The Timpson Lyceum · Linear Algebra (2026)

Solution & Applications of Linear Systems and Linear Independence – Lecture 3 (Linear Algebra)

Overview

This lecture covers three sections (1.5, 1.7 and 1.6). The first describes the solution sets of linear systems. A homogeneous system $A\mathbf{x} = \mathbf{0}$ always has the trivial solution, and it has others exactly when a free variable appears. Writing those solutions in parametric vector form shows them geometrically, as a line or plane through the origin. A consistent nonhomogeneous system $A\mathbf{x} = \mathbf{b}$ has the same solution set shifted by one particular solution. The second section introduces linear independence: a set of vectors is independent when the only way to combine them into $\mathbf{0}$ is the trivial one. It then develops fast tests for independence, including one or two vectors, more vectors than entries, and sets containing the zero vector. The last section applies homogeneous systems to two problems: balancing a chemical equation and finding equilibrium prices in a closed economy.

Key Concepts

Homogeneous systems

A system is homogeneous if it can be written $A\mathbf{x} = \mathbf{0}$ , so that every equation has $0$ on the right-hand side.
$A\mathbf{x} = \mathbf{0}$ always has the trivial solution $\mathbf{x} = \mathbf{0}$ , so it is always consistent.
Any linear system has no solution, exactly one, or infinitely many. A homogeneous system already has one, so a nontrivial solution means infinitely many.
$A\mathbf{x} = \mathbf{0}$ has a nontrivial solution if and only if the equation has at least one free variable, meaning some column of $A$ is not a pivot column.
From here on, context decides whether a symbol is a vector or a number: in $A\mathbf{x} = \mathbf{0}$ , $A$ is a matrix and $\mathbf{x}$ and $\mathbf{0}$ are vectors.

Parametric vector form

Solving a system and writing $\mathbf{x}$ as a sum of vectors, one for each free variable, each weighted by that free variable, gives the parametric vector form:

\mathbf{x} = s\mathbf{u} + t\mathbf{v} \qquad (s, t \in \mathbb{R})

With one free variable the solution set is $\operatorname{Span}\{\mathbf{v}\}$ , a line through the origin. With two, it is $\operatorname{Span}\{\mathbf{u}, \mathbf{v}\}$ , typically a plane through the origin. Every such solution set contains $\mathbf{0}$ , because the trivial solution always works.
One equation with two free variables gives a plane. For $10x_1 - 3x_2 - 2x_3 = 0$ , solving for $x_1$ gives $x_1 = 0.3x_2 + 0.2x_3$ with $x_2, x_3$ free:

\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = x_2\begin{bmatrix} 0.3 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} 0.2 \\ 0 \\ 1 \end{bmatrix}

This is a plane through the origin in $\mathbb{R}^3$ . In general a single linear equation in three variables describes a plane. The one exception is the degenerate equation $0x_1 + 0x_2 + 0x_3 = 0$ , whose solution set is all of $\mathbb{R}^3$ .

Solutions of nonhomogeneous systems

Keep the same coefficient matrix and change only the right-hand side. Row reduction performs the same operations, so only the last column of the augmented matrix changes. The free variables stay free, and each basic variable picks up a constant.
In parametric vector form the result is

\mathbf{x} = \mathbf{p} + t\mathbf{v}

where $t\mathbf{v}$ describes the homogeneous solutions and $\mathbf{p}$ is a constant vector. Geometrically, the solution set is the homogeneous solution set translated by $\mathbf{p}$ : a line or plane parallel to the one through the origin, now passing through $\mathbf{p}$ .

Loosely, this resembles $y = mx + b$ . The homogeneous part carries the shape, and $\mathbf{p}$ fixes where it sits.

Theorem 6 (solutions of nonhomogeneous systems). Suppose $A\mathbf{x} = \mathbf{b}$ is consistent, and let $\mathbf{p}$ be any one solution, called a particular solution. Then the solution set of $A\mathbf{x} = \mathbf{b}$ is the set of all vectors $\mathbf{w} = \mathbf{p} + \mathbf{v}_h$ , where $\mathbf{v}_h$ is any solution of $A\mathbf{x} = \mathbf{0}$ . Equivalently, $\mathbf{v}$ solves $A\mathbf{x} = \mathbf{0}$ if and only if $\mathbf{v} + \mathbf{p}$ solves $A\mathbf{x} = \mathbf{b}$ .

