Lecture 12 Linear Systems and Echelon Forms

Tags: #linear-algebra #matrices #row-reduction #echelon-form

1. The Augmented Matrix

To solve a system efficiently, we strip away the variables and extract the numbers into a matrix.

Coefficient Matrix: Contains only the coefficients of the variables on the left side.
Augmented Matrix: Includes the constants from the right-hand side of the equal signs, usually separated by a line.

Example:

{\begin{cases} 1 x_{1} + 2 x_{2} = 5 \\ 3 x_{1} + 4 x_{2} = 11 \end{cases} ⟹ [\begin{array}{ccc} 1 & 2 & 5 \\ 3 & 4 & 11 \end{array}]

Elementary Row Operations

We simplify matrices using three allowed operations that do not change the solution set:

Replacement: Add a multiple of one row to another row.
Interchange: Swap two rows.
Scaling: Multiply all entries in a row by a non-zero constant.

2. Echelon Forms

The goal of row operations is to reach one of these two forms. Think of them as the "messy but solved" form and the "perfectly clean" form.

Feature	Row Echelon Form (REF)	Reduced Row Echelon Form (RREF)
Shape	"Staircase" pattern of zeros.	"Staircase" pattern of zeros.
Pivots	Leading entries (first non-zero in a row).	Every leading entry (pivot) is exactly 1.
Zeros	Entries below pivots are 0.	Entries below AND above pivots are 0.
Uniqueness	One matrix has many valid REFs.	One matrix has exactly one unique RREF.

3. Basic vs. Free Variables

Once in Echelon Form, look at your columns to categorize your variables ( $x_{1}, x_{2}, \dots$ ):

Basic Variables: Variables whose columns contain a pivot position.
Free Variables: Variables whose columns do NOT contain a pivot position.

4. Existence and Uniqueness (Consistency)

Before solving, we check if a solution even exists.

The "No Solution" Row

A system is Inconsistent (has zero solutions) if you find a contradictory row in Echelon Form:

[0 0 \dots 0 ∣ c] where c \neq 0

This implies the impossible equation $0 = c$ .

The "Solution Count" Rule

Assuming the system is Consistent (no contradictory rows exist):

Unique Solution: There are no free variables. (Every column has a pivot).
Infinitely Many Solutions: There is at least one free variable.

5. The Solving Workflow

Forward Phase: Use row operations to reach Row Echelon Form.
Consistency Check: Stop and look for contradictory rows ( $0 = c$ ). If you find one, the system has no solution.
Backward Phase: If consistent, continue row operations upward to reach Reduced Row Echelon Form.
Parametric Solution: If free variables exist, express the basic variables in terms of the free variables (e.g., $x_{1} = 4 - 2 x_{3}$ ).

Insight: Matrix Dimensions and Free Variables

A $3 \times 5$ coefficient matrix (3 equations, 5 variables) can have at most 3 pivots (one per row).

Since there are 5 variables but only 3 possible pivots, you will always be left with at least 2 free variables ( $5 - 3 = 2$ ).
Conclusion: If a wide matrix system like this is consistent, it must have infinitely many solutions!