What Free Variables Record
When a consistent system has more variables than pivot positions, some variables are not determined by their own leading equations. Those variables are free variables: their values may be chosen arbitrarily, and the remaining variables must then adjust to satisfy the equations. A complete solution description therefore does two things: it assigns a parameter to each free variable, and it expresses every pivot variable in terms of those parameters.
Assume throughout that the system is over \(\mathbb{R}\), has \(n\) variables \(x_1,\dots,x_n\), and has been written as an augmented matrix in row echelon form. A pivot is the first nonzero entry of a nonzero row; its column is a pivot column, and the variable in that column is a pivot variable. Variables in the other columns are free variables. A row of zeros in the coefficient part is allowed in a consistent system only when its augmented entry is also zero. We will start from echelon systems, so the task here is to read and describe their solutions, not to perform the reduction that produced them.
Parameters are not extra unknowns in the original system. They are labels for all possible choices of free-variable values. For example, if \(x_2\) is free, writing \(x_2=s\) means that \(x_2\) may be any real number \(s\). The pivot equations then determine what values of the other variables are compatible with that choice.
A Reliable Reading Procedure
Read the nonzero rows from bottom to top. In the bottom nonzero row, the pivot variable can be solved in terms of the variables to its right. Move upward and substitute the expressions already found into each earlier pivot equation. Do not assign new parameters to pivot variables: doing so would make them appear arbitrary even though the equations constrain them.
Each pivot identifies a variable that must be solved from its row.
Give each nonpivot variable a distinct real parameter, such as \(s\) or \(t\).
Starting at the last nonzero row, solve for pivot variables in terms of the parameters and any pivot variables already found.
Give the variables together as a parametric vector, then substitute into the nonzero equations.
If a pivot entry is not \(1\), divide by that nonzero coefficient when solving its equation. A consistent all-zero row contributes no equation and imposes no additional restriction. In particular, if there are no pivot columns, all variables are free; consistency then requires that the equations impose no contradictory condition.
Why the Parameter Description Gives Every Solution
Proof. Let the nonzero rows be ordered from top to bottom, and consider them from bottom to top. In the last nonzero row, the pivot coefficient is nonzero, and every variable to its right is either free or a pivot variable in a later column. Once the free values are assigned, any later pivot variables have already been determined, so division by the pivot coefficient determines the row’s pivot variable. Repeat this argument upward: at each row, variables to the right already have values, so its pivot variable is determined. Thus the procedure constructs values satisfying every nonzero row; the zero rows impose no further conditions because the system is consistent. Conversely, any solution must use its own values for the free variables. In the bottom nonzero row, its pivot value is forced by that row; proceeding upward, every pivot value is forced in turn. Therefore the procedure reproduces that solution. Finally, two different assignments differ in at least one free-variable coordinate, so they cannot give the same solution vector. If there are no nonzero rows, all coordinates are free and the same statements hold directly. \(\square\)
Worked Example: Two Free Variables
Describe all solutions of the echelon system with augmented matrix
The columns for \(x_1\) and \(x_3\) contain pivots, so \(x_1,x_3\) are pivot variables. The columns for \(x_2,x_4\) have no pivots, so assign \(x_2=s\) and \(x_4=t\), where \(s,t\in\mathbb{R}\). The nonzero rows represent
Start with the lower equation. Substituting \(x_4=t\) gives \(x_3+3t=2\), hence \(x_3=2-3t\). Substitute this and \(x_2=s,\ x_4=t\) into the first equation:
Thus the complete solution is
Check: In the second equation, \(2-3t+3t=2\). In the first, the left side becomes \((7-2s-4t)+2s-(2-3t)+t=5\). Both equations hold for every real \(s,t\).
Parameter Form as an Affine Expression
A parametric vector can also be written as one fixed solution plus parameter-dependent changes. This form makes the effect of changing each free variable easier to read. It does not introduce a new method for solving: it is a rearrangement of the expressions obtained by back substitution.
