Matrices & Determinants
Determinant Properties
nta_pyq_2025_apr
Grade 12
Question:
Let $A$ be a square matrix of order 3 such that $\det(A) = -2$ and $\det(3\,\text{adj}(-6\,\text{adj}(3A))) = 2^{m+n} \cdot 3^{mn}$, $m > n$. Then $4m + 2n$ is equal to ___
Step-by-Step Solution
Key Concept: Use $\det(\lambda M) = \lambda^n \det(M)$ for $n \times n$ matrix, and $\det(\text{adj}(M)) = (\det M)^{n-1}$. Simplify step by step from the inside out.
$\det(3\,\text{adj}(-6\,\text{adj}(3A))) = 3^3 \det((-6)^2 \text{adj}(3A))^2... = 3^3(-6)^6|3A|^4 = 3^3 \cdot 6^6 \cdot 3^4 \cdot 16 = 3^{21} \cdot 2^{10}$. Comparing: $m+n = 10, mn = 21 \Rightarrow m=7, n=3$. $4m+2n = 28+6 = 34$.
Correct Answer: 34