos alpha cos beta, 0]]

A_((alpha))=[{:(cos alpha,-sin alpha),(sin alpha,cos alpha):}],A_((beta))=[{:(cos beta,-sin beta),(sin beta,cos beta):}] Which one of the following is correct ?

If A=[(cos^(2) alpha, cos alpha sin alpha),(cos alpha sin alpha,sin^(2)alpha)] and B=[(cos^(2) beta, cos beta sin beta),(cos beta sin beta, sin^(2) beta)] are two matrices such that the product AB is the null matrix then alpha-beta is

If A=[(cos^(2)alpha, cos alpha sin alpha),(cos alpha sin alpha, sin^(2)alpha)] and B=[(cos^(2)betas,cos beta sin beta),(cos beta sin beta, sin^(2) beta)] are two matrices such that the product AB is null matrix, then alpha-beta is

If A=[(cos^(2)alpha, cos alpha sin alpha),(cos alpha sin alpha, sin^(2)alpha)] and B=[(cos^(2)betas,cos beta sin beta),(cos beta sin beta, sin^(2) beta)] are two matrices such that the product AB is null matrix, then alpha-beta is

If A=[(cos^(2)alpha, cos alpha sin alpha),(cos alpha sin alpha, sin^(2)alpha)] and B=[(cos^(2)betas,cos beta sin beta),(cos beta sin beta, sin^(2) beta)] are two matrices such that the product AB is null matrix, then alpha-beta is

sin alpha + sin beta = a, cos alpha + cos beta = b rArr sin (alpha + beta)

If alpha-beta=(2n+1)(pi)/2, n epsilon Z then [(cos^(2) alpha, cos alpha sin alpha),(cos alpha sin alpha, sin^(2)alpha)][(cos^(2) beta, cos beta sin beta),(cos beta sin beta, sin^(2) beta)]=