Question

If $A\left(\theta \right)=\left[\begin{array}{cc}\sin \theta & i\cos \theta \\ i\cos \theta & \sin \theta \end{array}\right]$, then which of the following is not true?

• A. $A(\theta {)}^{-1}=A(\pi -\theta )$
• B. $A\left(\theta \right)+A(\pi +\theta )$ is a null matrix
• C. $A\left(\theta \right)$ is invertible for all $\theta \in R$
• D. $A(\theta {)}^{-1}=A(-\theta )$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $A(\theta {)}^{-1}=A(\pi -\theta )$
B $A\left(\theta \right)+A(\pi +\theta )$ is a null matrix
C $A\left(\theta \right)$ is invertible for all $\theta \in R$

Finding inverse of the matrix $A\left(\theta \right)=\left[\begin{array}{cc}\sin \theta & i\cos \theta \\ i\cos \theta & \sin \theta \end{array}\right]$

Determinant of $A\left(\theta \right)$ is $|A\left(\theta \right)|={\sin }^2\theta -i^2{\cos }^2\theta$

$={\sin }^2\theta +{\cos }^2\theta$

$=1$

Therefore $A\left(\theta \right)$ is a non-singular matrix. So , it is invertible of all $\theta \in R$

$A(\theta {)}^{-1}=\left[\begin{array}{cc}\sin \theta & -i\cos \theta \\ -i\cos \theta & \sin \theta \end{array}\right]$

Now. $A(\pi -\theta )=\left[\begin{array}{cc}\sin (\pi -\theta )& i\cos (\pi -\theta )\\ i\cos (\pi -\theta )& \sin (\pi -\theta )\end{array}\right]$

$=\left[\begin{array}{cc}\sin \theta & -i\cos \theta \\ -i\cos \theta & \sin \theta \end{array}\right]$

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

Now, $A(\pi +\theta )=\left[\begin{array}{cc}\sin (\pi +\theta )& i\cos (\pi +\theta )\\ i\cos (\pi +\theta )& \sin (\pi +\theta )\end{array}\right]$

$=\left[\begin{array}{cc}-\sin \theta & -i\cos \theta \\ -i\cos \theta & -\sin \theta \end{array}\right]$

$=-A\left(\theta \right)$

Therefore, $A\left(\theta \right)+A(\pi +\theta )=0$.

Hence, the correct options are $\left(A\right),\left(B\right)$ and $\left(C\right)$.