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, $A=\left[\begin{array}{cc}cos2\theta & -sin2\theta \\ <br/><br/>sin2\theta & cos2\theta \end{array}\right]$
And $A^{-1}=\frac{Adj.A}{|A|}$
Here $|A|=co{s}^{2}2\theta -\left(-{\sin }^{2}2\theta \right)={\cos }^{2}2\theta +{\sin }^{2}2\theta$
$=1\left(\because {\sin }^2\theta +{\cos }^2\theta =1\right)$
And, Adj $A=\left[\begin{array}{cc}{A}_{11}& A_{12}\\ A_{21}& A_{22}<br/><br/>\end{array}\right]$
Where, $A_{11}$ cofactor
and, $A_{11}=(-1{)}^{i+i}\cdot \cos 2\theta =\cos 2\theta$
$A_{12}=(-1{)}^{1+2}\cdot \sin 2\theta =-\sin 2\theta$
$A_{21}=(-1{)}^{2+1}\cdot (-\sin 2\theta )=+\sin 2\theta$
$A_{22}=(-1{)}^{2+2}\cos 20=\cos 20$
Hence, Adj A $={\left[\begin{array}{cc}cc\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]}^T$
Thus, Adj ( A ) $=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$
$\Rightarrow A^{-1}=\left[\begin{array}{cc}\cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{array}\right]$