Question

If $A\left(\alpha \right)=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$, then $A\left(\alpha \right)A\left(\beta \right)=$ ____

• A. $A\left(\alpha \right)-A\left(\beta \right)$
• B. $A\left(\alpha \right)+A\left(\beta \right)$
• C. $A(\alpha -\beta )$
• D. $A(\alpha +\beta )$

Answer 1

The correct option is D $A(\alpha +\beta )$

(D) $A\left(\alpha \right).A\left(\beta \right)=\left[\begin{array}{cc}cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{array}\right]\left[\begin{array}{cc}cos\beta & sin\beta \\ -sin\beta & cos\beta \end{array}\right]$
$=\left[\begin{array}{cc}cos\alpha .cos\beta -sin\alpha sin\beta & cos\alpha .sin\beta +sin\alpha cos\beta \\ -sin\alpha .cos\beta -cos\alpha sin\beta & cos\alpha .cos\beta -sin\alpha sin\beta \end{array}\right]$
$=\left[\begin{array}{cc}cos(\alpha +\beta )& sin(\alpha +\beta )\\ -sin(\alpha +\beta )& cos(\alpha +\beta )\end{array}\right]$
$=A(\alpha +\beta )$