MATH1131 Chap.5 Linear Systems and Gaussian Elimination
Linear Systems and Gaussian Elimination
A system of linear equations has either no solution, exactly one, or infinitely many, and geometry shows why: lines in the plane or planes in space can miss each other, cross at a point, or overlap along a line or plane. Gaussian elimination decides which case holds and finds the solutions.
The system is written as an augmented matrix, and three elementary row operations, swapping rows, adding to one row a multiple of a different row, and scaling a row by a non-zero number, reduce it to row-echelon form without changing the solution set. In echelon form the leading entries form a staircase.
A row reading zero equals a non-zero number means the system is inconsistent; otherwise leading variables are solved by back-substitution and every non-leading variable becomes a free parameter, so the number of parameters is the number of unknowns minus the number of leading entries.
The chapter also treats systems with an unknown right-hand side, where the reduction yields conditions on b for consistency, systems containing a constant k that must be split into cases, and applications to geometry, chemical equations and economics.
What this chapter covers
- 01
Linear equations and their solution sets
- 02
Augmented matrices
- 03
Elementary row operations
- 04
Row-echelon and reduced row-echelon form
- 05
Back-substitution and free parameters
- 06
Deducing consistency from echelon form
- 07
Conditions on an unknown right-hand side
- 08
Applications to geometry, chemistry and economics
Worked example · free
A three-variable system with a unique solution
- 1R2 - 2R1 gives (0, -3, -1 | -9); R3 - R1 gives (0, 1, -2 | -4).
- 1Swap R2 and R3 for a convenient pivot: (0, 1, -2 | -4) and (0, -3, -1 | -9).
- 1R3 + 3R2 gives (0, 0, -7 | -21), so z = 3.
- 1Back-substitute: y - 2(3) = -4 gives y = 2; x + 2 + 3 = 6 gives x = 1.
- 1Check in the second equation: 2(1) - 2 + 3 = 3.
Key terms
- Augmented matrix
- The coefficient matrix of a linear system with the right-hand-side column attached after a bar.
- Row-echelon form
- A matrix whose leading entries move strictly right going down, with zero rows at the bottom.
- Leading entry
- The first non-zero entry in a row of a matrix in echelon form.
- Free parameter
- A real variable assigned to a non-leading unknown to describe infinitely many solutions.
- Inconsistent system
- A system of equations with no solution, signalled by a row reading zero equals a non-zero number.
- Elementary row operation
- A row swap, a non-zero row scaling, or adding to one row a multiple of a different row.
Linear Systems and Gaussian Elimination FAQ
How can I tell from echelon form that a system has no solution?
Look for a row whose coefficient entries are all zero but whose right-hand entry is not. That row states zero equals a non-zero number, which no choice of the unknowns can satisfy.
How many parameters will an infinite solution set need?
One for each non-leading column of the coefficient matrix, so the number of unknowns minus the number of leading entries, provided the system is consistent.
Do I need reduced row-echelon form or is echelon form enough?
Echelon form is enough to decide consistency and count parameters, and back-substitution from it is usually quicker by hand. Use reduced form if the question asks for it or when reading the solution directly is clearer.
What changes when the system contains a constant like k?
Reduce as usual but avoid dividing by any expression in k. At the end, split into cases according to whether the critical expression is zero, and state the conclusion for each case in words.
Why do row operations not change the solutions?
Each operation can be undone by another row operation, and each replaces the equations with consequences of the originals. So any solution of the old system solves the new one and vice versa.
What is a homogeneous system and why is it always consistent?
A homogeneous system has every right-hand side equal to zero. Setting all unknowns to zero satisfies it, so it always has at least that trivial solution, and the only question is whether other solutions exist.
Exam move
Practise elimination on small systems until the arithmetic is reliable, then focus on reading the echelon form: consistent or not, unique or not, and how many parameters. Write general solutions in vector form, a point plus parameter multiples of directions, so they connect with lines and planes from the first chapters.
Work several questions with a parameter k and with an indeterminate right-hand side, because these test understanding rather than arithmetic. Finish each problem with a substitution check into the original equations. Time yourself on a full three-by-three reduction; if it takes longer than a few minutes, look for a better first pivot, such as a row whose leading coefficient is already one.
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