From a Homogeneous System to Parameters
A homogeneous matrix equation has the form \(\mathbf{A}\mathbf{x}=\mathbf{0}\), where \(\mathbf{A}\in\mathbb{R}^{m\times n}\), \(\mathbf{x}\in\mathbb{R}^n\), and \(\mathbf{0}\in\mathbb{R}^m\). It always has at least one solution: \(\mathbf{x}=\mathbf{0}\). The goal here is to describe all its solutions, including any that are nonzero. Row reduction turns the equation into a system where some variables are determined by others, and those undetermined variables become parameters.
Write the entries of \(\mathbf{x}\) as \(x_1,\dots,x_n\). A variable corresponding to a pivot column in the reduced matrix is a pivot variable. A variable corresponding to a column without a pivot is a free variable. A free variable is not unconstrained in every possible system; rather, in the reduced homogeneous system it can be assigned any real value, after which the pivot variables are determined. We use letters such as \(s\) and \(t\) for these parameter values.
To get the formula, reduce \(\mathbf{A}\) to reduced row echelon form (RREF) and write the equations represented by its nonzero rows. Assign a separate parameter to each free variable. Then solve the pivot equations in terms of those parameters. Finally, gather the coordinates into a column vector. If there are no free variables, there are no parameters to choose.
A Complete Description From RREF
For a homogeneous system, row reduction is useful because it does not change which vectors solve the equation. This fact can be justified directly for each elementary row operation. Swapping two rows only changes the order of two equations. Multiplying a row by a nonzero scalar replaces an equation by an equivalent one, since it can be reversed by multiplying by the reciprocal. Adding a multiple of one row to another also preserves the solutions, because the operation can be reversed by subtracting that same multiple. Thus the original and reduced homogeneous systems have exactly the same solutions.
Proof. First, the row operations used to obtain RREF preserve the solution set: a row swap is reversible by swapping back; multiplying a row by a nonzero scalar is reversible by multiplying by its reciprocal; and adding a multiple of one row to another is reversible by subtracting that multiple. Therefore, \(\mathbf{A}\mathbf{x}=\mathbf{0}\) and the RREF system have the same solutions. In RREF, each pivot row has a \(1\) in its pivot column and \(0\) in every other pivot column. Its equation therefore expresses its pivot variable as a linear combination of the free variables. Any solution of the RREF system must have some values \(t_j\) for its free variables, and its pivot variables must then be the values specified by those equations. So every solution is included in the formula. Conversely, choose any real values for the parameters and assign the pivot variables by the RREF equations. Each pivot equation is then satisfied, and any zero row gives the equation \(0=0\). Thus the resulting vector solves the RREF system and, by preservation of solutions, the original system. \(\square\)
This theorem gives a practical completeness check: do not stop after finding a solution or a few examples. The formula must account for every free variable and determine every pivot variable. The parameters should range independently over all real numbers, with no extra restrictions, because the system is homogeneous.
Worked Parametrizations
Worked Example: Two Free Variables
Describe every solution of \(\mathbf{A}\mathbf{x}=\mathbf{0}\) for
Replace row 2 by row 2 minus twice row 1:
The only nonzero equation is \(x_1+2x_2-x_3=0\). Column 1 is a pivot column, so \(x_1\) is a pivot variable; \(x_2\) and \(x_3\) are free. Set \(x_2=s\) and \(x_3=t\), where \(s,t\in\mathbb{R}\). Solving the equation gives \(x_1=-2s+t\), so
Check: Multiplication by the original matrix gives \(x_1+2x_2-x_3=(-2s+t)+2s-t=0\) in row 1, and \(2x_1+4x_2-2x_3=2(-2s+t)+4s-2t=0\) in row 2. The parameters can be chosen freely, and the RREF equation forces \(x_1=-2x_2+x_3\), so the formula gives every solution.