Proof. Both directions use the linearity of matrix–vector multiplication.

\begin{aligned} (\Rightarrow)\quad A\mathbf{v} = \mathbf{0}:&\qquad A(\mathbf{v} + \mathbf{p}) = A\mathbf{v} + A\mathbf{p} = \mathbf{0} + \mathbf{b} = \mathbf{b} \\ (\Leftarrow)\quad A(\mathbf{v} + \mathbf{p}) = \mathbf{b}:&\qquad A\mathbf{v} + A\mathbf{p} = \mathbf{b} \;\Rightarrow\; A\mathbf{v} + \mathbf{b} = \mathbf{b} \;\Rightarrow\; A\mathbf{v} = \mathbf{0} \end{aligned}

The theorem requires $A\mathbf{x} = \mathbf{b}$ to be consistent. An inconsistent system has no particular solution to shift by.

Linear independence

An indexed set $\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$ in $\mathbb{R}^n$ is linearly independent if the vector equation

x_1\mathbf{v}_1 + x_2\mathbf{v}_2 + \cdots + x_p\mathbf{v}_p = \mathbf{0}

has only the trivial solution. It is linearly dependent if there are weights $c_1, \dots, c_p$ , not all zero, with $c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p = \mathbf{0}$ . Such an equation is a linear dependence relation.

$p$ counts the vectors, and $n$ counts the entries in each vector.
Intuition: a set is independent when no vector in it can be built from the others. For example, $(1, 0)$ and $(0, 1)$ are independent. Adding $(1, 1)$ makes the set dependent, since $(1, 0) + (0, 1) - (1, 1) = \mathbf{0}$ .
A dependent set contains a vector lying in the span of the others. An independent set does not.
Matrix form. The columns of $A$ $A$ are linearly independent if and only if $A\mathbf{x} = \mathbf{0}$ $A x = 0$ has only the trivial solution, which means every column of $A$ $A$ is a pivot column.
- For $A = \begin{bmatrix} 0 & 1 & 4 \\ 1 & 2 & -1 \\ 5 & 8 & 0 \end{bmatrix}$ , row reduction yields a pivot in every column. The only solution is $\mathbf{x} = \mathbf{0}$ , so the columns are independent.

Sets of one or two vectors

A set containing one vector $\mathbf{v}$ is linearly independent if and only if $\mathbf{v} \neq \mathbf{0}$ . The zero vector alone is dependent, since $c\,\mathbf{0} = \mathbf{0}$ for every $c$ .
A set of two vectors is linearly dependent if and only if at least one of them is a scalar multiple of the other. Geometrically, dependent means the two vectors lie on the same line through the origin.
- $(3, 1)$ and $(6, 2)$ are dependent, because $2(3, 1) - (6, 2) = \mathbf{0}$ .
- $(3, 2)$ and $(6, 2)$ are independent, because neither is a multiple of the other.
Why (for two nonzero vectors): suppose $c\mathbf{v}_1 + d\mathbf{v}_2 = \mathbf{0}$ with $c, d$ not both zero. If $c = 0$ , then $d\mathbf{v}_2 = \mathbf{0}$ forces $d = 0$ , since $\mathbf{v}_2 \neq \mathbf{0}$ . So $c \neq 0$ , and dividing by $c$ gives $\mathbf{v}_1 = -\tfrac{d}{c}\mathbf{v}_2$ , a scalar multiple of $\mathbf{v}_2$ .
Adding a vector from the span. $\mathbf{u} = (3, 1, 0)$ and $\mathbf{v} = (1, 6, 0)$ are independent, and their span is a plane through the origin. A vector $\mathbf{w}$ in that plane, such as $(4, 7, 0) = \mathbf{u} + \mathbf{v}$ , makes $\{\mathbf{u}, \mathbf{v}, \mathbf{w}\}$ dependent. With $\mathbf{u}, \mathbf{v}$ independent, $\{\mathbf{u}, \mathbf{v}, \mathbf{w}\}$ is dependent if and only if $\mathbf{w} \in \operatorname{Span}\{\mathbf{u}, \mathbf{v}\}$ .

