Adjoint And Its Determinant

The determinant of the adjoint of an n by n matrix satisfies det(adj A) = (det A)^{n-1}, an identity derived from A·adj(A) = det(A)·I and central to JEE Advanced matrix theory. With n = 3 and det(A) = 4, we get det(adj A) = (det A)^{3-1} = 4^2 = 16. The exponent n - 1 = 2 comes from taking determinants of both sides of A·adj(A) = det(A)·I, where the right side gives (det A)^3 and the left side gives det(A)·det(adj A). Option 4 wrongly equates det(adj A) with det(A). Option 64 uses the exponent 3 instead of n - 1 = 2. Option 12 has no structural basis. Hence det(adj A) = 16. Plausibility check: from det(A)·det(adj A) = (det A)^3, we get det(adj A) = (det A)^2 = 16, and substituting back gives 4·16 = 64 = 4^3, internally consistent and confirming the value. The adjoint identity A times adj(A) equal to det(A) times the identity is the engine behind both the inverse formula and the determinant-of-adjoint relation, so deriving one quickly recovers the other. Taking determinants across that identity and matching powers of det(A) is the standard manoeuvre, and it generalizes cleanly to the determinant of the adjoint of an adjoint by iterating the exponent.