JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Determinants

Let n be the number obtained on rolling a fair die. If the probability that the system

x - n y + z = 6

x + (n - 2) y + (n + 1) z = 8

(n - 1) y + z = 1

has a unique solution is

\frac{k}{6}

, then the sum of k and all possible values of n is :

Answer: (D)

Solution

A unique solution requires the coefficient determinant to be non-zero. Expanding

110 - n n - 2 n - 1 1 n + 1 1 = - (n^{2} - 3 n + 2) = - (n - 1) (n - 2) .

This is non-zero when

n \neq = 1

and

n \neq = 2

. Since a fair die gives

n \in {1, 2, 3, 4, 5, 6}

, the favourable values are

n = 3, 4, 5, 6

, so the probability is

\frac{4}{6}

and

k = 4

. The required sum is

k + (3 + 4 + 5 + 6) = 4 + 18 = 22

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Single correct option