JEE Main 2026 — Jan 22, Shift 2 — Mathematics solution

JEE Main2026Jan 22, Shift 2Mathematics

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 :

Solution

A unique solution exists when the coefficient determinant is non-zero:

$Δ = 1 - n 1 1 n - 2 n + 1 0 n - 11 \neq = 0$

Expanding gives $n^{2} - 3 n + 2 \neq = 0$ , so $n \neq = 1$ and $n \neq = 2$ .

From the rolls ${1, 2, 3, 4, 5, 6}$ , the favourable values are $n \in {3, 4, 5, 6}$ (4 outcomes).

Probability $= \frac{4}{6}$ , so $k = 4$ .

Required sum $= k + (3 + 4 + 5 + 6) = 4 + 18 = 22$ .

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