A square matrix A is called orthogonal if (A^T)A = I_n.(a) Show that A = [[cos(theta), -sin(theta)],[sin(theta), cos(theta)]] is orthogonal.

Asked on by bogshow24

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

sorry , in secod row and first element I slipped.that should be

`A^TA=[[cos^2(theta)+sin^2(theta),-cos(theta).sin(theta)+sin(theta).cos(theta)],[-sin(theta)cos(theta)+cos(theta)sin(theta),sin^2(theta)+cos^2(theta)]]`

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

Given  matrix A. We wish to prove  A is  orthogonal.
By def of orthogonal matrix ,we have

`A^TA=[[1,0],[0,1]]`  , A is matrix of order 2 ,so `I_2=[[1,0],[0,1]]`

Now consider

`A=[[cos(theta),-sin(theta)],[sin(theta),cos(theta)]]`

`A^T=[[cos(theta),sin(theta)],[-sin(theta),cos(theta)]]`

`A^TA=[[cos(theta),sin(theta)],[-sin(theta),cos(theta)]][[cos(theta),-sin(theta)],[sin(theta),cos(theta)]]`

`=[[cos^2(theta)+sin^2(theta),-cos(theta)sin(theta)+sin(theta)cos(theta)],[-sin(theta)+cos(theta)sin(theta),sin^2(theta)+cos^2(theta)]]`

=`[[1,0],[0,1]]`

``  because `cos^2(theta)+sin^2(theta)=1`

Thus A is an orthogonal matrix.

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