If $$A = \begin{bmatrix} cos\: \alpha & sin\: \alpha \\ -sin\: \alpha & cos\: \alpha \end{bmatrix}$$, then show that AA' = I.

Toolbox:
• If A_{i,j} be a matrix m*n matrix , then the matrix obtained by interchanging the rows and column of A is called as transpose of A.
• If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
• $cos\:^2 \alpha+sin^2\alpha=1$
• An identity matrix or unit matrix of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. An identity matrix of order 2, $I_{2}= \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}$
Step1:
Given
$A = \begin{bmatrix} cos\: \alpha & sin\: \alpha \\ -sin\: \alpha & cos\: \alpha \end{bmatrix}$
Transpose can be obtained by changing the rows into column.
$A' = \begin{bmatrix} cos\: \alpha & -sin\: \alpha \\ sin\: \alpha & cos\: \alpha \end{bmatrix}$
Step2:
$AA'= \begin{bmatrix} cos\: \alpha & sin\: \alpha \\ -sin\: \alpha & cos\: \alpha \end{bmatrix}\begin{bmatrix} cos\: \alpha & -sin\: \alpha \\ sin\: \alpha & cos\: \alpha \end{bmatrix}$
$\Rightarrow \begin{bmatrix} cos\:^2 \alpha+sin^2\alpha & cos\alpha(-sin\alpha)+sin\: \alpha cos\alpha \\ -sin\: \alpha cos\alpha+cos\alpha sin\alpha & sin^2\alpha+cos\: ^2\alpha \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 1 & -cos\alpha sin\alpha+sin \alpha cos\alpha \\ -sin \alpha cos\alpha+cos\alpha sin\alpha & 1 \end{bmatrix}$
$\Rightarrow \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$\Rightarrow AA'=I$