Solutions of Homogeneous Systems of Equations
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Direct answer
A homogeneous linear system AX = 0 always has at least one solution — the trivial one, X = 0 — and the entire theory asks when it has more. For a square system (n equations, n unknowns), non-trivial solutions exist exactly when det(A) = 0; if det(A) ≠ 0, the trivial solution is unique. With det(A) = 0 the solution set becomes infinite: all points on a line through the origin for a 3 × 3 system of rank 2, or a plane through the origin for rank 1. Geometrically, every solution of a homogeneous 3-variable system passes through the origin — the normals of the planes all belong to a common plane, forcing the intersection to be a subspace, never a shifted line.
What you must remember
- Existence: X = 0 always solves AX = 0; uniqueness of the trivial solution holds iff det(A) ≠ 0 (square case).
- Non-trivial condition: det(A) = 0 is necessary and sufficient for infinitely many solutions in a square homogeneous system — the single most used test.
- Rank counting: with rank r of an n-unknown system, the solution space has dimension n − r: rank 3 in three unknowns gives only the origin; rank 2 gives a line through the origin; rank 1 gives a plane through the origin.
- Two-equation line: for a₁x + b₁y + c₁z = 0 and a₂x + b₂y + c₂z = 0, the solution line has direction (b₁c₂ − b₂c₁, c₁a₂ − c₂a₁, a₁b₂ − a₂b₁) — the cross product of the normals, exactly the 3D "cross-multiplication method" from school algebra.
- Parameter count: express the general solution with (n − r) parameters; the parametric ratios are what JEE asks for.
- Rouché–Capelli bridge: for non-homogeneous AX = B, consistency means rank(A) = rank([A|B]); the homogeneous companion system AX = 0 governs the "infinite solutions" case.
- Plane-through-origin reading: any equation ax + by + cz = 0 passes through the origin; two such planes intersect in a line through the origin unless identical or parallel-in-coincident sense.
When planes pass through the origin
Ask when the system x + 2y − z = 0, 2x + y + λz = 0, 3x − y + z = 0 has a non-trivial solution. The determinant of coefficients must vanish. Expanding along the first row: 1·(1·1 − λ·(−1)) − 2·(2·1 − λ·3) + (−1)·(2·(−1) − 1·3) = (1 + λ) − (4 − 6λ) + 5 = 7λ + 2. Setting 7λ + 2 = 0 gives λ = −2/7. For every other λ the origin stands alone; at λ = −2/7 the three planes share a full line through the origin, whose direction is (1, 2, −1) × (2, 1, −2/7) = (3/7, −12/7, −3) ∝ (1, −4, −7), so the solution is x = t, y = −4t, z = −7t. Substituting into the third (unused) equation confirms the line: 3t + 4t − 7t = 0 for all t. Matrix problems phrase the same content as "find k so that the columns of A are linearly dependent" — det(A) = 0 in disguise, since a non-trivial solution is exactly a dependence relation among columns.
Geometry behind the algebra
JEE Main keeps determinant computation at the centre; JEE Advanced prefers the geometric phrasing — "these planes intersect in a line; find its direction ratios" — or eigenvalue-adjacent questions where λ makes a system degenerate. The classic trap is declaring "no solution" for a homogeneous system: it never happens, the trivial solution is always there, so options offering "no solution" are decoys. The second trap is stopping at det = 0 without producing the solution line — many questions ask for the ratio x : y : z, and cross-multiplication (a₂b₃ − a₃b₂ form, read directly off the two equations) is the fastest legal route. When the question shifts to non-homogeneous, det ≠ 0 still means a unique solution, but det = 0 demands the augmented-rank check before any conclusion.
Frequently asked questions
When does a homogeneous square system have non-trivial solutions?
Exactly when the determinant of the coefficient matrix is zero; det ≠ 0 forces the trivial solution alone.
What is the dimension of the solution space of AX = 0?
n − r, where r is the rank of A and n the number of unknowns: a line for rank 2 in three unknowns, a plane for rank 1.
How do you write the solution line of two homogeneous equations in three variables?
Take the cross product of the two normals (or cross-multiply coefficients) to get direction ratios, then parametrise as (x, y, z) = t·(direction).
Can a homogeneous system ever have no solution?
Never — the zero vector always solves it; "inconsistent" is impossible for AX = 0.
How does the homogeneous theory enter non-homogeneous systems?
Through Rouché–Capelli: AX = B has solutions iff rank(A) = rank of the augmented matrix, and the associated AX = 0 then carries the structure of the infinite-solution case.