Worked Example: Four Variables and Two Parameters
Find a parametric vector form for the solutions of
Subtract row 1 from row 3, then subtract row 2 from the new row 3. Next subtract row 2 from row 1:
The equations are \(x_1-x_3+x_4=0\) and \(x_2+x_3+x_4=0\). Columns 1 and 2 are pivot columns, so set the free variables \(x_3=s\) and \(x_4=t\). The equations give \(x_1=s-t\) and \(x_2=-s-t\). Therefore
Check: For the first parameter vector, the rows of the original matrix give \(1-1=0\), \(-1+1=0\), and \(1-1=0\). For the second, they give \(-1-1+2=0\), \(-1+1=0\), and \(-1-1+2=0\). Thus both parameter directions are solutions, so every linear combination of them also solves the equation. The reduced equations show that every solution has exactly the displayed coordinate formula.
Worked Example: No Free Variables
Describe the solutions for the matrix
Subtract three times row 1 from row 2, scale row 2 by \(-\tfrac12\), then clear the entries above and below its pivot:
Both variables are pivot variables, and there are no free variables. The reduced equations are \(x_1=0\) and \(x_2=0\), so the only solution is \(\mathbf{x}=\mathbf{0}\).
Check: The original matrix multiplied by the proposed answer is \(\mathbf{A}\mathbf{0}=\mathbf{0}\). Since the reduced system forces both coordinates to be zero, there is no other solution.
What the Parameters Tell Us
Each free variable contributes one parameter to the formula. When the RREF has no free columns, the theorem shows that the zero vector is the only solution. When there is at least one free variable, assigning a nonzero value to one parameter and zero to the others gives a nonzero solution: the corresponding free coordinate is nonzero. These statements concern the number of free variables in the reduced system, not the number of rows that were originally present.
The vector form makes the structure visible. In the four-variable example, every solution is \(s\mathbf{v}+t\mathbf{w}\), where \(\mathbf{v}\) and \(\mathbf{w}\) are the two displayed direction vectors. Choosing \(s=t=0\) gives the zero solution; other choices give the rest of the solutions. The direction vectors can be checked against the original matrix, as in the example, which is a useful safeguard against sign errors.
Proof. If there are no free variables, every variable is a pivot variable. Each pivot equation has right-hand side zero and no other pivot variables in it, so every pivot variable equals zero. Conversely, if there is a free variable \(x_j\), set \(x_j=1\) and set every other free variable to \(0\). The pivot equations determine the pivot variables, and the resulting vector is a solution by the parametrization theorem. Its \(j\)th coordinate is \(1\), so it is nonzero. \(\square\)
Common Pitfall: One Solution Is Not the Full Formula
A frequent mistake is to set every free variable equal to zero and then report the resulting zero vector as the complete answer. In the first worked example, that choice gives \(\mathbf{x}=(0,0,0)^{\top}\), which is indeed a solution, but it misses the solutions obtained by other parameter values. For instance, take \(s=0\) and \(t=1\). The formula gives \(\mathbf{x}=(1,0,1)^{\top}\), and direct substitution yields
The correction is to assign parameters to all free variables, not just test one convenient choice. A second error is to treat a pivot variable as free. The pivot equations constrain it: in the first example, choosing \(x_2=s\) and \(x_3=t\) requires \(x_1=-2s+t\). An arbitrary value for \(x_1\) may fail the equation.
When checking a proposed formula, use two tests. First, substitute the formula into the original equation to confirm that all its parameter choices produce solutions. Second, return to the reduced equations and confirm that every solution must have the proposed form. The first test checks validity; the second checks completeness.
Check Your Understanding
For each problem, identify the free variables and give a formula for every solution. Check the formula by substitution.
- Row-reduce \(\begin{bmatrix}1&-1&2\\2&-2&4\end{bmatrix}\) and describe the solutions of the corresponding homogeneous system.
- For \(\begin{bmatrix}1&0&3\\0&1&-2\end{bmatrix}\mathbf{x}=\mathbf{0}\), write the solution as a sum of parameter direction vectors.
- How many free parameters occur for \(\begin{bmatrix}1&2&0&1\\0&0&1&-3\end{bmatrix}\mathbf{x}=\mathbf{0}\)? Give the solution formula.
- Explain why assigning all free variables the value zero does not, by itself, describe every solution when free variables are present.
- What conclusion about the solutions follows when every column of the RREF is a pivot column?