Theorem 7 (characterization of linearly dependent sets). An indexed set $S = \{\mathbf{v}_1, \dots, \mathbf{v}_p\}$ of two or more vectors is linearly dependent if and only if at least one vector in $S$ is a linear combination of the others.

Proof. Suppose first that some $\mathbf{v}_j$ is a combination of the others. Moving it across gives a relation whose weight on $\mathbf{v}_j$ is $-1$ , so the relation is nontrivial:

\begin{aligned} \mathbf{v}_j &= c_1\mathbf{v}_1 + \cdots + c_{j-1}\mathbf{v}_{j-1} + c_{j+1}\mathbf{v}_{j+1} + \cdots + c_p\mathbf{v}_p \\ \mathbf{0} &= c_1\mathbf{v}_1 + \cdots + (-1)\mathbf{v}_j + \cdots + c_p\mathbf{v}_p \end{aligned}

Conversely, suppose $S$ is dependent, so $c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p = \mathbf{0}$ with some $c_j \neq 0$ . Solve for $\mathbf{v}_j$ by moving its term across and dividing by $-c_j$ :

\mathbf{v}_j = -\frac{c_1}{c_j}\mathbf{v}_1 - \cdots - \frac{c_{j-1}}{c_j}\mathbf{v}_{j-1} - \frac{c_{j+1}}{c_j}\mathbf{v}_{j+1} - \cdots - \frac{c_p}{c_j}\mathbf{v}_p

This expresses $\mathbf{v}_j$ as a linear combination of the others. The theorem does not say that every vector in a dependent set is a combination of the others.

Theorem 8 (more vectors than entries). If a set contains more vectors than there are entries in each vector ( $p > n$ ), then the set is linearly dependent. Any set $\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$ in $\mathbb{R}^n$ with $p > n$ is dependent.

Proof. Let $A = \begin{bmatrix} \mathbf{v}_1 & \cdots & \mathbf{v}_p \end{bmatrix}$ , an $n \times p$ matrix. The equation $A\mathbf{x} = \mathbf{0}$ is $n$ equations in $p > n$ unknowns. There are at most $n$ pivots, one per row, so some column has no pivot. That free variable gives a nontrivial solution.

For example, $(2, 1)$ , $(4, -1)$ and $(-2, 2)$ must be dependent: three vectors in $\mathbb{R}^2$ . Put another way, if two of them are independent, they already span $\mathbb{R}^2$ , so the third lies in their span.

Theorem 9 (the zero vector). If a set $S = \{\mathbf{v}_1, \dots, \mathbf{v}_p\}$ in $\mathbb{R}^n$ contains the zero vector, then $S$ is linearly dependent.

Proof. If $\mathbf{v}_j = \mathbf{0}$ , then $0\mathbf{v}_1 + \cdots + 1\mathbf{v}_j + \cdots + 0\mathbf{v}_p = \mathbf{0}$ is a nontrivial relation. Any nonzero weight on $\mathbf{v}_j$ works just as well.

Deciding by inspection

$\begin{bmatrix} 1 \\ 7 \\ 6 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 9 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ 5 \end{bmatrix}, \begin{bmatrix} 4 \\ 1 \\ 8 \end{bmatrix}$ : dependent, since four vectors in $\mathbb{R}^3$ fall under Theorem 8.
$\begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ 8 \end{bmatrix}$ : dependent, since the set contains the zero vector (Theorem 9).
$\begin{bmatrix} -2 \\ 4 \\ 6 \\ 10 \end{bmatrix}, \begin{bmatrix} 3 \\ -6 \\ -9 \\ 15 \end{bmatrix}$ : independent. Multiplying the first vector by $-\tfrac{3}{2}$ matches the first three entries of the second but gives $-15$ , not $15$ , in the last. Neither vector is a multiple of the other.

Applications of linear systems

Balancing chemical equations. Represent each molecule by a vector counting its atoms. The unknown coefficients then satisfy a homogeneous system. Its free variable reflects the fact that any whole-number multiple of a balanced equation is also balanced. Choose the free variable to clear fractions, giving the smallest whole-number coefficients.
Equilibrium prices. In a closed economy, each sector's output is consumed entirely by the sectors. The equilibrium price of each sector's output equals its total expenditures on inputs. Each sector gives one linear equation, and moving everything to one side produces a homogeneous system. A free variable remains, because only the ratios of the prices are determined.
The hard part of an applied problem is usually translating it into a linear system. Once it is a matrix, solving it is routine, and it is exactly the kind of computation computers do extremely fast.

