JEE Main 2021 · 25 February · evening · Single correct · +4 / −1
The following system of linear equations
2x + 3y + 2z = 9
3x + 2y + 2z = 9
x - y + 4z = 8
- (a) does not have any solution
- (b) has a solution (α, β, γ) satisfying α + β² + γ³ = 12
- (c) has a unique solution
- (d) has infinitely many solutions
Answer and solution
Answer (c) has a unique solution
Solution
Maths sir · 4 steps
Check the coefficient determinant first: if it is not zero, there is exactly one solution.
- Cramer's test. For three equations in three unknowns, look at Δ, the determinant of the coefficients. If Δ ≠ 0, the system has exactly one solution.
Δ = |2, 3, 2; 3, 2, 2; 1, -1, 4| = 2(8+2) - 3(12-2) + 2(-3-2) = -20
- Find the solution. Subtracting the second equation from the first gives -x + y = 0, so x = y. Then the third equation becomes 4z = 8, so z = 2, and the first gives 5x + 4 = 9, so x = y = 1.
- Test option (b). The solution is (1, 1, 2), so α + β² + γ³ = 1 + 1 + 8 = 10, not 12.
- So. Δ = -20 ≠ 0, so the system has a unique solution, (1, 1, 2). The answer is (c).
The idea to keep. For a square system, Δ ≠ 0 means exactly one solution. Only when Δ = 0 do you check Δₓ, Δ_y, Δ_z to decide between no solution and infinitely many.
Answer from the NTA final answer key. Solution by PrepWiser.