Matrices & Determinants
General
Grade 12
Question:
Let $A = \begin{bmatrix} 0 & 2q & r \\ p & q & -r \\ p & -q & r \end{bmatrix}$. If $AA^T = I^3$, then $|p|$ is :
$\frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{5}}$
$\frac{1}{\sqrt{6}}$
$\frac{1}{\sqrt{3}}$
Step-by-Step Solution
Key Concept: General
$A$ is orthogonal matrix $\Rightarrow 0^2 + p^2 + p^2 = 1 \Rightarrow |p| = \frac{1}{\sqrt{2}}$
Correct Answer: A