Question

If $A = \begin{bmatrix} 1 & \tan x \\ -\tan x & 1 \end{bmatrix}$, then $A^T A^{-1} =$}

• A. $\begin{bmatrix} \cos 2x & -\sin 2x \\ -\sin 2x & \cos 2x \end{bmatrix}$
• B. $\begin{bmatrix} \cos 2x & -\sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}$
• C. $\begin{bmatrix} -\cos 2x & \sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}$
• D. $\begin{bmatrix} -\cos 2x & \sin 2x \\ -\sin 2x & \cos 2x \end{bmatrix}$

Hint:

Note that in this specific case $A^T = adj(A)$, so $A^T A^{-1} = \frac{1}{|A|} (A^T)^2$.

Answer 1

The Correct Option is B

Step 1: Concept
Find $A^T$ and $A^{-1}$ using the formula $A^{-1} = \frac{1}{|A|} adj(A)$.

Step 2: Meaning
$|A| = 1 - (-\tan^2 x) = 1 + \tan^2 x = \sec^2 x$. $A^T = \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$. $A^{-1} = \frac{1}{\sec^2 x} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} = \cos^2 x \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix}$.

Step 3: Analysis
Multiply $A^T A^{-1}$: $\cos^2 x \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} \begin{bmatrix} 1 & -\tan x \\ \tan x & 1 \end{bmatrix} = \cos^2 x \begin{bmatrix} 1-\tan^2 x & -2\tan x \\ 2\tan x & 1-\tan^2 x \end{bmatrix}$. Using double angle formulas: $\cos^2 x(1 - \tan^2 x) = \cos 2x$ and $\cos^2 x(2 \tan x) = \sin 2x$.

Step 4: Conclusion
The resulting matrix is $\begin{bmatrix} \cos 2x & -\sin 2x
\sin 2x & \cos 2x \end{bmatrix}$. Final Answer: (B)