Two fair dice are thrown. The numbers on them are taken as λ and μ, and a system of linear equations
x + y + z = 5
x + 2y + 3z = μ
x + 3y + λz = 1
is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then :

Solution

Maths sir · 4 steps

The determinant here is just λ - 5, so only the die showing λ = 5 can spoil uniqueness.

1. Find the determinant. A system has a unique solution exactly when its coefficient determinant is non-zero. Expand along the first row.

   Δ = |1, 1, 1; 1, 2, 3; 1, 3, λ| = (2λ - 9) - (λ - 3) + (3 - 2) = λ - 5

2. Unique solution. Δ ≠ 0 means λ ≠ 5. The first die can show 5 values other than 5, and μ can be anything, so p = 5/6.
3. When λ is 5. Subtract the first equation from the other two. This gives y + 2z = μ - 5 and 2y + 4z = -4, that is y + 2z = -2. They agree only if μ = 3 (infinitely many solutions). For any other μ they contradict, so there is no solution.
4. So. No solution needs λ = 5 and μ ≠ 3: that is 1 × 5 = 5 outcomes out of 36. So q = 5/36 and p = 5/6, which is option (b).

The idea to keep. Unique solution means Δ ≠ 0. When Δ = 0, reduce the equations and check them: consistent means infinitely many solutions, contradictory means no solution. Count those two cases separately.

Answer from the NTA final answer key. Solution by PrepWiser.