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: Understanding the Concept:
We need to calculate the matrix product of the transpose of A and the inverse of A.
First, find $A^T$. Then find the determinant $|A|$ and adjugate matrix $\text{adj}(A)$ to compute $A^{-1}$. Finally, multiply the two matrices.
Step 2: Key Formula or Approach:
For a $2 \times 2$ matrix $A = \begin{bmatrix} a & b
c & d \end{bmatrix}$:
Transpose $A^T = \begin{bmatrix} a & c
b & d \end{bmatrix}$.
Determinant $|A| = ad - bc$.
Inverse $A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b
-c & a \end{bmatrix}$.
Trigonometric identities:
$\frac{1 - \tan^2 x}{1 + \tan^2 x} = \cos 2x$
$\frac{2 \tan x}{1 + \tan^2 x} = \sin 2x$
Step 3: Detailed Explanation:
Let $t = \tan x$ for simplicity.
$A = \begin{bmatrix} 1 & t
-t & 1 \end{bmatrix}$
The transpose is:
$A^T = \begin{bmatrix} 1 & -t
t & 1 \end{bmatrix}$
The determinant is:
$|A| = (1)(1) - (t)(-t) = 1 + t^2$
The inverse is:
$A^{-1} = \frac{1}{|A|} \begin{bmatrix} 1 & -t
-(-t) & 1 \end{bmatrix} = \frac{1}{1 + t^2} \begin{bmatrix} 1 & -t
t & 1 \end{bmatrix}$
Notice that $A^{-1} = \frac{1}{1+t^2} A^T$.
Now, calculate the product $A^T A^{-1}$:
\[ A^T A^{-1} = A^T \left( \frac{1}{1 + t^2} A^T \right) = \frac{1}{1 + t^2} (A^T A^T) \]
Let's multiply $A^T$ by itself:
\[ A^T A^T = \begin{bmatrix} 1 & -t
t & 1 \end{bmatrix} \begin{bmatrix} 1 & -t
t & 1 \end{bmatrix} \]
\[ = \begin{bmatrix} (1)(1) + (-t)(t) & (1)(-t) + (-t)(1)
(t)(1) + (1)(t) & (t)(-t) + (1)(1) \end{bmatrix} \]
\[ = \begin{bmatrix} 1 - t^2 & -2t
2t & 1 - t^2 \end{bmatrix} \]
Now, multiply by the scalar $\frac{1}{1+t^2}$:
\[ A^T A^{-1} = \frac{1}{1 + t^2} \begin{bmatrix} 1 - t^2 & -2t
2t & 1 - t^2 \end{bmatrix} = \begin{bmatrix} \frac{1 - t^2}{1 + t^2} & \frac{-2t}{1 + t^2}
\frac{2t}{1 + t^2} & \frac{1 - t^2}{1 + t^2} \end{bmatrix} \]
Substitute $t = \tan x$ back into the matrix:
\[ A^T A^{-1} = \begin{bmatrix} \frac{1 - \tan^2 x}{1 + \tan^2 x} & -\frac{2\tan x}{1 + \tan^2 x}
\frac{2\tan x}{1 + \tan^2 x} & \frac{1 - \tan^2 x}{1 + \tan^2 x} \end{bmatrix} \]
Using the double angle trigonometric identities:
\[ A^T A^{-1} = \begin{bmatrix} \cos 2x & -\sin 2x
\sin 2x & \cos 2x \end{bmatrix} \]
Step 4: Final Answer:
The resulting matrix is $\begin{bmatrix} \cos 2x & -\sin 2x
\sin 2x & \cos 2x \end{bmatrix}$.