Matrices & Determinants
Nested adjugate determinant
nta_pyq_2025_apr
Grade 12
Question:
Let $A$ be a $3 \times 3$ matrix such that $|\text{adj}(\text{adj}(\text{adj}A))| = 81$. If $S = \{n \in \mathbb{Z} : (|\text{adj}(\text{adj}A)|)^{2(n-1)} = |A|^{2(3n^2-5n-4)}\}$, then $\sum_{n \in S} |n^2 + n|$ is equal to
Step-by-Step Solution
Key Concept: Apply the matrix property for nested adjugate determinant and reduce it to determinant or parameter equations.
$|\text{adj}(\text{adj}(\text{adj}A))| = 81$
$|\text{adj}A|^4 = 81 \Rightarrow |\text{adj}A| = 3$
$|A|^2 = 3 \Rightarrow |A| = \sqrt{3}$
$(|\text{adj}(\text{adj}A)|)^{2(n-1)} = |A|^{2(3n^2-5n-4)}$
$3^{2(n-1)} = (\sqrt{3})^{2(3n^2-5n-4)}$
$3^{2(n-1)} = 3^{3n^2-5n-4}$
$2(n-1) = 3n^2 - 5n - 4$
$2n - 2 = 3n^2 - 5n - 4$
$0 = 3n^2 - 7n - 2$
$0 = 3n^2 - 5n - 4$
$(n-3)(n+2) = 0$
$n = 3, -2$
$\sum_{n \in S} |n^2 + n| = |3^2 + 3| + |(-2)^2 + (-2)| = |9 + 3| + |4 - 2| = 12 + 2 = 732$
Correct Answer: 4