Question

Given $a_i^2 + b_i^2 + c_i^2 = 1$ and $a_i a_j + b_i b_j + c_i c_j = 0$ for $i \neq j$, and
\[ A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} \] Then $\det(AA^T) = $ ?

• A. 0
• B. 1
• C. -1
• D. 3

Hint:

For orthogonal matrices, $AA^T = I$, so $\det(AA^T) = 1$

Answer 1

The Correct Option is B

Given vectors are orthonormal (dot products zero for $i \ne j$ and each has magnitude 1), hence matrix $A$ is orthogonal.
So $AA^T = I \Rightarrow \det(AA^T) = \det(I) = 1$