Homogeneous Linear Systems
Source inspiration: (Mathew 2000-2019).
Background
A homogeneous linear system has the form
\[ A\mathbf{x}=\mathbf{0}. \tag{1} \]
The vector \(\mathbf{x}=\mathbf{0}\) always solves (1), so every homogeneous system has at least the trivial solution.
A matrix is in reduced row echelon form (RREF) if:
- Every nonzero row has leading entry \(1\).
- Each leading \(1\) is the only nonzero entry in its column.
- Leading \(1\) positions move to the right as you move down rows.
- Any zero rows are grouped at the bottom.
The rank of a matrix \(A\), written \(\operatorname{rank}(A)\), is the number of nonzero rows in its RREF.
Because an augmented homogeneous matrix is
\[ M=[A\mid \mathbf{0}], \tag{2} \]
the last column is always zero, so
\[ \operatorname{rank}(A)=\operatorname{rank}(M). \tag{3} \]
For \(A\in\mathbb{R}^{n\times n}\):
\[ \operatorname{rank}(A)=n \Longleftrightarrow \text{the only solution is }\mathbf{x}=\mathbf{0}, \tag{4} \]
\[ \operatorname{rank}(A)<n \Longleftrightarrow \text{there are infinitely many nontrivial solutions}. \tag{5} \]
For square matrices, this is equivalent to:
\[ \det(A)\neq 0 \Rightarrow \text{unique trivial solution}, \qquad \det(A)=0 \Rightarrow \text{infinitely many solutions}. \tag{6} \]
Algorithm
- Form \([A\mid\mathbf{0}]\).
- Row reduce to RREF.
- Identify pivot variables and free variables.
- Express pivot variables in terms of free variables.
- Write the parametric vector form and verify by substitution into \(A\mathbf{x}=\mathbf{0}\).
Example 1 - Unique Trivial Solution
Solve
\[ \begin{aligned} x_1+x_2-2x_3&=0,\\ 3x_1+2x_2+4x_3&=0,\\ 4x_1+3x_2+3x_3&=0. \end{aligned} \]
Step 1: Build the coefficient matrix and row reduce.
\[ A= \begin{pmatrix} 1&1&-2\\ 3&2&4\\ 4&3&3 \end{pmatrix} \longrightarrow \operatorname{RREF}(A)=I_3. \]
Step 2: Read off solution.
\[ x_1=x_2=x_3=0, \qquad \mathbf{x}=\begin{pmatrix}0\\0\\0\end{pmatrix}. \]
Verification:
| Row | Substitution result | Target |
|---|---|---|
| 1 | \(0+0-2(0)=0\) | \(0\ \checkmark\) |
| 2 | \(3(0)+2(0)+4(0)=0\) | \(0\ \checkmark\) |
| 3 | \(4(0)+3(0)+3(0)=0\) | \(0\ \checkmark\) |
Since \(\det(A)=-1\neq0\), only the trivial solution exists.
Example 2 - One Free Variable
Solve
\[ \begin{aligned} x_1+x_2-2x_3&=0,\\ 3x_1+2x_2+4x_3&=0,\\ 4x_1+3x_2+2x_3&=0. \end{aligned} \]
Step 1: Row reduce.
\[ \operatorname{RREF}(A)= \begin{pmatrix} 1&0&8\\ 0&1&-10\\ 0&0&0 \end{pmatrix}. \]
So
\[ x_1+8x_3=0, \qquad x_2-10x_3=0. \]
Step 2: Let the free variable \(x_3=t\).
\[ x_1=-8t, \qquad x_2=10t, \qquad x_3=t. \]
Hence
\[ \mathbf{x}(t)=t \begin{pmatrix} -8\\10\\1 \end{pmatrix}, \qquad t\in\mathbb{R}. \]
Verification:
| Row | Substitution with \((-8t,10t,t)\) | Target |
|---|---|---|
| 1 | \(-8t+10t-2t=0\) | \(0\ \checkmark\) |
| 2 | \(3(-8t)+2(10t)+4t=0\) | \(0\ \checkmark\) |
| 3 | \(4(-8t)+3(10t)+2t=0\) | \(0\ \checkmark\) |
Example 3 - Two Free Variables
Solve
\[ \begin{aligned} 4x_1+3x_2+2x_3&=0,\\ 4x_1+3x_2+2x_3&=0,\\ 4x_1+3x_2+2x_3&=0. \end{aligned} \]
Step 1: Row reduce.
\[ \operatorname{RREF}(A)= \begin{pmatrix} 1&\tfrac34&\tfrac12\\ 0&0&0\\ 0&0&0 \end{pmatrix}. \]
So
\[ x_1+\tfrac34x_2+\tfrac12x_3=0. \]
Step 2: Let \(x_2=s\) and \(x_3=t\) be free.
\[ x_1=-\tfrac34 s-\tfrac12 t, \qquad x_2=s, \qquad x_3=t. \]
Parametric vector form:
\[ \mathbf{x}(s,t)= s\begin{pmatrix}-\tfrac34\\1\\0\end{pmatrix} + t\begin{pmatrix}-\tfrac12\\0\\1\end{pmatrix}. \]
Verification:
| Row | Substitution result | Target |
|---|---|---|
| 1 | \(4(-\tfrac34 s-\tfrac12 t)+3s+2t=0\) | \(0\ \checkmark\) |
| 2 | same as row 1 | \(0\ \checkmark\) |
| 3 | same as row 1 | \(0\ \checkmark\) |
When \(\operatorname{rank}(A)<n\), free variables are unavoidable. The solution set is a subspace (a line, plane, or higher-dimensional analogue through the origin), not a single point.
Application - Balancing Propane Combustion
Write
\[ x_1\,\mathrm{C_3H_8}+x_2\,\mathrm{O_2} \rightarrow x_3\,\mathrm{CO_2}+x_4\,\mathrm{H_2O}. \]
Atom-balance equations give the homogeneous system
\[ \begin{aligned} 3x_1-x_3&=0,\\ 8x_1-2x_4&=0,\\ 2x_2-2x_3-x_4&=0. \end{aligned} \]
From RREF,
\[ x_1=\tfrac14x_4, \qquad x_2=\tfrac54x_4, \qquad x_3=\tfrac34x_4. \]
Let \(x_4=t\):
\[ \mathbf{x}(t)= \begin{pmatrix} \tfrac14 t\\ \tfrac54 t\\ \tfrac34 t\\ t \end{pmatrix}. \]
Choose \(t=4\) for the smallest positive integers:
\[ (x_1,x_2,x_3,x_4)=(1,5,3,4). \]
Balanced equation:
\[ \mathrm{C_3H_8}+5\mathrm{O_2} \rightarrow 3\mathrm{CO_2}+4\mathrm{H_2O}. \]
Verification (atom counts):
| Element | Left side | Right side |
|---|---|---|
| C | \(3(1)=3\) | \(1(3)=3\ \checkmark\) |
| H | \(8(1)=8\) | \(2(4)=8\ \checkmark\) |
| O | \(2(5)=10\) | \(2(3)+1(4)=10\ \checkmark\) |
Summary
Homogeneous systems are classified by rank (or determinant in square cases):
- Full rank gives only the trivial solution.
- Rank deficiency creates free variables and infinitely many solutions.
- RREF exposes the null-space basis directly, which is useful both in algebra and in applications such as chemical equation balancing.