Question

If $A=\left[\begin{array}{cc}1& tanx\\ -tanx& 1\end{array}\right]$, then $A^T{A}^{-1}$ is

• A. $\left[\begin{array}{cc}-cos2x& sin2x\\ -sin2x& cos2x\end{array}\right]$
• B. $\left[\begin{array}{cc}cos2x& -sin2x\\ sin2x& cos2x\end{array}\right]$
• C. $\left[\begin{array}{cc}cos2x& cos2x\\ cos2x& sin2x\end{array}\right]$
• D. none of these

Answer 1

The correct option is B $\left[\begin{array}{cc}cos2x& -sin2x\\ sin2x& cos2x\end{array}\right]$

$A=\left[\begin{array}{cc}1& tanx\\ -tanx& 1\end{array}\right]$
$\Rightarrow |A|={\sec }^{2}x$
$adjA=C^T={\left[\begin{array}{cc}1& \tan x\\ -\tan x& 1\end{array}\right]}^T$
$\Rightarrow adjA=\left[\begin{array}{cc}{\cos }^{2}x& -\sin x\cos x\\ \sin x\cos x& {\cos }^{2}x\end{array}\right]$
$A^{-1}=\frac{adjA}{|A|}$
$=\left[\begin{array}{cc}{\cos }^{2}x& -\sin x\cos x\\ \sin x\cos x& {\cos }^{2}x\end{array}\right]$
Now, $A^T{A}^{-1}=\left[\begin{array}{cc}1& -\tan x\\ \tan x& 1\end{array}\right]\left[\begin{array}{cc}{\cos }^{2}x& -\sin x\cos x\\ \sin x\cos x& {\cos }^{2}x\end{array}\right]$
$\Rightarrow A^T{A}^{-1}=\left[\begin{array}{cc}\cos 2x& -\sin 2x\\ \sin 2x& \cos 2x\end{array}\right]$