Let A be a 3 times 3 matrix and det(A)=2. If | vedclass

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(A) Given that $A$ is a $3 	imes 3$ matrix,so $|A| = 2$. The formula for the determinant of the $k$-th iterated adjoint of $A$ is given by $\det(\ope

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$7$

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(A) Given that $A$ is a $3 	imes 3$ matrix,so $|A| = 2$. The formula for the determinant of the $k$-th iterated adjoint of $A$ is given by $\det(\operatorname{adj}^k(A)) = |A|^{(n-1)^k}$,where $n$ is the order of the matrix. Here,$n=3$ and $k=2024$. Thus,$n = |A|^{(3-1)^{2024}} = 2^{2^{2024}}$. We need to find $2^{2^{2024}} \pmod{9}$. By Euler's totient theorem,$\phi(9) = 9(1 - 1/3) = 6$. We find $2^{2024} \pmod{6}$. $2^{2024} \equiv 0 \pmod{2}$ and $2^{2024} = (2^2)^{1012} = 4^{1012} \equiv 1^{1012} \equiv 1 \pmod{3}$. By Chinese Remainder Theorem,$2^{2024} \equiv 4 \pmod{6}$. So,$2^{2024} = 6k + 4$ for some integer $k$. Then $n = 2^{6k+4} = (2^6)^k \cdot 2^4 = 64^k \cdot 16$. Since $64 \equiv 1 \pmod{9}$,we have $n \equiv 1^k \cdot 16 \equiv 16 \equiv 7 \pmod{9}$. Therefore,the remainder is $7$.

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

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