DPST1013 Chap.4 Linear Systems and Gaussian Elimination
Linear Systems and Gaussian Elimination
Gaussian elimination replaces a linear system with reversible equations that expose its solution structure. Echelon form separates pivot and free variables, while the augmented column reveals contradictions. The chapter classifies unique, infinite and absent solutions before using back-substitution or parameters.
What this chapter covers
- 01
Augmented matrices
- 02
Elementary row operations
- 03
Row-echelon form
- 04
Pivot and free variables
- 05
Consistency and contradiction rows
- 06
Parameterised solution sets
Worked example · free
Solve a system with one free variable
- 1Eliminate x from the second and third rows.
- 1The two new rows are identical, so subtraction creates a zero row without contradiction.
- 1Let z = t and solve y = (7 - t)/3.
- 1Use the first equation to obtain x = (5 - 2t)/3 and state t is real.
Key terms
- Augmented matrix
- A matrix that appends the right-hand side to the system's coefficient matrix.
- Pivot column
- A column containing a leading entry that determines a leading variable.
- Free variable
- A variable associated with a non-pivot column in a consistent system.
- Contradiction row
- A row representing zero equal to a non-zero constant, proving inconsistency.
Linear Systems and Gaussian Elimination FAQ
What does a zero row mean?
A completely zero row says one equation was redundant and adds no new restriction. It does not prove inconsistency. A row with zero coefficients and a non-zero augmented entry is the contradiction that proves no solution.
When should I introduce free parameters?
First reduce the system, confirm it is consistent and identify every non-pivot column. Then assign parameters to the free variables and back-substitute. Introducing them earlier can hide a contradiction or miss a pivot.
How do I handle a parameter in the coefficient matrix?
Eliminate symbolically but avoid dividing by an expression that might be zero. When a potential pivot depends on the parameter, split into the zero and non-zero cases because the rank and solution count may change.
Which row operations are allowed during elimination?
You may swap two rows, scale a row by a non-zero number, or replace one row by itself plus a scaled copy of a different row. Each operation produces an equivalent system because it can be reversed. Multiplying by zero is forbidden because it destroys information. Record operations explicitly when arithmetic is long, especially if a later inconsistency suggests that an augmented entry or sign was copied incorrectly.
What is the practical difference between echelon and reduced echelon form?
Echelon form has pivots stepping to the right with zeros below them, so back-substitution can solve the system. Reduced echelon form also makes every pivot one and clears its entire column, allowing solutions to be read directly. Either may be efficient depending on the question. Continue to reduced form when classification or a parameterisation would otherwise remain ambiguous.
How does rank predict the number of solutions?
Compare the rank of the coefficient matrix with the rank of the augmented matrix. Different ranks mean a contradiction and no solution. Equal ranks give a consistent system; if that common rank equals the number of variables, the solution is unique, while a smaller rank leaves free variables and infinitely many solutions. This summary should follow, not replace, evidence from valid row reduction.
How should an infinite solution set be written?
Assign independent parameters to the free variables and express every pivot variable in terms of them. Then present the solution as an ordered tuple or as a particular vector plus a linear combination of direction vectors. State that the parameters range over the real numbers. Substitute the parameterisation into the original system symbolically, or test several values, to check that no equation was lost.
Why is choosing a convenient pivot mathematically safe?
A row swap merely reorders equations, so moving a simple non-zero entry into the pivot position preserves the solution set. Choosing one or minus one can avoid fractions and reduce error. The convenience is conditional: if a prospective pivot contains a parameter, you cannot assume it is non-zero. Separate that exceptional value before dividing, then analyse both branches of the system.
Exam move
Practise row operations as reversible equation moves rather than visual symbol pushing. Write the operation beside each matrix, preserve the augmented column and inspect every final row before solving. Classify the system as inconsistent, uniquely determined or parameterised.
State the solution as an ordered vector with parameter domains, then substitute a simple parameter value or the unique solution into the original equations as an audit. Alternate between equation form and augmented-matrix form until each row operation has a clear algebraic meaning. Aim first for echelon structure, not decorative fractions; a sensible pivot choice can reduce arithmetic.
After reduction, circle pivot columns and mark free columns before writing any variables. For parameter-dependent systems, create a case table at each pivot that might vanish. Finish practice sets by comparing rank, number of variables and the geometry of the solution set.
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