Question

Which one of the following matrices is orthogonal?

• A. $\begin{bmatrix} \tfrac{1}{2} & -\tfrac{\sqrt{3}}{2}
  -\tfrac{\sqrt{3}}{2} & \tfrac{1}{2} \end{bmatrix}$
• B. $\begin{bmatrix} \tfrac{1}{2} & -\tfrac{\sqrt{3}}{2}
  \tfrac{\sqrt{3}}{2} & \tfrac{1}{2} \end{bmatrix}$
• C. $\begin{bmatrix} \tfrac{1}{\sqrt{2}} & -\tfrac{\sqrt{3}}{2}
  -\tfrac{\sqrt{3}}{2} & \tfrac{1}{2} \end{bmatrix}$
• D. $\begin{bmatrix} \tfrac{1}{\sqrt{2}} & -\tfrac{\sqrt{3}}{2}
  \tfrac{\sqrt{3}}{2} & -\tfrac{1}{\sqrt{2}} \end{bmatrix}$

Hint:

To test orthogonality quickly, check if the column vectors are unit length and mutually perpendicular (dot product zero). This is often faster than multiplying $A^T A$.

Answer 1

The Correct Option is B

Step 1: Recall condition for orthogonality.
A matrix $A$ is orthogonal if $A^T A = I$. This means that the columns (and rows) of $A$ are orthonormal vectors. Step 2: Check Option (A).
\[ A = \begin{bmatrix} \tfrac{1}{2} & -\tfrac{\sqrt{3}}{2}
-\tfrac{\sqrt{3}}{2} & \tfrac{1}{2} \end{bmatrix} \] Column 1 = $\left(\tfrac{1}{2}, -\tfrac{\sqrt{3}}{2}\right)$ has length \[ \left(\tfrac{1}{2}\right)^2 + \left(-\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{1}{4} + \tfrac{3}{4} = 1. \] Column 2 = $\left(-\tfrac{\sqrt{3}}{2}, \tfrac{1}{2}\right)$ also has length $1$. Dot product = $\tfrac{1}{2}\left(-\tfrac{\sqrt{3}}{2}\right) + \left(-\tfrac{\sqrt{3}}{2}\right)\left(\tfrac{1}{2}\right) = -\tfrac{\sqrt{3}}{2} \neq 0$. Hence not orthogonal. Step 3: Check Option (B).
\[ B = \begin{bmatrix} \tfrac{1}{2} & -\tfrac{\sqrt{3}}{2}
\tfrac{\sqrt{3}}{2} & \tfrac{1}{2} \end{bmatrix} \] Column 1 = $\left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right)$ length = $\tfrac{1}{4} + \tfrac{3}{4} = 1$. Column 2 = $\left(-\tfrac{\sqrt{3}}{2}, \tfrac{1}{2}\right)$ length = $1$. Dot product = $\tfrac{1}{2}\left(-\tfrac{\sqrt{3}}{2}\right) + \tfrac{\sqrt{3}}{2}\left(\tfrac{1}{2}\right) = 0$. Hence columns are orthonormal ⇒ Matrix is orthogonal. Step 4: Quickly check C and D.
- (C) Column lengths not equal to 1 ⇒ Not orthogonal.
- (D) Column lengths not unit ⇒ Not orthogonal. \[ \boxed{\text{Correct orthogonal matrix is Option (B)}} \]