Solve Matrix Inverse with Trigonometric Values

[nk735]

Homework Statement

Show that the matrix [cos(theta), -sin(theta); sin(theta), cos(theta)] is invertible, regardless of the value of theta

Homework Equations

Identity matrix, elementary row operations

The Attempt at a Solution

I have the basic idea as to how to go about this; (let the above matrix = A)

- form an augmented matrix with the identity matrix, eg. [A|I]

- perform row operations (forward and backwards elimination) until matrix looks like [I|A^-1]

However, I'm at a loss as to how to perform these operations with the trigonometric values instead of numbers.

Just a push in the right direction would be greatly appreciated, i'd like to solve this myself

 

[morphism]

Try to think about this matrix geometrically. Can you guess what a possible inverse of it is? Verify that your guess is an actual inverse.

 

[HallsofIvy]

The problem doesn't actually ask you to find the inverse- just to show that it exists. Are you aware that a matrix is invertible if and only if its determinant is not 0? What is the determinant of this matrix?

Actually, it's not that hard to find the inverse they way you are doing it- just tedious. It turns out to be surprisingly easy and morphism's suggestion shows why.

 

[tiny-tim]

… just a gentle push …

Just a push in the right direction would be greatly appreciated, i'd like to solve this myself

Hi nk735!

Try:
(cosA sinA (cosB sinB
-sinA cosA) x -sinB cosB).

The (1,1) term will be cosAcosB + sinA(-sinB), = … ?

So the whole matrix is … ?