Worked Example

A nontrivial solution in parametric vector form. Determine whether the homogeneous system

\begin{aligned} 3x_1 + 5x_2 - 4x_3 &= 0 \\ -3x_1 - 2x_2 + 4x_3 &= 0 \\ 6x_1 + x_2 - 8x_3 &= 0 \end{aligned}

has a nontrivial solution, and describe its solution set. Row reduce the augmented matrix:

\begin{aligned} \left[\begin{array}{ccc|c} 3 & 5 & -4 & 0 \\ -3 & -2 & 4 & 0 \\ 6 & 1 & -8 & 0 \end{array}\right] &\sim \left[\begin{array}{ccc|c} 3 & 5 & -4 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & -9 & 0 & 0 \end{array}\right] && R_2 \to R_2 + R_1,\; R_3 \to R_3 - 2R_1 \\ &\sim \left[\begin{array}{ccc|c} 3 & 5 & -4 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] && R_3 \to R_3 + 3R_2,\; R_2 \to \tfrac{1}{3}R_2 \\ &\sim \left[\begin{array}{ccc|c} 1 & 0 & -\tfrac{4}{3} & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] && R_1 \to \tfrac{1}{3}(R_1 - 5R_2) \end{aligned}

Column 3 has no pivot, so $x_3$ is free and a nontrivial solution exists. The system reads $x_1 = \tfrac{4}{3}x_3$ , $x_2 = 0$ , so

\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} \tfrac{4}{3}x_3 \\ 0 \\ x_3 \end{bmatrix} = x_3\begin{bmatrix} \tfrac{4}{3} \\ 0 \\ 1 \end{bmatrix} = x_3\mathbf{v}

The solution set is $\operatorname{Span}\{\mathbf{v}\}$ , a line through the origin. Taking $x_3 = 1$ gives the nontrivial solution $(\tfrac{4}{3}, 0, 1)$ .

The same matrix with a nonzero right-hand side. For $A\mathbf{x} = \mathbf{b}$ with $\mathbf{b} = (7, -1, -4)$ , the same row operations give

\left[\begin{array}{ccc|c} 1 & 0 & -\tfrac{4}{3} & -1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]

so $x_1 = -1 + \tfrac{4}{3}x_3$ , $x_2 = 2$ , and $x_3$ is free:

\mathbf{x} = \begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} \tfrac{4}{3} \\ 0 \\ 1 \end{bmatrix} = \mathbf{p} + x_3\mathbf{v}

This is the homogeneous line shifted to pass through the particular solution $\mathbf{p} = (-1, 2, 0)$ , as Theorem 6 predicts.

Testing a set for independence. Are $\mathbf{v}_1 = (1, 2, 3)$ , $\mathbf{v}_2 = (4, 5, 6)$ , $\mathbf{v}_3 = (2, 1, 0)$ linearly independent?

\begin{aligned} \left[\begin{array}{ccc|c} 1 & 4 & 2 & 0 \\ 2 & 5 & 1 & 0 \\ 3 & 6 & 0 & 0 \end{array}\right] &\sim \left[\begin{array}{ccc|c} 1 & 4 & 2 & 0 \\ 0 & -3 & -3 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] && \text{(echelon form)} \\ &\sim \left[\begin{array}{ccc|c} 1 & 0 & -2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] && \text{(reduced echelon form)} \end{aligned}

Once the echelon form shows that column 3 has no pivot, the set is known to be dependent. The reduced form gives $x_1 = 2x_3$ , $x_2 = -x_3$ . Taking $x_3 = 1$ gives the dependence relation

2\mathbf{v}_1 - \mathbf{v}_2 + \mathbf{v}_3 = \begin{bmatrix} 2 - 4 + 2 \\ 4 - 5 + 1 \\ 6 - 6 + 0 \end{bmatrix} = \mathbf{0}