Proof. Each free-variable coordinate is either one parameter or, if it is the coordinate for \(t_j\), the value \(t_j\). For a pivot coordinate, its row expresses it as a constant plus a linear combination of variables to its right: divide the row by its nonzero pivot coefficient and move the other terms to the other side. Starting at the last nonzero row and proceeding upward, substitution shows that every pivot coordinate is also a constant plus a linear combination of \(t_1,\dots,t_k\). Consequently, coordinate by coordinate, the solution has the form \(\mathbf{x}_0+\sum_{j=1}^k t_j\mathbf{d}_j\): its constant coordinates form \(\mathbf{x}_0\), and the coefficient of \(t_j\) in each coordinate forms \(\mathbf{d}_j\). Setting all parameters to zero gives \(\mathbf{x}_0\); setting \(t_j=1\) and the rest to zero shows that the change from \(\mathbf{x}_0\) is exactly \(\mathbf{d}_j\). \(\square\)
For the preceding example, setting \(s=t=0\) gives \(\mathbf{x}_0=(7,0,2,0)^{\top}\). The coefficient of \(s\) is \((-2,1,0,0)^{\top}\), and the coefficient of \(t\) is \((-4,0,-3,1)^{\top}\). Therefore the same solution set is
Worked Examples with Different Pivot Patterns
Worked Example: A Pivot Coefficient Other Than One
Find all solutions of
The pivots are in columns 1 and 3. Thus \(x_2\) is free; set \(x_2=s\). The lower nonzero row gives \(4x_3=8\), so \(x_3=2\). The first row gives
Hence
Check: The second equation is \(4(2)=8\). In the first, \(2(\frac{13}{2}-\frac{s}{2})+s-3(2)=13-s+s-6=7\), as required.
Worked Example: Free Variables Before and Between Pivots
Describe all solutions of
The pivots are in columns 2 and 4. Therefore \(x_1\) and \(x_3\) are free. Set \(x_1=s\) and \(x_3=t\). The second row says \(x_4=3\). Substituting that into the first row gives \(x_2+2t-3=1\), so \(x_2=4-2t\). The full description is
Check: The second equation reads \(x_4=3\), which holds. The first becomes \((4-2t)+2t-3=1\). The zero row contributes \(0=0\), so it imposes no restriction on \(s\) or \(t\).
Common Mistakes and a Useful Check
Do not assign parameters to pivot variables. In the first worked example, a mistaken approach might set \(x_1=s\), \(x_2=t\), and \(x_4=u\) as if all three could be chosen freely. But the lower equation forces \(x_3=2-3u\), and the first then forces \(x_1=7-2t-4u\). The proposed arbitrary \(s\) is not valid unless it happens to equal \(7-2t-4u\). The correction is to parameterize only nonpivot columns, then solve for every pivot variable.
A second common error is to stop after finding the pivot variables without stating which parameter values are allowed. For a consistent echelon system over \(\mathbb{R}\), each free variable can take any real value: the bottom-up substitutions divide only by nonzero pivot coefficients. State the parameter domain explicitly, and verify every nonzero row after substitution. Checking just one chosen parameter value does not verify a formula intended to describe all solutions.
Parameters provide a compact record of the system’s freedom: the free coordinates can be chosen first, and the pivot coordinates respond to those choices. This description is useful whenever a system has infinitely many solutions, because it specifies the entire family rather than a single example from it. It also gives a direct way to test a proposed answer: substitute the parameterized coordinates into each equation and simplify.
Check Your Understanding
For each system, identify the free variables and give a complete parametric solution. Check the result in the equations.
- For \(\begin{bmatrix}1&-2&1&4\\0&0&1&-1\end{bmatrix}\), which variables are free, and what is the solution set?
- In an echelon system, a pivot row is \(3x_2-x_4=6\), and \(x_4\) is free. Solve for \(x_2\) using a parameter \(t\) for \(x_4\).
- Explain why a zero row with zero augmented entry gives no restriction on the parameters.
- Write \(\begin{bmatrix}2+s\\-1+3s\\s\end{bmatrix}\) as a fixed vector plus a scalar multiple of a direction vector.
- Why must two different assignments to the free variables produce different solution vectors?
