Let A be a 3× 3 matrix and det(A)=2. If n= det(underset2024- times underset⏟ adj(adj(....(adjA))))), then the remainder when n is divided by 9 is…

Given,

$$|A|=2$$∣A∣=2

And $n=\mathrm{det}(\underset{2024\mathrm{times}}{\underset{⏟}{adj(adj(adj...(A)))}})$n=det(2024times⏟adj(adj(adj...(A)))​​)

Now, using the formula $\underset{r\mathrm{times}}{\underset{⏟}{|adj(adj(adj...(A)))|}}={|A|}^{{(n-1)}^{r}}$rtimes⏟∣adj(adj(adj...(A)))∣​​=∣A∣(n−1)r we get,

$$\underset{2024\mathrm{times}}{|\underset{⏟}{adj(adj(adj...(A)))}|}={|A|}^{{(n-1)}^{2024}}$$2024times∣⏟adj(adj(adj...(A)))​∣​=∣A∣(n−1)2024

$$\Rightarrow n={|A|}^{{2}^{2024}}$$⇒n=∣A∣22024

$\Rightarrow n={2}^{{2}^{2024}}$⇒n=222024 ${\text{as}|A|=2}$as∣A∣=2

Now, solving

$${2}^{2024}={4}^{1012}={(3+1)}^{1012}=3k+1$$22024=41012=(3+1)1012=3k+1

$$\Rightarrow {2}^{{2}^{2024}}={2}^{3k+1}$$⇒222024=23k+1

$$\Rightarrow {2}^{{2}^{2024}}=2\cdot {2}^{3k}$$⇒222024=2⋅23k

$$\Rightarrow {2}^{{2}^{2024}}=2\cdot {(9-1)}^{k}$$⇒222024=2⋅(9−1)k

$$\Rightarrow {2}^{{2}^{2024}}=9m-2$$⇒222024=9m−2

$$\Rightarrow {2}^{{2}^{2024}}=9t-2+9$$⇒222024=9t−2+9

$$\Rightarrow {2}^{{2}^{2024}}=9m+7$$⇒222024=9m+7

Hence, the remainder is $7$7