Systems of linear equations

IB Mathematics: Analysis and Approaches HL· 45 min read

1. Matrix Representation and Geometric Interpretation★★☆☆☆⏱ 15 min

📘 Definition

Augmented Matrix

$aug (A ∣ b)$

For a system $A x = b$ , the augmented matrix combines the coefficient matrix $A$ and constant vector $b$ for row operations.

Example:

For $2 x + y = 5, x - 3 y = - 1$ , $aug (A ∣ b) = (21 1 - 3 5 - 1)$

Each linear equation in $n$ variables corresponds to a hyperplane in $n$ -dimensional space. For 2 and 3 variables, this translates to lines (2D) and planes (3D), where the solution set is the intersection of these shapes.

Unique solution: All shapes intersect at exactly one point
Infinite solutions: Shapes intersect along a line or coincide completely
No solution: There is no common intersection point for all equations

📐 Worked Example

Describe the geometric meaning of an inconsistent 2-variable system of linear equations.

1
Each linear equation in 2 variables represents a straight line in 2D space.
2
An inconsistent system has no solutions, so no point lies on both lines.
3
This only occurs when the two lines are parallel and distinct, with the same gradient but different intercepts.

Exam tip:

Always link algebraic solution type to geometric interpretation when asked — this is often a separate marking point.

2. Solution Methods: Inverse Matrices and Cramer's Rule★★★☆☆⏱ 20 min

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For square systems ( $n$ equations, $n$ variables) $A x = b$ , if $A$ is invertible ( $det (A) \neq = 0$ ), the system has a unique solution that can be found directly with inverse matrices or Cramer's rule.

📘 Definition

Cramer's Rule

For a 3x3 system $A x = b$ , replace the $i$ -th column of $A$ with $b$ to get $A_{i}$ . The solution is $x_{i} = \frac{d e t ( A _{i} )}{d e t ( A )}$ , for $det (A) \neq = 0$ .

📐 Worked Example

Solve $2 x + y = 7$ , $x - 2 y = 6$ using Cramer's rule.

1
Write coefficient matrix $A$ and constant vector $b$ :
2
$A = \begin{pmatrix} 2 & 1 \\ 1 & -2 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 7 \\ 6 \end{pmatrix}$
3
Calculate $det (A)$ :
4
$\det(A) = (2)(-2) - (1)(1) = -5 \neq 0$
5
Create $A_{1}$ (replace column 1 with $b$ ) and $A_{2}$ (replace column 2 with $b$ ):
6
$A_1 = \begin{pmatrix} 7 & 1 \\ 6 & -2 \end{pmatrix}, \quad A_2 = \begin{pmatrix} 2 & 7 \\ 1 & 6 \end{pmatrix}$
7
Calculate determinants:
8
$\det(A_1) = -20, \quad \det(A_2) = 5$
9
Find solutions:
10
$x = \frac{-20}{-5} = 4, \quad y = \frac{5}{-5} = -1$

3. Row Reduction (Gaussian Elimination)★★★☆☆⏱ 25 min

Gaussian elimination is the most flexible method for any system, including non-square systems and systems with no or infinite solutions. It uses elementary row operations to reduce the augmented matrix to row-echelon form, followed by back substitution to find solutions.

📐 Worked Example

Find all solutions of $x + 2 y - z = 3$ , $2 x - y + 2 z = 1$ , $3 x + y + z = 4$ using Gaussian elimination.

1
Write the augmented matrix:
2
$\begin{pmatrix} 1 & 2 & -1 & 3 \\ 2 & -1 & 2 & 1 \\ 3 & 1 & 1 & 4 \end{pmatrix}$
3
Eliminate $x$ from rows 2 and 3: $R_{2} = R_{2} - 2 R_{1}$ , $R_{3} = R_{3} - 3 R_{1}$ :
4
$\rightarrow \begin{pmatrix} 1 & 2 & -1 & 3 \\ 0 & -5 & 4 & -5 \\ 0 & -5 & 4 & -5 \end{pmatrix}$
5
Eliminate $y$ from row 3: $R_{3} = R_{3} - R_{2}$ :
6
$\rightarrow \begin{pmatrix} 1 & 2 & -1 & 3 \\ 0 & -5 & 4 & -5 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
7
The last row is all zeros, so we have one free variable. Let $z = t$ , $t \in R$ .
8
Back substitute for $y$ : $- 5 y + 4 t = - 5 ⟹ y = 1 - \frac{4}{5} t$
9
Back substitute for $x$ : $x + 2 (1 - \frac{4}{5} t) - t = 3 ⟹ x = 1 + \frac{13}{5} t$
10
The general infinite solution set is:
11
$x = 1 + \frac{13}{5}t, \quad y = 1 - \frac{4}{5}t, \quad z = t, \quad t \in \mathbb{R}$

Exam tip:

Always clearly state the free parameter and full general solution for infinite solution sets to earn full marks.

