Finding inverse of matrix with trig values

[fend]

Find the inverse of the rotation matrix

cos(theta) -sin(theta)
sin(theta) cos(theta)

I have no problems finding the inverse of a standard 2x2 matrix, however I am confused by the trig functions.

I guess the determanent would be 2cos(theta) and I would end up with

1/2cos(theta)*(the above 2x2 matrix)

Any assistance is greatly appreciated.

 

[weejee]

I think you made a sign error in applying the determinant formula.

 

[HallsofIvy]

Frankly, it appears you have no clue what you are doing. Pretty much you say is wrong. The determinant is NOT "2 cos(theta)", it is (cos(theta))(cos(theta))- (sin(theta))(-sin(theta))= cos^2(theta)+ sin^2(theta)= 1.

And the inverse of a matrix is NOT the determinant time the "above matrix", it is the determinant times the matrix made of the minors of the original matrix.

For this particular problem, the simplest way to find the inverse is to recognize that this matrix represents rotation, about the origin, through an angle theta. And that the inverse ("reverse") matrix is rotation about the origin through angle -theta.

 

[Mentallic]

Well it's no different to doing this for any other matrix

For a 2x2 matrix

a b
c d

the determinant is ad-bc, yes?

So for your example you should have $$sin^2\theta+cos^2\theta=1$$ so actually, the inverse of your matrix is exactly the same. (edit: but you switch the signs for the $$sin\theta$$'s.)

 

[fend]

Thank you for all the responses. The subject has been clarified and I appreciate the assistance provided.