Consider the matrix Aα = `[(cos alpha, - sin alpha),(sin alpha, cos alpha)]` Show that `"A"_alpha "A"_beta = "A"_((alpha + beta))`
Solution
`"A"_alpha "A"_beta = [(cos alpha, - sin alpha),(sin alpha, cos alpha)] [(cos beta, - sin beta),(sin beta, cos beta)]`
= `[(cos alpha cos beta - sin alpha sin beta, - cos alpha sin beta - sin alpha cos beta),(sin alpha cos beta + cos alpha sin beta, - sin alpha sin beta + cos alpha cos beta)]`
= `[(cos alpha cos beta - sin alpha sin beta, -(sinalpha cos beta + cos alpha sin beta)),(sin alpha cos beta + cos alpha sin beta, cos alpha cos beta - sin alpha sin beta)]`
`"A"_alpha "A"_beta = [(cos(alpha + beta), - sin(alpha + beta)),(sin(alpha + beta) , cos(alpha + beta))]`
From equation (1), (2) and (3)
`"A"_alpha "A"_beta = "A"_((alpha + beta))`