Let \(\mathrm{A}=\left[\begin{array}{ccc}1 & 1 & 2 \\ -2 & 0 & 1 \\ 1 & 3 & 5\end{array}\right]\). Then the sum of all elements of the matrix \(\operatorname{adj}\left(\operatorname{adj}\left(2(\operatorname{adj} \mathrm{~A})^{-1}\right)\right)\) is equal to :

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Matrices

Let

A = 1 - 2 1 103215

. Then the sum of all elements of the matrix

adj (adj (2 (adj A)^{- 1}))

is equal to :

Answer: (D)

∣ A ∣ = 1 (0 - 3) - 1 (- 10 - 1) + 2 (- 6 - 0) = - 3 + 11 - 12 = - 4

. Since

(adj A)^{- 1} = \frac{A}{∣ A ∣}

, the inner matrix is

B = 2 (adj A)^{- 1} = \frac{2 A}{- 4} = - \frac{A}{2}

, with

∣ B ∣ = (- \frac{1}{2})^{3} ∣ A ∣ = \frac{- ( - 4 )}{8} = \frac{1}{2}

. For a

3 \times 3

matrix

adj (adj B) = ∣ B ∣^{3 - 2} B = ∣ B ∣ B = \frac{1}{2} (- \frac{A}{2}) = - \frac{A}{4}

. The sum of the entries of

A

1 + 1 + 2 - 2 + 0 + 1 + 1 + 3 + 5 = 12

, so the sum of the entries of

- \frac{A}{4}

- \frac{12}{4} = - 3

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option