1. Tardigrade
  2. Question
  3. Mathematics
  4. If Aα=[ cos α sin α - sin α cos α], then Aα Aβ is equal to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $A_\alpha=\begin{bmatrix}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{bmatrix}$, then $A_\alpha A_\beta$ is equal to

Matrices

Solution:

We have
$A_\alpha A_\beta =\begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}\begin{bmatrix} \cos \beta & \sin \beta \\ -\sin \beta & \cos \beta \end{bmatrix}$
$=\begin{bmatrix} \cos \alpha \cos \beta-\sin \alpha \sin \beta \cos \alpha \sin \beta+\sin \alpha \cos \beta \\ -\sin \alpha \cos \beta-\cos \alpha \sin \beta -\sin \alpha \sin \beta+\cos \alpha \cos \beta \end{bmatrix}$
$=\begin{bmatrix} \cos (\alpha+\beta) & \sin (\alpha+\beta) \\ -\sin (\alpha+\beta) & \cos (\alpha+\beta) \end{bmatrix}=A_{\alpha+\beta}$