Matrices & Determinants
Iterated Adjoint Determinant — Modular Arithmetic
nta_pyq_2024_jan
Grade 12
Question:
Let $A$ be a $3\times3$ matrix and $\det(A)=2$. If $n=\det(\underbrace{\text{adj}(\text{adj}(\cdots(\text{adj}\,A)\cdots))}_{\text{2024 times}})$, then the remainder when $n$ is divided by $9$ is equal to
Step-by-Step Solution
Key Concept: For an $n\times n$ matrix, $\det(\text{adj}(A))=\det(A)^{n-1}$. For $3\times3$: $\det(\text{adj}(A))=\det(A)^2$. Applying adjoint 2024 times: $\det = |A|^{2^{2024}}=2^{2^{2024}}$. For $3\times3$ matrix: $\det(\text{adj}^{(k)}(A))=|A|^{(n-1)^k}=2^{2^{2024}}$.
$n=2^{2^{2024}}$. $2^{2024}\bmod6=4$. $2^4=16\equiv7\pmod9$. Remainder $=7$.
Correct Answer: 7