An Infinite-Solution Reduction Workshop

From a Reduced System to a Family of Solutions

When a consistent system has free variables, a single answer cannot describe all its solutions. Instead, we describe a family: each free variable can be chosen, and the pivot variables are then determined by those choices. If there are multiple free variables, each needs its own independent parameter. This workshop focuses on carrying out that process carefully, from row reduction through a check in the original equations.

Throughout, the variables and parameters are real. An augmented matrix records the coefficients of a linear system together with its right-hand-side column. Row operations on an augmented matrix correspond to swapping equations, multiplying an equation by a nonzero number, or adding a multiple of one equation to another.

The reason row reduction is valid here is that each elementary row operation preserves the system's solutions. This lets us solve a simpler equivalent system instead of repeatedly manipulating the original equations.

Proof. Swapping two rows only changes the order of the equations, so it does not change which variable values satisfy them. Multiplying a row by a nonzero scalar preserves its equation because an equation \(L=b\) is equivalent to \(cL=cb\) when \(c\ne0\). Replacing one row by itself plus a multiple of another also preserves the solutions: any solution of the original equations satisfies the new equation, and the original row can be recovered by subtracting that same multiple. Thus each elementary row operation preserves the solution set, and any sequence of such operations does too. \(\square\)

A General Parametrization Rule

Suppose a consistent augmented matrix has been reduced to reduced row echelon form (RREF). Assume it has \(n\) variable columns, with pivot columns \(p_1,\dots,p_r\). The remaining column indices are the free-variable indices. Because the system is consistent, there is no row of the form \([0\ \cdots\ 0\mid b]\) with \(b\ne0\). Each nonzero row therefore gives an equation for one pivot variable.

Proof. In each nonzero row \(i\), let \(x_{p_i}\) be its pivot variable. The RREF conditions say that its coefficient is \(1\), and that every other pivot column has coefficient \(0\). Thus the equation in that row has the form \(x_{p_i}+\sum_{j\text{ free}}a_{ij}x_j=b_i\). Assign a real number \(t_j\) to each free variable \(x_j\). The equation forces \(x_{p_i}=b_i-\sum_{j\text{ free}}a_{ij}t_j\), so it determines one value for this pivot variable. Doing this for every nonzero row produces a solution: every pivot equation holds by construction, every free variable has its assigned value, and any zero row is \(0=0\) because the system is consistent. Conversely, any solution must satisfy each pivot equation, so its pivot variables must equal these same expressions in its free-variable values. It is therefore obtained by assigning its actual free-variable values as the parameters. Those parameter values are unique because each parameter equals its corresponding free variable. Finally, if there is a free variable, changing its real parameter while holding the others fixed changes that coordinate of the solution, giving infinitely many distinct solutions. \(\square\)

In practice, the rule is simple: read the free columns first, name their variables with different parameters, and then substitute those expressions into the pivot equations. The signs matter. A final substitution into the original equations checks both the reduction and the parametrization.

Worked Examples

Reading and Checking a Parametric Answer

A parametric answer is complete only if it includes every free variable and all pivot variables are correctly expressed in terms of them. The parameters do not need to appear in every pivot equation: in the second example, changing \(s\) changes \(x_1\), while the pivot variables do not depend on \(s\). That is still a genuine free choice.

A useful check has two parts. First, substitute the parameter expressions into the reduced equations; this confirms the algebra used to solve for the pivot variables. Then substitute into the original equations, as in the examples. Since the row operations preserve the solution set, a correct solution of the reduced system is a solution of the original one. The direct substitution is also a practical way to catch arithmetic mistakes.

Common Pitfall: Tying Parameters Together

In the first example, both \(x_2\) and \(x_4\) are free. A tempting but incorrect shortcut is to set \(x_2=t\) and \(x_4=t\), using the same parameter for both. This only describes the special choices where those two coordinates happen to be equal. For instance, \(x_2=1\) and \(x_4=0\) are allowed choices in the first example, but they cannot be represented by the tied assignment.

The correction is to use separate parameters: \(x_2=s\) and \(x_4=t\). They can take any real values independently. Another frequent error is to lose a sign while isolating a pivot variable. In the first example, \(x_1+2s+\frac13t=\frac{11}{3}\) gives \(x_1=\frac{11}{3}-2s-\frac13t\), not \(\frac{11}{3}+2s+\frac13t\). Substituting that incorrect expression into the left side gives \(\frac{11}{3}+4s+\frac23t\), which does not equal \(\frac{11}{3}\) for arbitrary \(s,t\). Solving and then substituting back is a reliable way to catch this mistake.