MTH2021 · Linear Algebra with Applications
Linear Systems & Gaussian Elimination
Week 1 of Monash MTH2021 opens with systems of linear equations written as an augmented matrix [A | b], the three elementary row operations, and the march to row-echelon and reduced-row-echelon form via Gaussian and Gauss–Jordan elimination. The examinable payoff is classifying the solution set — unique, infinitely many, or none — by reading the pivots. This material is drilled in Quiz 1 and forms half of Test 1 (Weeks 1–2), and consistency arguments reappear throughout the proof-heavy final.
What this chapter covers
- 01Linear equation a₁x₁ + … + aₙxₙ = b; matrix form Ax = b and the augmented matrix [A | b]
- 02Elementary row operations: Rᵢ → cRᵢ (c ≠ 0), Rᵢ ↔ Rⱼ, Rᵢ → Rᵢ + cRⱼ; row equivalence A ~ B
- 03Row-echelon form vs reduced-row-echelon form (RREF); pivots unique in number and position, RREF unique
- 04Gaussian elimination (to echelon + back-substitution) and Gauss–Jordan elimination (to RREF)
- 05Leading vs free variables; free variables ↔ non-pivot columns
- 06Consistency: r = n pivots ⇒ 0 or 1 solution; r < n ⇒ 0 or ∞; if a system has >1 solution it has infinitely many
- 07Homogeneous systems Ax = 0 are always consistent; more unknowns than equations ⇒ infinitely many solutions
- 08General solution = particular solution + solution of Ax = 0 (x = p + z)
Classifying a system by its parameters
- +1Write the augmented matrix and apply R2 → R2 − R1 and R3 → R3 − R1: from [[1,1,1|1],[1,2,3|2],[1,2,a|b]] you get [[1,1,1|1],[0,1,2|1],[0,1,a−1|b−1]].
- +1Clear the second column below the pivot with R3 → R3 − R2: [[1,1,1|1],[0,1,2|1],[0,0,a−3|b−2]]. The whole classification now sits in the bottom row (0 0 a−3 | b−2).
- +1Unique solution: if a − 3 ≠ 0, i.e. a ≠ 3, the third column has a pivot, so all three variables are leading and the system has exactly one solution — for ANY value of b.
- +1No solution: if a = 3 and b − 2 ≠ 0 (b ≠ 2), the bottom row reads (0 0 0 | b−2) with a nonzero right-hand side — an impossible equation 0 = b−2, so the system is inconsistent.
- +1Infinitely many: if a = 3 and b = 2, the bottom row is (0 0 0 | 0); z is a free variable, so there are infinitely many solutions (a one-parameter family x = p + z·(homogeneous)).
Key terms
- Augmented matrix
- The matrix [A | b] formed by appending the constant column b to the coefficient matrix A of the system Ax = b; row operations on it mirror the algebra on the equations.
- Elementary row operation
- One of: scale a row by c ≠ 0, swap two rows, or add a multiple of one row to another. Each preserves the solution set, giving a row-equivalent system A ~ B.
- Row-echelon form
- A matrix with all zero rows at the bottom and each pivot strictly to the right of the pivot above it. Reached by Gaussian elimination; not unique.
- Reduced row-echelon form (RREF)
- Echelon form in which every pivot is 1 and is the only nonzero entry in its column. It is unique for a given matrix; Gauss–Jordan elimination reaches it.
- Free variable
- A variable whose column has no pivot; it can take any value, and each free variable contributes one parameter to an infinite solution set.
- Consistent / inconsistent
- A system is consistent if it has at least one solution and inconsistent if it has none. Inconsistency shows up as a pivot in the constant column (a row 0 … 0 | nonzero).
Linear Systems & Gaussian Elimination FAQ
How do I read off unique / infinite / no solution from an echelon form?
Look at the pivots after elimination. A row (0 … 0 | nonzero) means no solution. Otherwise, if every variable column has a pivot the solution is unique; if some variable column has no pivot, that variable is free and there are infinitely many solutions. The number of free variables equals n − r, where r is the number of pivots.
Why can a linear system never have exactly two solutions?
If x₁ and x₂ both solve Ax = b, then x₁ + λ(x₁ − x₂) solves it for every scalar λ (linearity of A). So the moment there are two distinct solutions there are infinitely many — a system has 0, 1, or infinitely many solutions and nothing in between.
What is the difference between Gaussian and Gauss–Jordan elimination?
Gaussian elimination reduces to row-echelon form and then uses back-substitution to finish. Gauss–Jordan continues all the way to reduced-row-echelon form (pivots equal to 1 and cleared above and below), so you can read the solution directly. Both give the same answer; Gauss–Jordan is also the engine of the matrix inversion algorithm in Chapter 2.
Is a homogeneous system Ax = 0 ever inconsistent?
Never. x = 0 (the trivial solution) always works, so Ax = 0 is always consistent. If it has more unknowns than equations, or more generally any free variable, it has infinitely many solutions — a fact you use constantly when computing kernels and eigenspaces later.
How is this tested in MTH2021?
Row reduction and consistency appear in Quiz 1 and make up half of Test 1 (Weeks 1–2, closed book). In the final's handwritten section you may be asked to classify a parameterised system or prove the particular-plus-homogeneous structure of the general solution, so practise both the computation and the short argument.
Exam move
Drill row reduction until it is automatic, then practise reading the answer off the pivots: r = n means unique, a free column means infinitely many, a pivot in the constant column means none. Rehearse parameterised systems (a coefficient or a constant left as a letter) because that is the classic Test 1 and exam item, and always write the general solution as particular + homogeneous. Week-1 content is examined in Quiz 1 and Test 1 (closed book, Weeks 1–2) and underpins every later chapter. When a reduction goes wrong, ask Sia to check your row operations one step at a time.
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