A parameterised 3×3 system: when singularity means a line, not emptiness
A consulting exercise models three outputs (x, y, z) by x + y + z = 1, x + 2y + 3z = 4, x + 4y + αz = β, with tuning constants α and β. A candidate writes: 'three equations in three unknowns, so the coefficient matrix is invertible and the solution is unique for every α and β.' A colleague substitutes α = 7, β = 10 and finds that x = t − 2, y = 3 − 2t, z = t satisfies all three equations for every real t. Changing only β to 11 makes the third equation read 10 = 11. Unmoved, the candidate argues: 'a singular coefficient matrix always means no solution, and when a singular 3×3 system is consistent, at most one equation is redundant, so the solution set is at most a line.' The colleague wants a complete classification in terms of α and β, plus a verdict on both claims.
1. Classify the system for all real α and β (unique solution, infinitely many, none), showing the reduction.
Read the worked answer
Subtract row 1 from rows 2 and 3: y + 2z = 3, 3y + (α − 1)z = β − 1. Eliminate y by subtracting three times the first relation from the second: (9 − 6z) replaced appropriately gives 9 − 6z + (α − 1)z = β − 1, i.e. (α − 7)z = β − 10.
Cases: (i) α ≠ 7: z = (β − 10)/(α − 7), then y = 3 − 2z and x = 1 − y − z; a unique solution for every β. (ii) α = 7 and β = 10: the equation becomes 0 = 0, so z is free and the solutions are the line (−2, 3, 0) + t(1, −2, 1). (iii) α = 7 and β ≠ 10: 0 = β − 10, a contradiction, so there is no solution.
Cross-check: det of the coefficient matrix equals α − 7, so the matrix is singular exactly in cases (ii) and (iii), and in (iii) the row echelon form exposes the inconsistency.
2. For α = 7, β = 10 describe the solution set geometrically, relate the free direction to the null space, and verify rank plus nullity.
Read the worked answer
The rows of A are (1,1,1), (1,2,3), (1,4,7). Row 3 − row 1 = (0,3,6) = 3(row 2 − row 1), so rank A = 2 and nullity = 3 − 2 = 1.
Null space: solve x + y + z = 0 and y + 2z = 0. Then y = −2z and x = −y − z = z, so N(A) = span{(1, −2, 1)}.
Particular solution: (−2, 3, 0) checks out, since −2 + 3 + 0 = 1, −2 + 6 + 0 = 4, −2 + 12 + 0 = 10.
Hence the solution set is the affine line (−2, 3, 0) + t(1, −2, 1), t ∈ R. Geometrically the first two planes meet in a line, and the third plane contains that entire line, so the intersection of all three planes is a line rather than a point. Rank 2 with 3 unknowns leaves exactly one degree of freedom, consistent with the one-parameter family found by substitution.
3. Judge the candidate's two claims, with a counterexample or a proof for each.
Read the worked answer
'Singular ⇒ inconsistent' is false. At α = 7, β = 10 the coefficient matrix is singular and the system has infinitely many solutions. Singularity destroys uniqueness only; consistency is a separate condition on the right-hand side - here the compatibility relation β = 10, equivalently b orthogonal to the left null vector (1, −3, 1) up to scaling.
'A consistent singular 3×3 system has a solution set of dimension at most 1' is false as a general statement. Take A with all nine entries equal to 1 and b = (1, 1, 1). Then rank A = 1, nullity = 2, and the solution set is the plane x + y + z = 1. Combining the two correct facts - consistency is a condition on b, and the solution set is a translate of N(A) of dimension 3 − rank A - gives the full picture. The candidate's 'at most one redundant equation' is an accident of this particular matrix, not a theorem.