Let $A$ be a $3 \times 3$ matrix and $\det(A) = 2$. If $n = \underbrace{\text{adj}(\text{adj}(\ldots…
Let $A$ be a $3 \times 3$ matrix and $\det(A) = 2$. If $n = \underbrace{\text{adj}(\text{adj}(\ldots \text{adj}(A)\ldots))}_{2024 \text{ times}}$, then the remainder when $n$ is divided by $9$ is equal to __________.
Solution
Given, $|A|=2$
And $n=\underbrace{adj(adj(adj(...adj(A)...)))}_{2024 \text{ times}}$
Now, using the formula $\underbrace{adj(adj(adj(...adj(A)...)))}_{r \text{ times}} = |A|^{n-1^r}$ we get,
$\underbrace{|adj(adj(adj(...adj(A)...)))|}_{2024 \text{ times}} = |A|^{(n-1)^{2024}}$
$\Rightarrow n = |A|^{2^{2024}}$
$\Rightarrow n = 2^{2^{2024}}$ as $|A|=2$
Now, solving
$2^{2024} = 4^{1012} = (3+1)^{1012} = 3k+1$
$\Rightarrow 2^{2^{2024}} = 2^{3k+1}$
$\Rightarrow 2^{2^{2024}} = 2 \cdot 2^{3k}$
$\Rightarrow 2^{2^{2024}} = 2 \cdot (9-1)^k$
$\Rightarrow 2^{2^{2024}} = 9m-2$
$\Rightarrow 2^{2^{2024}} = 9t-2+9$
$\Rightarrow 2^{2^{2024}} = 9m+7$
Hence, the remainder is $7$