What Makes a System Homogeneous?
A homogeneous system is a system of linear equations whose right-hand sides are all zero. Its matrix form is \(\mathbf{A}\mathbf{x}=\mathbf{0}\), where \(\mathbf{A}\) has real entries, \(\mathbf{x}\in\mathbb{R}^n\) is a column of unknowns, and \(\mathbf{0}\) is the zero vector with as many entries as \(\mathbf{A}\) has rows. The system may have more than one equation, but every equation has zero on its right-hand side.
For example, the system \(x_1+2x_2=0\), \(3x_1+6x_2=0\) is homogeneous. In contrast, a system with even one nonzero right-hand side is not homogeneous. The special feature of a homogeneous system is that setting every unknown to zero always satisfies every equation.
Row reduction is a practical way to distinguish these cases. Write the system as an augmented matrix \([\mathbf{A}\mid\mathbf{0}]\) and reduce it. Since its right-hand column consists entirely of zeros, row operations keep it zero; the essential information lies in the coefficient columns. In the reduced system, a variable corresponding to a pivot is a pivot variable. A variable in a column without a pivot is a free variable: it can be assigned a value, and the pivot variables must then satisfy the reduced equations.
Why the Zero Solution Is Always Present
Proof. Let \(\mathbf{A}\) have \(m\) rows and \(n\) columns. The zero vector \(\mathbf{0}\in\mathbb{R}^n\) has each of its \(n\) entries equal to zero. By the definition of matrix-vector multiplication, each entry of \(\mathbf{A}\mathbf{0}\) is a sum of products of matrix entries with zero entries of \(\mathbf{0}\), so each entry is zero. Thus \(\mathbf{A}\mathbf{0}=\mathbf{0}\), and \(\mathbf{x}=\mathbf{0}\) is a solution. Since a solution exists, the system is consistent. \(\square\)
This result rules out one possible outcome: a homogeneous system cannot become inconsistent during correct row reduction. To see why the reduction preserves the solutions, consider the three permitted row operations. Swapping two equations keeps the same equations, just in a different order. Multiplying an equation by a nonzero scalar gives an equivalent equation, because we can divide by that scalar to recover the original. Adding a multiple of one equation to another also preserves the system: the operation can be undone by subtracting that same multiple. Therefore each row operation is reversible and preserves exactly the same solutions.
The zero solution being present does not mean it is the only solution. The reduced equations determine whether some variables can be assigned freely. The next result gives a precise test using the pivot positions.
Proof. Let the variables be \(x_1,\dots,x_n\). In RREF, each nonzero row has a leading \(1\), and its pivot column has zeros in every other row. The equation for a pivot variable expresses it as a linear combination of the free variables. If every variable column has a pivot, there are no free variables. Each pivot variable is then equal to zero in the homogeneous system, so the only solution is \(\mathbf{x}=\mathbf{0}\).
Now suppose a variable column has no pivot. Assign the corresponding free variable the value \(1\), and assign every other free variable the value \(0\). The equations for the pivot variables then determine their values, so this assignment extends to a solution of the reduced homogeneous system. That solution is nontrivial because one of its free-variable entries is \(1\). Since the row operations preserve solutions, the original system also has a nontrivial solution. These two cases prove the test. \(\square\)
A Homogeneous System with Free Variables
Worked Example: One Free Variable
Solve the homogeneous system \(x_1+2x_2=0\), \(3x_1+6x_2=0\), and write all of its solutions.
Its augmented matrix is \([\mathbf{A}\mid\mathbf{0}]\). Apply \(R_2\to R_2-3R_1\):
The second row says \(0=0\), so it adds no restriction. Column 1 has a pivot and column 2 does not. Thus \(x_1\) is a pivot variable and \(x_2\) is free. Set \(x_2=t\), where \(t\in\mathbb{R}\). The first equation becomes \(x_1+2t=0\), so \(x_1=-2t\). All solutions are
Check: For any real \(t\), the left-hand sides are \((-2t)+2t=0\) and \(3(-2t)+6t=-6t+6t=0\). Taking \(t=0\) gives the trivial solution; taking \(t=1\) gives the nontrivial solution \(\begin{bmatrix}-2\\1\end{bmatrix}\).
A free variable is not a variable that can be chosen without regard to the rest of the system. It is free in the sense that the reduced equations do not set its value directly. The pivot variables still depend on the chosen free-variable values. In the example, choosing \(x_2=t\) forces \(x_1=-2t\); choosing \(x_1\) and \(x_2\) independently would generally fail to satisfy the equation.
