Matrices & Determinants
Orthogonal matrix parameter condition
nta_pyq_2025_apr
Grade 12
Question:
Let $A = \begin{bmatrix} \cos\theta & 0 & -\sin\theta \\ 0 & 1 & 0 \\ \sin\theta & 0 & \cos\theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to _____.
Step-by-Step Solution
Key Concept: Apply the matrix property for orthogonal matrix parameter condition and reduce it to determinant or parameter equations.
Since $A$ is an orthogonal matrix, $A^T = A^{-1}$.
$A^2 = A^T$ implies $A^2 = A^{-1}$
$\Rightarrow A^3 = I$
Let $B = (A + I)^3 + (A - I)^3 - 6A$
$= 2(A^3 + 3A) - 6A$
$= 2A^3 + 6A - 6A$
$= 2I$
$B = 2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
Sum of diagonal elements $= 2 + 2 + 2 = 6$
Correct Answer: 6