Question

If \( A = \begin{bmatrix} 1 & \cot x \\ -\cot x & 1 \end{bmatrix} \), show that \( A^T A^{-1} = \begin{bmatrix} -\cos 2x & -\sin 2x \\ \sin 2x & -\cos 2x \end{bmatrix}. \)

Hint:

For matrices with trigonometric elements, use trigonometric identities like \( 1 - \cot^2 x = -\cos 2x \) to simplify determinant and inverse calculations.

Answer 1

Step 1: Compute the transpose of \( A \).
The matrix \( A \) is: \[ A = \begin{bmatrix} 1 & \cot x \\ -\cot x & 1 \end{bmatrix}. \] The transpose of \( A \) is: \[ A^T = \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}. \] Step 2: Compute the determinant of \( A \).
\[ \text{det}(A) = (1)(1) - (-\cot x)(\cot x) = 1 - \cot^2 x. \] Using the trigonometric identity \( 1 + \cot^2 x = \csc^2 x \), we get: \[ \text{det}(A) = \frac{-\cos 2x}{\sin^2 x}. \] Step 3: Compute the inverse of \( A \).
The formula for the inverse of a 2x2 matrix is: \[ A^{-1} = \frac{1}{\text{det}(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}. \] For \( A = \begin{bmatrix} 1 & \cot x \\ -\cot x & 1 \end{bmatrix} \), we have: \[ A^{-1} = \frac{1}{1 - \cot^2 x} \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}. \] Using the identity \( 1 - \cot^2 x = -\cos 2x \), the inverse becomes: \[ A^{-1} = \frac{1}{-\cos 2x} \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}. \] Step 4: Compute \( A^T A^{-1} \).
Substitute \( A^T \) and \( A^{-1} \): \[ A^T = \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}, \quad A^{-1} = \frac{1}{-\cos 2x} \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}. \] Multiply \( A^T \) and \( A^{-1} \): \[ A^T A^{-1} = \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix} \cdot \frac{1}{-\cos 2x} \begin{bmatrix} 1 & -\cot x \\ \cot x & 1 \end{bmatrix}. \] Step 5: Simplify the product.
Compute the product: \[ A^T A^{-1} = \frac{1}{-\cos 2x} \begin{bmatrix} 1 - \cot^2 x & -\cot x - \cot x \\ \cot x + \cot x & 1 - \cot^2 x \end{bmatrix}. \] Using \( 1 - \cot^2 x = -\cos 2x \) and simplifying: \[ A^T A^{-1} = \begin{bmatrix} -\cos 2x & -\sin 2x \\ \sin 2x & -\cos 2x \end{bmatrix}. \] Conclusion:
We have shown that: \[ \boxed{A^T A^{-1} = \begin{bmatrix} -\cos 2x & -\sin 2x \\ \sin 2x & -\cos 2x \end{bmatrix}}. \]