JEE Main 2021 · 26 February · evening · Single correct · +4 / −1
Consider the following system of equations :
x + 2y - 3z = a
2x + 6y - 11z = b
x - 2y + 7z = c,
where a, b and c are real constants. Then the system of equations :
- (a) has no solution for all a, b and c
- (b) has a unique solution when 5a = 2b + c
- (c) has infinite number of solutions when 5a = 2b + c
- (d) has a unique solution for all a, b and c
Answer and solution
Answer (c) has infinite number of solutions when 5a = 2b + c
Solution
Maths sir · 4 steps
The determinant is zero, so find which combination of the first two equations gives the third.
- Check the determinant. Expand the coefficient determinant along the first row. It comes out to zero, so a unique solution is never possible.
Δ = |1, 2, -3; 2, 6, -11; 1, -2, 7| = 1(42-22) - 2(14+11) - 3(-4-6) = 20 - 50 + 30 = 0
- Find the dependence. Look for p and q with p R₁ + q R₂ = R₃. From p + 2q = 1 and 2p + 6q = -2 we get q = -2, p = 5. The third coefficient also checks out: 5(-3) - 2(-11) = 7.
- Condition on constants. The left sides satisfy (equation 3) = 5 × (equation 1) - 2 × (equation 2). So the right sides must too: c = 5a - 2b, that is, 5a = 2b + c. Otherwise the equations contradict each other and there is no solution.
- So. When 5a = 2b + c, the third equation adds nothing new. That leaves two independent equations in three unknowns, so there are infinitely many solutions. The answer is (c).
The idea to keep. When Δ = 0, write one equation as a combination of the others. If the constants obey the same combination, you get infinitely many solutions. If they do not, you get none.
Answer from the NTA final answer key. Solution by PrepWiser.