The following system of linear equations
7x + 6y – 2z = 0
3x + 4y + 2z = 0
x – 2y – 6z = 0, has

Solution

Maths sir · 5 steps

All right sides are zero, so check the determinant, then write x and y in terms of z.

1. Homogeneous system. Every right side is 0, so x = y = z = 0 always works, and no solution is impossible. Non-zero solutions exist only when the coefficient determinant is 0.

   |7, 6, -2; 3, 4, 2; 1, -2, -6| = 7(-24+4) - 6(-18-2) - 2(-6-4) = -140 + 120 + 20 = 0

2. Infinitely many. Since Δ = 0, the system has infinitely many solutions. To find the relation they satisfy, take z as free and solve for x and y.
3. Solve using z. The third equation gives x = 2y + 6z. Put this in the second equation: 3(2y + 6z) + 4y + 2z = 10y + 20z = 0, so y = -2z. Then x = -4z + 6z = 2z.
4. Check. Substitute into the first equation: 7(2z) + 6(-2z) - 2z = 14z - 12z - 2z = 0. It holds for every z.
5. So. Every solution has the form (2z, -2z, z). That means x = 2z, which is option (c).

The idea to keep. For a homogeneous system, Δ ≠ 0 gives only the trivial solution and Δ = 0 gives infinitely many. Keep one variable free and write the others in terms of it.

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