Given,
∣A∣=2
And n=det(2024times⏟adj(adj(adj...(A))))
Now, using the formula rtimes⏟∣adj(adj(adj...(A)))∣=∣A∣(n−1)r we get,
2024times∣⏟adj(adj(adj...(A)))∣=∣A∣(n−1)2024
⇒n=∣A∣22024
⇒n=222024 as∣A∣=2
Now, solving
22024=41012=(3+1)1012=3k+1
⇒222024=23k+1
⇒222024=2⋅23k
⇒222024=2⋅(9−1)k
⇒222024=9m−2
⇒222024=9t−2+9
⇒222024=9m+7
Hence, the remainder is 7