Question

Inverse of the matrix $\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$ is :

• A. $\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$
• B. $\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{array}\right]$
• C. $\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$
• D. $\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$

Answer 1

Answer: (D)

Let, $A=\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]$
And $A^{-1}=\frac{AdjA}{|A|}$
Here $\left|A\right|={\cos }^{2}2\theta -\left(-{\sin }^{2}2\theta \right)$
$={\cos }^{2}2\theta +{\sin }^{2}2\theta$
$=1\left(\because {\sin }^2\theta +{\cos }^2\theta =1\right)$
And , $AdjA=\left|\begin{array}{cc}{A}_{11}& A_{12}\\ A_{21}& A_{22}\end{array}\right|$
Where, $A_{11}$ = cofactor
and , $A_{11}=(-1{)}^{1+}.\cos 2\theta =\cos 2\theta$
$A_{12}=(-1{)}^{1+2}.\sin 2\theta =-\sin 2\theta$
$A_{21}=(-1{)}^{2+1}.(-\sin 2\theta )=+\sin 2\theta$
$A_{22}=(-1{)}^{2+2}\cos 2\theta =\cos 2\theta$
Hence, Adj $A={\left[\begin{array}{cc}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]}^T$
Where 'T' denotes the transpose of the matrix. And the transpose of the matrix is obtained by interchanging the rows and columns of the given matrix.
Thus, Adj $(A)=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$
$\Rightarrow A^{-1}=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$