4. Classifying Solution Types with the Rouché–Capelli Theorem★★★★☆⏱ 20 min

📘 Definition

Rouché–Capelli Theorem

A system $A x = b$ has a solution if and only if $rank (A) = rank (aug (A ∣ b))$ . If ranks are equal: (1) $rank =$ number of variables → unique solution, (2) $rank <$ number of variables → infinite solutions. If ranks are unequal → no solution.

📐 Worked Example

Find the value of $k$ for which the system has no solution: $x + 2 y - 3 z = 4$ , $2 x + 3 y + 2 z = 2$ , $3 x + 5 y - k z = 8$ .

1
Write the augmented matrix:
2
$\begin{pmatrix} 1 & 2 & -3 & 4 \\ 2 & 3 & 2 & 2 \\ 3 & 5 & -k & 8 \end{pmatrix}$
3
Row reduce: $R_{2} = R_{2} - 2 R_{1}$ , $R_{3} = R_{3} - 3 R_{1}$ , then $R_{3} = R_{3} - R_{2}$ :
4
$\rightarrow \begin{pmatrix} 1 & 2 & -3 & 4 \\ 0 & -1 & 8 & -6 \\ 0 & 0 & 1 - k & 2 \end{pmatrix}$
5
For no solution, $rank (A) < rank (aug)$ . This occurs when the coefficient of $z$ is zero, but the constant term is non-zero.
6
Set $1 - k = 0 ⟹ k = 1$ . For $k = 1$ , the last row becomes $[000 ∣ 2]$ , which is a contradiction, so there is no solution.

5. Common Pitfalls

Wrong move:

Applying Cramer's rule when $det (A) = 0$

Why:

Cramer's rule only works for systems with unique solutions, it cannot find infinite solutions or confirm inconsistency

Correct move:

Check $det (A)$ first, use row reduction to classify solutions when $det (A) = 0$

Wrong move:

Assuming all systems with $det (A) = 0$ are inconsistent

Why:

$det (A) = 0$ only means there is no unique solution, the system can still have infinite solutions

Correct move:

Check the rank of the augmented matrix after row reduction to confirm inconsistency

Wrong move:

Not including the free parameter when writing infinite solution sets

Why:

Examiners require the full general solution, not just a single particular solution

Correct move:

Introduce a free parameter for each non-pivot column, express all variables in terms of parameters

Wrong move:

Multiplying a row by zero during row reduction

Why:

This erases all information from the row and changes the solution set of the system

Correct move:

Only multiply rows by non-zero constants when performing Gaussian elimination

Wrong move:

Skipping verification of solutions for Cramer's rule

Why:

Small sign errors when calculating determinants lead to wrong solutions that are easy to miss

Correct move:

Always substitute your solution back into the original equations to check

6. Quick Reference Cheatsheet

Solution Type	Rank Condition	3D Geometric Meaning
Unique solution	$rank (A) = rank (aug) = 3$	Three planes intersect at one point
Infinite solutions (line)	$rank (A) = rank (aug) = 2$	Three planes intersect along a common line
Infinite solutions (plane)	$rank (A) = rank (aug) = 1$	All three planes are coincident
No solution	$rank (A) < rank (aug)$	At least two planes are parallel or no common intersection

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Classify 3x3 system solution type
2023 · 2
Find parameter for no solution

What's Next

Systems of linear equations are the foundation for many further topics in IB AA HL, particularly vectors in 3D space, where you will use similar methods to find intersections of lines and planes. Solution type classification is also a core skill for problems involving eigenvalues and eigenvectors of matrices, and for solving systems of linear differential equations in later calculus units. Mastery of Gaussian elimination and rank-based classification will not only help you solve exam questions on this topic, but also save time on multi-step problems across the entire syllabus.