Question

Which one of the following matrices is orthogonal?

• A. \( \begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} \)
• B. \( \begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} \)
• C. \( \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} \)
• D. \( \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{\sqrt{2}} \end{bmatrix} \)

Hint:

For a matrix to be orthogonal, the rows (or columns) must be orthogonal to each other, and each row (or column) must have a magnitude of 1. To check this, compute the dot product of the rows and the dot product of each row with itself.

Answer 1

The Correct Option is B

A matrix is orthogonal if the transpose of the matrix is equal to its inverse, i.e., \( A^T A = I \), where \( A \) is the matrix and \( I \) is the identity matrix.

Let’s check each option to determine which matrix is orthogonal:

- Option (A): \[ A = \begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} \] The dot product of the rows of \( A \) is: \[ \left( \frac{1}{2} \times -\frac{\sqrt{3}}{2} \right) + \left( -\frac{\sqrt{3}}{2} \times \frac{1}{2} \right) = -\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = -\frac{\sqrt{3}}{2} \neq 0 \] Since the rows are not orthogonal, Option A is not orthogonal.

- Option (B): \[ B = \begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} \] The dot product of the rows of \( B \) is: \[ \left( \frac{1}{2} \times \frac{\sqrt{3}}{2} \right) + \left( -\frac{\sqrt{3}}{2} \times \frac{1}{2} \right) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 \] The rows are orthogonal. Also, the dot product of each row with itself is 1, so this matrix is orthogonal. Therefore, Option B is the correct answer.

- Option (C): \[ C = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} \] This matrix fails the orthogonality condition as the rows are not orthogonal. Hence, Option C is not orthogonal.

- Option (D): \[ D = \begin{bmatrix} \frac{1}{\sqrt{2}} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{\sqrt{2}} \end{bmatrix} \] This matrix also fails the orthogonality condition as the rows are not orthogonal. Hence, Option D is not orthogonal.

Thus, the correct option is Option B.