JEE Main 2021 · 25 July · morning · Single correct · +4 / −1
The values of a and b, for which the system of equations
2x + 3y + 6z = 8
x + 2y + az = 5
3x + 5y + 9z = b
has no solution, are :
- (a) a = 3, b ≠ 13
- (b) a ≠ 3, b ≠ 13
- (c) a ≠ 3, b = 3
- (d) a = 3, b = 13
Answer and solution
Answer (a) a = 3, b ≠ 13
Solution
Maths sir · 4 steps
Look at the left sides: the first two equations added together give the third one's coefficients.
- Find the determinant. No solution needs the coefficient determinant to be zero first. If it is not zero, the system has exactly one solution.
D = |2, 3, 6; 1, 2, a; 3, 5, 9| = 2(18-5a) - 3(9-3a) + 6(5-6) = 3 - a
- Fix a. D = 0 gives a = 3. Now add the first two equations: 3x + 5y + 9z = 13. That left side is the same as the third equation's left side.
- Compare the right sides. The third equation says 3x + 5y + 9z = b. If b = 13, it repeats information you already have, so there are infinitely many solutions. If b ≠ 13, the same expression would equal two different numbers, which cannot happen.
- So. The system has no solution when a = 3 and b ≠ 13, which is option (a).
The idea to keep. For no solution, first make D = 0. Then find how the equations depend on each other: matching constants means infinitely many solutions, clashing constants means none.
Answer from the NTA final answer key. Solution by PrepWiser.