Let $A$ be a $3 \times 3$ matrix such that $|\text{adj}(\text{adj}(\text{adj } A))|=81$. If $S=\{n \in \mathbb{Z}:(|\text{adj}(\text{adj } A)|)^{\frac{(n-1)^2}{2}}=|A|^{(3n^2-5n-4)}\}$, then $\sum_{n \in S}|A^{(n^2+n)}|$ is equal to

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Matrices

Let

A

be a

3 \times 3

matrix such that

∣ adj (adj (adj A)) ∣ = 81

. If

S = {n \in Z : (∣ adj (adj A) ∣)^{\frac{( n - 1 ) ^{2}}{2}} = ∣ A ∣^{(3 n^{2} - 5 n - 4)}}

, then

\sum_{n \in S} ∣ A^{(n^{2} + n)} ∣

is equal to

Answer: (D)

For a

3 \times 3

matrix

∣ adj A ∣ = ∣ A ∣^{2}

, so applying the adjugate three times gives

∣ adj (adj (adj A)) ∣ = ∣ A ∣^{8} = 81

, hence

∣ A ∣^{2} = 3

. The defining relation becomes

(∣ A ∣^{4})^{(n - 1)^{2} /2} = ∣ A ∣^{3 n^{2} - 5 n - 4}

, and equating exponents gives

2 (n - 1)^{2} = 3 n^{2} - 5 n - 4

, i.e.

n^{2} - n - 6 = 0

, so

n = 3

n = - 2

. Then

n \sum A^{n^{2} + n} = A^{12} + A^{2} = (∣ A ∣^{2})^{6} + ∣ A ∣^{2} = 3^{6} + 3 = 732

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Single correct option