Question

Let \( A \) be a \( 3 \times 3 \) matrix and \( \det(A) = 2 \). If \(n = \det(\text{adj}(\text{adj}(\ldots(\text{adj}(A))\ldots)))\) with adjoint applied 2024 times, then the remainder when \( n \) is divided by 9 is equal to \(\_\_\_\_\_.\)

Answer 1

Correct Answer: 7

Given:

\[ \underbrace{\text{adj(adj(adj... (A)))}}_{\text{2024 times}} = |A|^{(n-1)^{2024}} \]

\[ = |A|^{2024} \]

\[ = 2^{2024} \]

Step 1:

\[ 2^{2024} = (2^2)^{1012} = 4^{1012} \] \[ = 4 \times (8)^{674} = 4(9 - 1)^{674} \]

Step 2:

\[ \Rightarrow 2^{2024} \equiv 4 \pmod{9} \]

Step 3:

\[ \Rightarrow 2^{2024} \equiv 9m + 4, \quad m \text{ even} \]

Step 4:

\[ 2^{9m + 4} = 16 \cdot (2^3)^{3m} \equiv 16 \pmod{9} \]

\[ \Rightarrow 2^{2024} \equiv 7 \pmod{9} \]

Final Answer:

\[ \boxed{7} \]

Answer 2

\[ 2^{2024} = (2^2)^{2022} = 4 \cdot (8)^{674} = 4 \cdot (9 - 1)^{674}. \]

Applying modulo 9, we get:

\[ 2^{2024} \equiv 4 \pmod{9}. \]

Thus,

\[ 2^{2024} = 9m + 4, \quad m \text{ is even}. \]

Now, consider \(2^{9m+4}\):

\[ 2^{9m+4} = 16 \cdot (2^3)^{3m} \equiv 16 \pmod{9}. \]

Thus,

\[ = 7. \]

Therefore, the answer is:

\[ 7. \]