Lesson 2-2: Systems & Linear Combinations

Systems & Linear Combinations

Use augmented matrices and Gaussian elimination to solve linear systems, analyze ranks to determine whether solutions are unique, none, or infinitely many, and interpret Ax=b as a linear combination of columns of A equaling b.

Learning Objectives

Gaussian Elimination

Perform row operations to reach row-echelon or reduced row-echelon form

Rank & Solution Types

Use rank(A) and rank([A|b]) to classify solutions

Linear Combination View

Interpret Ax=b as combining columns of A to produce b

Pivot & Free Variables

Express solution sets using parameters for free variables

Core Knowledge Points

Elementary Row Operations

1. Row Swap (Ri ↔ Rj)

Interchange rows i and j. Changes determinant sign: $\det(A') = -\det(A)$

2. Row Scaling (kRi)

Multiply row i by scalar k ≠ 0. Scales determinant: $\det(A') = k \cdot \det(A)$

3. Row Addition (Ri + kRj)

Add k times row j to row i. Preserves determinant: $\det(A') = \det(A)$

Rank & Solution Classification

Consistency Test

System $Ax = b$ is consistent if and only if: $\text{rank}(A) = \text{rank}([A|b])$

Unique Solution

If consistent and $\text{rank}(A) = n$ (number of unknowns), then solution is unique.

Infinite Solutions

If consistent and $\text{rank}(A) < n$ , then $n - \text{rank}(A)$ free variables yield infinitely many solutions.

No Solution

If $\text{rank}(A) \neq \text{rank}([A|b])$ , then the system is inconsistent.

Linear Combination Interpretation

Column Space

For matrix $A = [a_1, a_2, \ldots, a_n]$ , the column space is: $\text{Col}(A) = \text{span}\{a_1, a_2, \ldots, a_n\}$

Linear Combination Form

System $Ax = b$ can be written as: $x_1a_1 + x_2a_2 + \cdots + x_na_n = b$

Solvability Condition

$Ax = b$ is solvable if and only if $b \in \text{Col}(A)$

Detailed Worked Example

Solve the system using Gaussian elimination

$\begin{cases} x + 2y - z = 3 \\ 2x + 5y + z = 12 \\ -x + y + 2z = 1 \end{cases}$

Step 1: Form the augmented matrix [A|b]

$[A|b] = \begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 2 & 5 & 1 & | & 12 \\ -1 & 1 & 2 & | & 1 \end{pmatrix}$

Step 2: Forward elimination to REF

R2 ← R2 - 2R1:

$\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 3 & | & 6 \\ -1 & 1 & 2 & | & 1 \end{pmatrix}$

R3 ← R3 + R1:

$\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 3 & | & 6 \\ 0 & 3 & 1 & | & 4 \end{pmatrix}$

R3 ← R3 - 3R2:

$\begin{pmatrix} 1 & 2 & -1 & | & 3 \\ 0 & 1 & 3 & | & 6 \\ 0 & 0 & -8 & | & -14 \end{pmatrix}$

Step 3: Back substitution

From R3: $-8z = -14 \Rightarrow z = \frac{7}{4}$

From R2: $y + 3z = 6 \Rightarrow y = 6 - 3 \cdot \frac{7}{4} = \frac{3}{4}$

From R1: $x + 2y - z = 3 \Rightarrow x = 3 - 2 \cdot \frac{3}{4} + \frac{7}{4} = \frac{13}{4}$

Step 4: Solution verification

Final solution: $x = \frac{13}{4}, y = \frac{3}{4}, z = \frac{7}{4}$

Since rank(A) = rank([A|b]) = 3 = number of unknowns, the solution is unique.

Practice Problems

Problem 1: Gaussian Elimination

Solve the following system using Gaussian elimination:

$\begin{cases} 2x + 3y - z = 7 \\ x - y + 2z = 1 \\ 3x + 2y + z = 8 \end{cases}$

Classify the solution type and verify your answer.

Problem 2: Rank Analysis

Given the matrix and vector:

$A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 1 & 0 \end{pmatrix}, \quad b = \begin{pmatrix} 3 \\ 6 \\ 2 \end{pmatrix}$

Compute rank(A) and rank([A|b]) to determine if the system Ax = b is consistent.

Problem 3: Linear Combination

For the matrix A and vector b from Problem 2:

Express b as a linear combination of the columns of A, or explain why it's impossible.

If possible, find the coefficients x₁, x₂, x₃ such that x₁a₁ + x₂a₂ + x₃a₃ = b.

Problem 4: Free Variables

Solve the system with infinitely many solutions:

$\begin{cases} x + y + z = 3 \\ 2x + 2y + 2z = 6 \\ x - y + z = 1 \end{cases}$

Express the solution set in parametric form using free variables.

System Types & Examples

Unique Solution

$\begin{cases} x + y = 3 \\ 2x - y = 0 \end{cases}$

rank(A) = rank([A|b]) = n

Infinite Solutions

$\begin{cases} x + y + z = 3 \\ 2x + 2y + 2z = 6 \end{cases}$

rank(A) = rank([A|b]) < n

No Solution

$\begin{cases} x + y = 3 \\ x + y = 5 \end{cases}$

rank(A) ≠ rank([A|b])

Key Takeaways

Row-Reduction Workflow

Pivot selection, elimination below/above, scaling pivots to 1

Rank & Consistency

Compare rank(A) vs rank([A|b]) to determine solution existence

Free Variables

Parameterize infinite solution sets using non-pivot columns

Geometric View

Ax=b solvable iff b in span(columns of A)

Strategy Playbook

Pivot selection to avoid zeros and reduce round-off
Forward elimination to row-echelon form (REF)
Back substitution or continue to reduced REF (RREF)
Classify solutions by ranks and free variables

Row-Reduction Templates

Unique Solution

rank(A)=rank([A|b])=n → all variables pivoted

Infinite Solutions

rank(A)=rank([A|b])<n → free variables parameterize solutions

No Solution

rank(A)≠rank([A|b]) → inconsistent rows appear

Span Test

b in span(columns of A) if and only if system is consistent

Additional Practice

Set C: Rank & Consistency

C1. Compute rank via RREF for 3×4 matrix
C2. Determine consistency for random [A|b]
C3. Identify pivot and free columns

Set D: Parametric Solutions

D1. Express solution set using parameters t, s
D2. Write solution as particular + homogeneous
D3. Interpret solution set geometrically

FAQ

What if a pivot is zero?

Swap with a lower row having nonzero candidate or use partial pivoting.

Why avoid explicit A^-1?

Numerically unstable and more expensive; elimination/factorization is preferred.

How to read solution geometry?

Lines/planes intersect; free variables trace subspaces; inconsistent means parallel disjoint objects.

Project Ideas

Build a solver UI that row-reduces steps and highlights pivots
Visualize span(columns of A) and position of b
Benchmark elimination vs computing A^-1 across sizes

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