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ECON1202 Chap.7 Row Reduction and Linear Systems

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Chapter 7 of 12 · ECON1202

Row Reduction and Linear Systems

Gaussian elimination uses reversible row operations to expose the structure of a linear system. A row swap reorders equations, a non-zero row scaling multiplies one equation by a reversible constant, and a row replacement adds a multiple of another equation. Every operation must reach the augmented constant column.

In row-echelon form, pivots move right as rows descend, entries below each pivot are zero, and zero rows appear at the bottom. A zero candidate pivot requires a search and possible row swap, not an immediate singularity claim. Classification follows a strict order. First look for a contradiction row with zero coefficients and a non-zero constant; that means no solution.

If the system is consistent, compare rank with the number of unknowns. Equal values give a unique solution, while fewer pivots leave free variables and infinitely many solutions. Free variables receive independent parameters. Applied systems may then impose non-negativity or integer restrictions and optimise over the feasible parameter values. Final verification substitutes a candidate into every original equation.

In this chapter

What this chapter covers

  • 01

    Augmented matrices

  • 02

    Elementary row operations

  • 03

    Pivot selection

  • 04

    Row-echelon form

  • 05

    Rank from pivots

  • 06

    Contradiction rows

  • 07

    Free-variable parameterisation

  • 08

    Feasible applied systems

Worked example · free

Classify and solve a three-variable system

Q [4 marks]. A reduced system has equations x+2z=7 and y-z=1, with no third pivot equation. Describe the full solution set. This weighting is not a published University assessment scheme; it rewards the free-variable structure and verification in this exercise only.
Classify in the right order.Classify in the right order. Test consistency before comparing rank with the number of unknowns; rank alone cannot rescue a contradictory augmented row.row-echelon augmented matrixcontradiction row [0 ... 0 | nonzero]?no solutionrank = unknownsone solutionrank < unknownsinfinitely manyyesno, full rankno, free columns
Classify in the right order. Test consistency before comparing rank with the number of unknowns; rank alone cannot rescue a contradictory augmented row.
  • +1There is no contradiction, so the system is consistent.
  • +1There are two pivots for three unknowns, so one variable is free.
  • +1Let z=t. Then x=7-2t and y=1+t.
  • +1The solution family is (x,y,z)=(7-2t,1+t,t) for any real t, verified in both equations.
The system has infinitely many solutions with one free parameter: (7-2t,1+t,t). A particular t may be chosen to check both original equations.
Sia tip — Scan the reduced matrix from the bottom. Mark contradictions before counting pivots or beginning back-substitution.
Glossary

Key terms

Augmented matrix
A coefficient matrix joined to the right-hand-side constants of a linear system.
Elementary row operation
A reversible swap, non-zero scaling or row replacement applied across an entire augmented row.
Pivot
The leading non-zero entry used to clear entries below in its column.
Row-echelon form
A staircase arrangement with pivots moving right, zeros below pivots and zero rows at the bottom.
Free variable
An unknown in a non-pivot column that receives an independent parameter in a consistent system.
Contradiction row
A reduced row stating zero equals a non-zero constant, which makes the system inconsistent.
FAQ

Row Reduction and Linear Systems FAQ

Which row operations preserve the solution set?

You may swap rows, multiply a row by a non-zero constant, or replace a row by itself plus a multiple of another row. Apply the operation to every entry including the augmented constant. Column operations change variables and are not permitted as row operations.

Does a zero row mean there is no solution?

Not by itself. A row of all zeros including the augmented entry is redundant and can occur in a consistent system. A row of zero coefficients followed by a non-zero constant is a contradiction and means no solution.

How does rank determine the solution count?

After confirming consistency, compare coefficient rank with the number of unknowns. Equal values give a unique solution. Lower rank leaves at least one free variable and produces infinitely many solutions, with one parameter per pivot-free column.

Why should the answer be checked in the original system?

Back-substitution and row arithmetic can contain sign errors even when the reduced form looks plausible. Substituting the candidate into every original equation independently verifies that no condition was dropped.

What does a complete row-reduction argument contain?

Start with a correctly ordered augmented matrix and record each elementary row operation beside the row it creates. Choose pivots, clear entries without disturbing completed pivot columns, and carry the augmented entry through every operation. Before solving numerically, inspect the echelon form for a contradiction row.

If none appears, compare pivot count with the number of variables to classify the system as unique or as a family with free variables. Parameterise any free variables explicitly. Then apply contextual restrictions such as non-negativity, integrality or capacity bounds to the algebraic family rather than assuming every algebraic point is feasible.

Finally, substitute the unique or parametric answer into the original equations, where a propagated row-arithmetic error can be detected independently.

Study strategy

Exam move

Practise one row operation at a time and write its multiplier so sign errors remain visible. Keep the augmentation bar prominent and reject any step that changes only the coefficient side. Use the staircase test to identify pivots and rank, but always inspect the augmented constants for a contradiction before classifying.

Pair a unique-solution system with a zero-row system and a contradictory system so the three outcomes are compared directly. For free-variable cases, mark pivot and non-pivot columns, assign one parameter to each free variable and express every pivot variable in terms of them. In applied questions, add non-negativity and integrality after finding the algebraic family, then evaluate any cost only on feasible parameter values.

Verify the final candidate or a sample family member in every original equation. Stop a pure classification question once consistency and rank have established the outcome.

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