A System with Only the Trivial Solution
Worked Example: A Pivot in Every Variable Column
Solve the system \(x_1+x_2=0\), \(x_2+x_3=0\), \(x_1+x_3=0\).
Reduce its augmented matrix. First apply \(R_3\to R_3-R_1\), then \(R_3\to R_3+R_2\):
Scale \(R_3\) by \(\tfrac12\), then apply \(R_2\to R_2-R_3\) and \(R_1\to R_1-R_2\), in that order:
Every variable column has a pivot, and the reduced equations are \(x_1=0\), \(x_2=0\), \(x_3=0\). Hence the only solution is \(\mathbf{x}=\mathbf{0}\).
Check: Substituting \(\mathbf{x}=\begin{bmatrix}0\\0\\0\end{bmatrix}\) into the original equations gives \(0+0=0\), \(0+0=0\), and \(0+0=0\). The pivot test shows that there are no other solutions.
In this example, the last nonzero row gave \(2x_3=0\), which forces \(x_3=0\). The second equation then forces \(x_2=0\), and the first forces \(x_1=0\). This is the same conclusion as reading the final RREF, where each variable has its own pivot equation.
Two Free Variables and a Parametric Description
Worked Example: A Family of Solutions
Solve the system with coefficient rows \((1,2,1,0)\), \((0,1,-1,2)\), and \((1,3,0,2)\). In equation form, it is \(x_1+2x_2+x_3=0\), \(x_2-x_3+2x_4=0\), and \(x_1+3x_2+2x_4=0\).
Start with the augmented matrix. Apply \(R_3\to R_3-R_1\), then \(R_3\to R_3-R_2\):
Next apply \(R_1\to R_1-2R_2\):
Columns 1 and 2 contain pivots, while columns 3 and 4 do not. Set \(x_3=s\) and \(x_4=t\), with \(s,t\in\mathbb{R}\). The first reduced equation gives \(x_1+3s-4t=0\), so \(x_1=-3s+4t\). The second gives \(x_2-s+2t=0\), so \(x_2=s-2t\). Therefore all solutions are
Check: In the first original equation, the left-hand side becomes \((-3s+4t)+2(s-2t)+s=-3s+4t+2s-4t+s=0\). In the second it becomes \((s-2t)-s+2t=0\). In the third it becomes \((-3s+4t)+3(s-2t)+2t=-3s+4t+3s-6t+2t=0\). Thus every parameter choice gives a solution. The two free variables also give nontrivial solutions; for example \(s=1,t=0\) gives \(\begin{bmatrix}-3\\1\\1\\0\end{bmatrix}\).
A Common Mistake
A frequent error is to reason that because the right-hand side is zero, every variable must be zero. That conclusion confuses the trivial solution—which always exists—with the only solution—which need not be the case. For the first worked example, the incorrect calculation might say: “The right-hand side is zero, so \(x_2=0\), and then \(x_1=0\).” But the reduced equation is \(x_1+2x_2=0\), not \(x_2=0\). It permits \(x_2=1\) and \(x_1=-2\).
The correction is to read the reduced equations and identify which variables are actually constrained. If a variable has no pivot, assign it a parameter and solve for the pivot variables. Then check that the resulting values satisfy the original equations. A zero row such as \(0=0\) does not set any variable to zero; it simply contributes no additional equation.
Why These Solutions Matter
Solving \(\mathbf{A}\mathbf{x}=\mathbf{0}\) asks which coefficient vectors \(\mathbf{x}\) make the matrix-vector product equal zero. Row reduction organizes the answer: pivot variables are determined by the free variables, and each allowed choice of free-variable values produces a solution. When every column is a pivot column, there are no choices to make and only the trivial solution remains. When a free variable occurs, setting it to a nonzero value produces a nontrivial solution.
For a small system, keep a reliable sequence in mind: form the coefficient matrix and zero right-hand side; row-reduce using reversible row operations; identify pivot and free variables; assign a parameter to each free variable; solve for the pivot variables; and substitute the resulting vector into the original equations. This separates the always-present zero solution from the additional solutions that the system may allow.
Check Your Understanding
For each system, reduce its augmented matrix and explain what the pivot and free variables say about its solutions.
- Does \(x_1-3x_2=0,\;2x_1-6x_2=0\) have nontrivial solutions? Write all solutions.
- What is the solution of \(x_1+x_2=0,\;x_1-x_2=0\)? Does the reduced system have a free variable?
- For \(x_1+2x_2-x_3=0,\;x_2+x_3=0\), identify the free variable and give a parametric description of all solutions.
- Explain why a homogeneous system cannot produce a row of the form \(0=1\) when row-reduced correctly.
- In your own words, distinguish the trivial solution from a nontrivial solution.
