If the system of linear equations
2x + 2ay + az = 0
2x + 3by + bz = 0
2x + 4cy + cz = 0,
where a, b, c ∈ R are non-zero distinct; has a non-zero solution, then:

Solution

Maths sir · 4 steps

Homogeneous system with a non-zero solution: set the determinant to zero and simplify.

1. Condition to use. All three right sides are 0, so x = y = z = 0 always works. A non-zero solution exists only when the coefficient determinant is 0.
2. Row operations. Take 2 out of the first column, then do R₂ → R₂ - R₁ and R₃ → R₃ - R₁ to create zeros below the top entry.

   |2, 2a, a; 2, 3b, b; 2, 4c, c| = 2|1, 2a, a; 0, 3b-2a, b-a; 0, 4c-2a, c-a| = 0

3. Expand. Expanding along the first column gives (3b-2a)(c-a) - (b-a)(4c-2a) = 0. The products are 3bc - 3ab - 2ac + 2a² and 4bc - 2ab - 4ac + 2a², and their difference is 2ac - ab - bc = 0, that is, 2ac = ab + bc.
4. So. Divide by abc, which is non-zero: 2/b = 1/a + 1/c. So 1/a, 1/b, 1/c are in A.P., which is option (a).

The idea to keep. Non-zero solution of a homogeneous system means determinant is zero. Make zeros with row operations, then expand. Seeing 2ac = ab + bc? Divide by abc: the reciprocals are in A.P.

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