Prerequisite · Linear Algebra

Linear Systems and Gauss's Method

18 min read
By the end of this reading you will be able to:
  • Translate a word problem into an augmented matrix and classify the system as consistent or inconsistent
  • Apply Gaussian elimination to bring a matrix to echelon form, recording each row operation
  • Identify pivot variables and free variables from echelon form and state what each implies about the solution set
  • Determine existence and uniqueness of solutions directly from echelon form without back-substituting

Linear Systems

A linear equation in variables x1,x2,,xnx_1, x_2, \ldots, x_n has the form

a1x1+a2x2++anxn=ba_1 x_1 + a_2 x_2 + \cdots + a_n x_n = b

where the coefficients aia_i and the right-hand side bb are constants. A linear system is a collection of mm such equations that must hold simultaneously.

Three outcomes are possible for any linear system:

  • Unique solution — exactly one assignment of values satisfies all equations
  • No solution — the equations are inconsistent (a contradictory equation appears during reduction)
  • Infinitely many solutions — at least one variable is free to range over R\mathbb{R}

No other outcome is possible. This trichotomy is a theorem, not an assumption.

Matrix Form

A linear system is compactly written as Ax=bA\vec{x} = \vec{b}, where AA is the m×nm \times n coefficient matrix, x\vec{x} is the column vector of unknowns, and b\vec{b} is the column vector of right-hand sides.

The augmented matrix [Ab][A \mid \vec{b}] packages the coefficient matrix and right-hand side together:

[a11a1nb1am1amnbm]\left[\begin{array}{ccc|c} a_{11} & \cdots & a_{1n} & b_1 \\ \vdots & & \vdots & \vdots \\ a_{m1} & \cdots & a_{mn} & b_m \end{array}\right]

All row operations are performed on the augmented matrix — we never need to write the variable names explicitly during elimination.

Row Operations

Three elementary row operations transform a system without changing its solution set:

Operation Notation Effect
Swap two rows ρiρj\rho_i \leftrightarrow \rho_j Reorders equations
Scale a row by k0k \neq 0 kρik\rho_i Multiplies equation by a nonzero constant
Add a multiple of one row to another kρi+ρjρjk\rho_i + \rho_j \to \rho_j Combines two equations

The swap and scale operations are preparatory. The workhorse is the row combination: adding a multiple of one row to another is what actually eliminates variables.

Gauss's Method (Row Reduction)

Gauss's method applies row combinations to produce a staircase pattern where each row's first nonzero entry — called the pivot or leading entry — is strictly to the right of the pivot in the row above. This form is called echelon form (or row echelon form).

Algorithm

  1. Find the leftmost column with a nonzero entry. Swap rows if needed to place a nonzero entry at the top of that column.
  2. Use row combinations to zero out all entries below the pivot.
  3. Cover the top row and repeat on the remaining submatrix.
  4. Stop when no rows remain or all remaining rows are zero.

Example

Solve the 3×33 \times 3 system:

x+y+z=62x+yz=1xy+2z=5\begin{array}{rcrcrcl} x & + & y & + & z & = & 6 \\ 2x & + & y & - & z & = & 1 \\ x & - & y & + & 2z & = & 5 \end{array}

Augmented matrix and reduction:

[111621111125]2ρ1+ρ2,  ρ1+ρ3[1116013110211]\left[\begin{array}{rrr|r} 1 & 1 & 1 & 6 \\ 2 & 1 & -1 & 1 \\ 1 & -1 & 2 & 5 \end{array}\right] \xrightarrow{-2\rho_1 + \rho_2,\;-\rho_1 + \rho_3} \left[\begin{array}{rrr|r} 1 & 1 & 1 & 6 \\ 0 & -1 & -3 & -11 \\ 0 & -2 & 1 & -1 \end{array}\right]

2ρ2+ρ3[11160131100721]\xrightarrow{-2\rho_2 + \rho_3} \left[\begin{array}{rrr|r} 1 & 1 & 1 & 6 \\ 0 & -1 & -3 & -11 \\ 0 & 0 & 7 & 21 \end{array}\right]

This is echelon form. Back-substitution:

  • Row 3: 7z=21z=37z = 21 \Rightarrow z = 3
  • Row 2: y3(3)=11y=2-y - 3(3) = -11 \Rightarrow y = 2
  • Row 1: x+2+3=6x=1x + 2 + 3 = 6 \Rightarrow x = 1

Solution: (x,y,z)=(1,2,3)(x, y, z) = (1, 2, 3).

Pivot Variables and Free Variables

After reduction to echelon form:

  • A pivot variable (or leading variable) is one whose column contains a leading entry. It can be solved for in terms of the free variables.
  • A free variable corresponds to a column with no leading entry. It can take any value, generating infinitely many solutions.

Decision rule:

  • No free variables + no contradictory row → unique solution
  • Any free variable + no contradictory row → infinitely many solutions
  • Any row of the form [0    0c][0 \;\cdots\; 0 \mid c] with c0c \neq 0 → no solution

Reduced Echelon Form and Gauss-Jordan

Reduced row echelon form (RREF) adds two requirements to echelon form:

  • Each pivot equals 1
  • All entries above each pivot are also zero

Gauss-Jordan reduction achieves RREF by additionally eliminating upward (above each pivot) after the forward pass. The result directly reads off the solution without back-substitution. For large systems, Gauss-Jordan is less numerically preferred than plain Gaussian elimination followed by back-substitution — but it is conceptually clean and useful for computing matrix inverses.

References
Hefferon 2020 — Linear Algebra, Ch. One §I: Solving Linear Systems