If `A=[cosalphasinalpha-sinalphacosalpha]` , then verify that `A^T\ A=I_2` .

Explanation

Taking L.H.S.

`A^T A`

Given

`A=[[cos alpha,sin alpha],[-sin alpha,cos alpha]]`

So, `A^T=[[cos alpha,-sin alpha],[sin alpha,cos alpha]]`

`A^T \ A=[[cos alpha,-sin alpha],[sin alpha,cos alpha]]`

`=[[(cos alpha*sin alpha)+(-sin alpha) (-sin alpha),(cos alpha * sin alpha) +(-sin alpha)*cos alpha],[sin alpha * cos alpha + cos alpha (-sin alpha),sin alpha* sin alpha + cos alpha*cos alpha]]`

`=[[cos^2 alpha + sin^2 alpha, sin alpha cos alpha – sin alpha con alpha], [sin alpha cos alpha – sin alpha cos alpha, sin^2 alpha + cos^2 alpha]]`
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