Question

For real constants $a$ and $b$, let \[ M = \begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} \\ a & b \end{bmatrix} \] be an orthogonal matrix. Then which of the following statements is/are always TRUE?

• A. $a + b = 0$
• B. $b = \sqrt{1 - a^2}$
• C. $ab = -\dfrac{1}{2}$
• D. $M^2 = I_2$

Hint:

For a matrix to be orthogonal, its columns (or rows) must be orthonormal — both unit length and mutually perpendicular.

Answer 1

The Correct Option is A, C

Step 1: Orthogonality condition. 
For $M$ to be orthogonal, $M^T M = I_2$. That is, \[ \begin{bmatrix} \dfrac{1}{\sqrt{2}} & a \\ \dfrac{1}{\sqrt{2}} & b \end{bmatrix} \begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} \\ a & b \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. \]

Step 2: Compute the product.
\[ M^T M = \begin{bmatrix} \dfrac{1}{2} + a^2 & \dfrac{1}{2} + ab \\ \\ \dfrac{1}{2} + ab & \dfrac{1}{2} + b^2 \end{bmatrix}. \]

Step 3: Apply orthogonality equations.
From $M^T M = I_2$, we get: \[ \dfrac{1}{2} + a^2 = 1 \text{and} \dfrac{1}{2} + ab = 0. \] Thus: \[ a^2 = \dfrac{1}{2}, ab = -\dfrac{1}{2}. \] This implies $b = -a$. Hence, $a + b = 0$.

Step 4: Conclusion.
\[ \boxed{a + b = 0 \text{ and } ab = -\dfrac{1}{2}}. \]