Question

The multiplicative inverse of $A=\left[\begin{array}{cc}cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}\right]$ is

• A. $\left[\begin{array}{cc}-cos\theta & sin\theta \\ -sin\theta & -cos\theta \end{array}\right]$
• B. $\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]$
• C. $\left[\begin{array}{cc}-cos\theta & -sin\theta \\ sin\theta & -cos\theta \end{array}\right]$
• D. $\left[\begin{array}{cc}cos\theta & sin\theta \\ sin\theta & -cos\theta \end{array}\right]$

Answer 1

Answer: (B)

$|A|={\cos }^2\theta +{\sin }^2\theta =1$
adj (A) =$\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]$
$A^{-1}=\frac{adj(A)}{|A|}=\left[\begin{array}{cc}cos\theta & sin\theta \\ -sin\theta & cos\theta \end{array}\right]$