2x2 Matrix Question how to show $A(-\theta)$

Question

The question is as follows -
If $A(\theta)= $$\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$$ $ show that $A^{-1}(\theta)=A(-\theta)$

I've already worked out that $A^{-1}(\theta)=$$$\begin{pmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{pmatrix}$$ So ofcourse $A(-\theta)$ is equal to same as above but I don't know how to mathematically show it

Answer 1

Note thta $\sin(-\theta)=-\sin\theta$ and $\cos(-\theta)=\cos\theta$

$$A(\theta)A(-\theta)=\begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix}\begin{bmatrix}\cos(-\theta) &-\sin(-\theta) \\ \sin(-\theta) & \cos(-\theta) \end{bmatrix}=\begin{bmatrix} \cos^2\theta+\sin^2\theta &\cos\theta\sin\theta-\cos\theta\sin\theta \\ \sin\theta\cos\theta-\sin\theta\cos\theta & \sin^2\theta+\cos^2\theta \end{bmatrix}=\begin{bmatrix}1&0\\0&1 \end{bmatrix} $$

Therefore, $A^{-1}(\theta)=A(-\theta)$

It is another method to prove $A^{-1}(\theta)=A(-\theta)$.

Answer 2

This is simply a rotation matrix (used pretty heavily in applications), sometimes also known as the Givens rotation (named after Wallace Givens). The inverse of a rotation matrix is its transpose, which is also a rotation matrix.

For a rotation matrix $A$,

\begin{equation*} \left(A^{T}\right)^{T} A^{T} = A A^{T} =\mathbb{I} \end{equation*}

Here the transpose is also reflection (hence $A(-\theta)=\left(A(\theta)\right)^{T}$) By the way, this can be generalized to any dimension. This (Givens rotation concept) is sometimes used in QR decomposition as an efficient implementation algorithm.

Also see: https://en.wikipedia.org/wiki/Givens_rotation