Balancing the combustion of propane. Balance $x_1\,\text{C}_3\text{H}_8 + x_2\,\text{O}_2 \to x_3\,\text{CO}_2 + x_4\,\text{H}_2\text{O}$ . Record each molecule as a vector of (carbon, hydrogen, oxygen) counts, then move everything to one side:

x_1\begin{bmatrix} 3 \\ 8 \\ 0 \end{bmatrix} + x_2\begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix} - x_3\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} - x_4\begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} = \mathbf{0}

Row reducing the coefficient matrix gives

\begin{aligned} \begin{bmatrix} 3 & 0 & -1 & 0 \\ 8 & 0 & 0 & -2 \\ 0 & 2 & -2 & -1 \end{bmatrix} &\sim \begin{bmatrix} 1 & 0 & 0 & -\tfrac{1}{4} \\ 0 & 1 & 0 & -\tfrac{5}{4} \\ 0 & 0 & 1 & -\tfrac{3}{4} \end{bmatrix} \end{aligned}

so $x_1 = \tfrac{1}{4}x_4$ , $x_2 = \tfrac{5}{4}x_4$ , $x_3 = \tfrac{3}{4}x_4$ , with $x_4$ free. Choosing $x_4 = 4$ clears the fractions:

\text{C}_3\text{H}_8 + 5\,\text{O}_2 \to 3\,\text{CO}_2 + 4\,\text{H}_2\text{O}

Check: carbon $3 = 3$ , hydrogen $8 = 8$ , oxygen $10 = 6 + 4$ . Doubling every coefficient, giving 2, 10, 6 and 8, is also balanced.

Equilibrium prices. An economy has three sectors (coal, electric and steel), and each sector's output is split among the three as follows:

Output of →	Coal	Electric	Steel
purchased by Coal	0.0	0.4	0.6
purchased by Electric	0.6	0.1	0.2
purchased by Steel	0.4	0.5	0.2

At equilibrium each sector's income equals its expenses. Setting up those equations and moving everything to one side gives the homogeneous system:

\begin{aligned} p_C &= 0.4p_E + 0.6p_S \\ p_E &= 0.6p_C + 0.1p_E + 0.2p_S \\ p_S &= 0.4p_C + 0.5p_E + 0.2p_S \end{aligned} \quad\Longrightarrow\quad \left[\begin{array}{ccc|c} 1 & -0.4 & -0.6 & 0 \\ -0.6 & 0.9 & -0.2 & 0 \\ -0.4 & -0.5 & 0.8 & 0 \end{array}\right]

Row reduction leaves $p_S$ free, with $p_C = \tfrac{31}{33}p_S \approx 0.94\,p_S$ and $p_E = \tfrac{28}{33}p_S \approx 0.85\,p_S$ :

\mathbf{p} = \begin{bmatrix} p_C \\ p_E \\ p_S \end{bmatrix} \approx p_S\begin{bmatrix} 0.94 \\ 0.85 \\ 1 \end{bmatrix}

Only the ratios are determined. For example, if steel's output is priced at $100 million, coal's is about $94 million and electric's about $85 million. Ordering the variables differently would leave a different price free, giving an equivalent answer.

Summary

A homogeneous system $A\mathbf{x} = \mathbf{0}$ always has the trivial solution. It has a nontrivial solution, and hence infinitely many, exactly when it has a free variable.
Parametric vector form writes the solutions as combinations of one vector per free variable. The homogeneous solution set is a span through the origin: a line for one free variable, typically a plane for two.
If $A\mathbf{x} = \mathbf{b}$ is consistent with particular solution $\mathbf{p}$ , its solutions are exactly $\mathbf{p} + \mathbf{v}_h$ , the homogeneous solution set translated by $\mathbf{p}$ (Theorem 6).
A set is linearly independent when $x_1\mathbf{v}_1 + \cdots + x_p\mathbf{v}_p = \mathbf{0}$ has only the trivial solution. The columns of $A$ are independent exactly when every column of $A$ is a pivot column.
A set is dependent exactly when some vector in it is a combination of the others (Theorem 7). Two vectors are dependent exactly when one is a multiple of the other. More vectors than entries (Theorem 8) and containing $\mathbf{0}$ (Theorem 9) each force dependence.
Balancing chemical equations and finding equilibrium prices both reduce to homogeneous systems. The free variable reflects scale: whole-number multiples of a reaction, or the overall level of prices.