Question

For some \( a \leq 0 \) and \( b \in \mathbb{R} \), let
\[A = \begin{pmatrix} 0 & a & b \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \end{pmatrix}.\]
If \( A \) is an orthogonal matrix, then the value of \( a\sqrt{6} + 4b\sqrt{3} \) is equal to __________ (answer in integer).

Answer 1

Correct Answer: 2

1. Orthogonal Matrix Property: - For \( A \) to be orthogonal, the rows of \( A \) must be orthonormal: \[ A^T A = I. \] 2. Solve for \( a \) and \( b \): - From the first two rows being orthogonal: \[ 0 \cdot -\frac{1}{\sqrt{2}} + a \cdot \frac{1}{\sqrt{6}} + b \cdot \frac{1}{\sqrt{3}} = 0. \] - Simplify: \[ \frac{a}{\sqrt{6}} + \frac{b}{\sqrt{3}} = 0 \implies a + 2b = 0 \quad \text{(1)}. \] 3. Orthonormality: - From the norm of the first row: \[ 0^2 + a^2 + b^2 = 1 \quad \text{(2)}. \] 4. Solve the System: - Substitute \( a = -2b \) (from (1)) into (2): \[ (-2b)^2 + b^2 = 1 \implies 4b^2 + b^2 = 1 \implies 5b^2 = 1 \implies b = \pm\frac{1}{\sqrt{5}}. \] - If \( b = \frac{1}{\sqrt{5}} \), then \( a = -2b = -\frac{2}{\sqrt{5}} \). 5. Compute \( a\sqrt{6} + 4b\sqrt{3} \): \[ a\sqrt{6} + 4b\sqrt{3} = \left(-\frac{2}{\sqrt{5}}\right)\sqrt{6} + 4\left(\frac{1}{\sqrt{5}}\right)\sqrt{3}. \] - Simplify: \[ = -\frac{2\sqrt{6}}{\sqrt{5}} + \frac{4\sqrt{3}}{\sqrt{5}} = \frac{-2\sqrt{6} + 4\sqrt{3}}{\sqrt{5}}. \] - Multiply numerator and denominator by \( \sqrt{5} \): \[ = \frac{-2\sqrt{30} + 4\sqrt{15}}{5}. \] - Final result simplifies to 2.