Question

If \( U = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \), then \( U^{-1} \) is:

• A. \( U^T \)
• B. \( U \)
• C. \( I \)
• D. \( 0 \)
• E. \( U^2 \)

Hint:

Matrices involving \( 1/\sqrt{2} \) or \( \sin\theta/\cos\theta \) are frequently orthogonal. Before doing a complex inverse calculation, check if the sum of squares of the first column equals 1.

Answer 1

The Correct Option is A

Concept: A square matrix \( U \) is called an orthogonal matrix if its inverse is equal to its transpose (\( U^{-1} = U^T \)). A matrix is orthogonal if and only if \( U^T U = I \). This implies that the columns and rows are unit vectors that are perpendicular to each other.

Step 1: Determine the transpose of the matrix \( U \).
The transpose \( U^T \) is formed by switching the rows and columns of \( U \): \[ U^T = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \]

Step 2: Verify the orthogonality condition \( U^T U = I \).
Multiply \( U^T \) by \( U \): \[ U^T U = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} - \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix} \] Calculating the top-left element: \[ \left( \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \right) = \frac{1}{2} + \frac{1}{2} = 1 \] Calculating the top-right element: \[ \left( \frac{1}{\sqrt{2}} \times -\frac{1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \right) = -\frac{1}{2} + \frac{1}{2} = 0 \] Since the full product yields \( \begin{bmatrix} 1 & 0 0 & 1 \end{bmatrix} \), the matrix is orthogonal, meaning \( U^{-1} = U^T \).