Inverse matrix for a matrix with sinus and cosinus functions

Question

I have this matrix A: $$\left(\begin{array}{cc} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right)$$

and I need to create an inverse matrix for this matrix A. The sinus and cosinus functions in there makes me confused, I don't know how to start and proceed. To count a determinant from this matrix is kinda easy, but how to count an inverse matrix to this?

Thank you

Answer 1

If $M$ is a matrix where $$ M=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) $$ then inverse of $M$ is $$ M^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc} d & -b \\ -c & a \end{array}\right) $$ Similarly $$ A=\left(\begin{array}{cc} \cos x & -\sin x \\ \sin x & \cos x \end{array}\right) $$ then inverse of $A$ is \begin{align} A^{-1}&=\frac{1}{(\cos x)(\cos x)-(-\sin x)(\sin x)}\left(\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right)\\ &=\frac{1}{\cos^2 x+\sin^2 x}\left(\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right)\\ &=\left(\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right) \end{align} where $\cos^2 x+\sin^2 x=1$.

Answer 2

HINT

Your matrix represents an anti-clockwise rotation about the origin, through $x$-radians.

Geometrically, what is the inverse of that transformation?

Clockwise rotations are given by negative angles.

If you insist upon findind the inverse algebraically, note that $\det = \cos^2x + \sin^2x \equiv 1$.

Answer 3

Your matrix is a rotation by angle $x$. Therefore the inverse is the rotation by the angle $-x$, which has the same form except you just substitute $-x$ for $x$. Using $\cos^2(x) + \sin^2(x) = 1$ and $\cos(-x) = \cos(x)$ and $\sin(-x) = - \sin(x)$ you can numerically verify this matrix gives the inverse.