If $A = \begin{bmatrix} 1 & \tan x \\ - \tan x & 1 \end{bmatrix}$, show that $\;$ $A^t A^{-1} = \begin{bmatrix} \cos 2x & - \sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}$
$A = \begin{bmatrix}
1 & \tan x \\
- \tan x & 1
\end{bmatrix}$
Transpose of $A = A^t = \begin{bmatrix}
1 & \tan x \\
- \tan x & 1
\end{bmatrix}^t$
i.e. $\;$ $A^t = \begin{bmatrix}
1 & - \tan x \\
\tan x & 1
\end{bmatrix}$
i.e. $\;$ $A^t = \begin{bmatrix}
1 & \dfrac{- \sin x}{\cos x} \\
\dfrac{\sin x}{\cos x} & 1
\end{bmatrix}$
$A^{-1} = \dfrac{1}{\left|A\right|} \times adj \left(A\right)$
$\left|A\right| = \begin{vmatrix}
1 & \tan x \\
- \tan x & 1
\end{vmatrix} = 1 + \tan^2 x = \sec^2 x$
Cofactors of matrix $A$ are:
$A_{11} = 1$, $\;$ $A_{12} = \tan x$, $\;$ $A_{21} = - \tan x$, $\;$ $A_{22} = 1$
$\therefore \;$ $adj \left(A\right) = \begin{bmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{bmatrix}^t = \begin{bmatrix}
1 & \tan x \\
- \tan x & 1
\end{bmatrix}^t = \begin{bmatrix}
1 & - \tan x \\
\tan x & 1
\end{bmatrix}$
$\begin{aligned}
\therefore \; A^{-1} & = \dfrac{1}{\sec^2 x} \begin{bmatrix}
1 & - \tan x \\
\tan x & 1
\end{bmatrix} \\\\
& = \cos^2 x \begin{bmatrix}
1 & \dfrac{- \sin x}{\cos x} \\
\dfrac{\sin x}{\cos x} & 1
\end{bmatrix} \\\\
& = \begin{bmatrix}
\cos^2 x & - \sin x \cos x \\
\sin x \cos x & \cos^2 x
\end{bmatrix}
\end{aligned}$
$\begin{aligned}
\therefore \; A^t A^{-1} & = \begin{bmatrix}
1 & \dfrac{- \sin x}{\cos x} \\
\dfrac{\sin x}{\cos x} & 1
\end{bmatrix} \begin{bmatrix}
\cos^2 x & - \sin x \cos x \\
\sin x \cos x & \cos^2 x
\end{bmatrix} \\\\
& = \begin{bmatrix}
\cos^2 x - \dfrac{\sin x}{\cos x} \times \sin x \cos x & - \sin x \cos x - \dfrac{\sin x}{\cos x} \times \cos^2 x \\
\dfrac{\sin x}{\cos x} \times \cos^2 x + \sin x \cos x & \dfrac{\sin x}{\cos x} \times \left(- \sin x \cos x\right) + \cos^2 x
\end{bmatrix} \\\\
& = \begin{bmatrix}
\cos^2 x - \sin^2 x & - 2 \sin x \cos x \\
2 \sin x \cos x & \cos^2 x - \sin^2 x
\end{bmatrix} \\\\
& = \begin{bmatrix}
\cos 2x & - \sin 2x \\
\sin 2x & \cos 2x
\end{bmatrix}
\end{aligned}$