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 \(\_\_\_\_\_.\)
Correct Answer: 7
Approach Solution - 1
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:
\[ \boxed{7} \]
Approach Solution -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:
\[ 7. \]
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