27 exam-style questions on Systems of linear equations (up to 3 unknowns) — topic 1.16 of Number and Algebra in the IB Mathematics: Analysis and Approaches syllabus — with answers and explanations. Samples below; the full set is free with an account.
Solve the system of equations: x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.
Answer: (1, 2, 3)
Adding the first and third equations gives 2x + 3y = 8; adding the second and third gives 3x + y = 5. Solving these gives x = 1, y = 2, and then z = 6 − 1 − 2 = 3.
Consider the system x + y + z = 2, x + 2y + 3z = 5, x + 3y + 5z = k. Determine the value of k for which the system is consistent.
Answer: 8
R2 − R1 gives y + 2z = 3 and R3 − R2 gives y + 2z = k − 5, so consistency requires k − 5 = 3, i.e. k = 8. The system then has infinitely many solutions.
A system of three linear equations in three unknowns has no solution, yet every pair of the three corresponding planes intersects in a line. Describe the configuration of the planes.
Answer: A triangular prism: three parallel, distinct intersection lines
If two planes were parallel, that pair would not intersect at all; if the planes met in a common line there would be infinitely many solutions. The prism arrangement is the only one where each pair meets but no point lies on all three.
Row reduction of a system of three equations in x, y and z produces the final row 0x + 0y + 0z = 4. State what this shows about the system.
Answer: The system has no solution
The row reads 0 = 4, which is impossible, so the equations are inconsistent and the system has no solution. Do not confuse it with the row 0 = 0, which indicates redundancy.
After row reduction, a system of three equations in three unknowns has two non-zero rows and one row reading 0 = 0. Determine the nature of the solution set.
Answer: A one-parameter family of solutions (a line)
Two independent equations in three unknowns leave 3 − 2 = 1 free parameter, so the solutions form a line. Geometrically the three planes share a common line.
For which value of a does the system x + 2y = 4, 3x + ay = 12 have infinitely many solutions?
Answer: a = 6
With a = 6 the second equation is exactly 3 times the first (3x + 6y = 12), so the two lines coincide and every point on the line is a solution. For any a ≠ 6 the lines cross in exactly one point.
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