JEE Main 2021 · 25 February · morning · Numerical answer · +4 / −0
If the system of equations
kx + y + 2z = 1
3x - y - 2z = 2
-2x -2y -4z = 3
has infinitely many solutions, then k is equal to __________.
Answer and solution
Answer 21
Solution
Maths sir · 5 steps
Look at the y and z columns: y and 2z always come together, so treat y + 2z as one unknown.
- Spot the pattern. The coefficients of z are exactly twice those of y in every equation: (2, -2, -4) = 2 × (1, -1, -2). So the determinant of coefficients is 0 for every k, and D = 0 alone cannot fix k.
- Combine the unknowns. Put w = y + 2z. The system becomes three equations in only two unknowns: kx + w = 1, 3x - w = 2 and -2x - 2w = 3.
- Solve two equations. The second equation gives w = 3x - 2. Substituting into the third: -2x - 6x + 4 = 3, so x = 1/8 and w = 3/8 - 2 = -13/8.
- Force consistency. The first equation must also hold for these values: k/8 - 13/8 = 1, which gives k = 21. Then x = 1/8 and y + 2z = -13/8, and every pair (y, z) on that line works, so there are infinitely many solutions.
- So. k = 21.
The idea to keep. If two columns of the coefficient matrix are proportional, D = 0 for every value of the parameter. Merge those variables into one and demand that the smaller system be consistent.
Answer from the NTA final answer key. Solution by PrepWiser.