What a One-Parameter Description Says
A system of linear equations may have infinitely many solutions. When exactly one variable can be chosen freely and the other variables are then determined, we can describe all its solutions using one real parameter. The description must do two jobs: every parameter value must give a solution, and every solution of the system must arise from some parameter value. Checking only the first job is not enough to establish that we have described the entire solution set.
In this tutorial, a parameter is a real number used to generate solutions. We will call a variable free when it can be assigned any real value while the other variables are determined by the equations. This description applies to systems with exactly one free variable. It does not apply to every system with infinitely many solutions: for instance, \(0x+0y=0\) allows both \(x\) and \(y\) to vary independently, so one parameter cannot describe all points in \(\mathbb{R}^2\).
Variables in a solution are kept in their original order. For example, a solution to a system in \(x,y,z\) is written as the column \(\begin{bmatrix}x\\y\\z\end{bmatrix}\), not as an unordered collection of values. If we choose \(z=t\) and find \(x=3-t,\ y=3\), the ordered solution is \(\begin{bmatrix}3-t\\3\\t\end{bmatrix}\), where \(t\) may be any real number.
A useful way to organize a description is to separate a fixed part from the part that changes with \(t\). For a system written as \(\mathbf{A}\mathbf{x}=\mathbf{b}\), this form is \(\mathbf{x}=\mathbf{p}+t\mathbf{v}\), where \(\mathbf{p}\) is a fixed column and \(\mathbf{v}\) records how the coordinates change. This notation is convenient, but it does not replace solving the equations or checking the result.
When Does the Formula Give Solutions?
Here \(\mathbf{A}\) is the matrix of coefficients, \(\mathbf{x}\) is the column of variables, and \(\mathbf{b}\) is the column of constants. The equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is simply a compact way to write the original system. The next result gives a direct test for a proposed formula \(\mathbf{p}+t\mathbf{v}\).
Proof. Assume first that \(\mathbf{A}\mathbf{p}=\mathbf{b}\) and \(\mathbf{A}\mathbf{v}=\mathbf{0}\). For any real \(t\), distributivity of matrix multiplication gives \(\mathbf{A}(\mathbf{p}+t\mathbf{v})=\mathbf{A}\mathbf{p}+t\mathbf{A}\mathbf{v}=\mathbf{b}+t\mathbf{0}=\mathbf{b}\). Thus every parameter value gives a solution. Conversely, suppose \(\mathbf{p}+t\mathbf{v}\) solves the system for every real \(t\). Taking \(t=0\) gives \(\mathbf{A}\mathbf{p}=\mathbf{b}\). Taking \(t=1\) gives \(\mathbf{A}(\mathbf{p}+\mathbf{v})=\mathbf{b}\), so \(\mathbf{A}\mathbf{p}+\mathbf{A}\mathbf{v}=\mathbf{b}\). Since \(\mathbf{A}\mathbf{p}=\mathbf{b}\), subtraction yields \(\mathbf{A}\mathbf{v}=\mathbf{0}\). Both conditions follow. \(\square\)
This result checks that all generated columns satisfy the equations. To prove completeness as well, we must also show that the equations force every solution to have the stated form. In practice, we do this by choosing the free variable and solving the remaining equations, then checking that the resulting formulas account for all possibilities.
Solving for One Parameter
Worked Example: A Line of Solutions to Two Equations
Describe all solutions of the system as an ordered column depending on one real parameter, and verify the description in the original equations.
Add the left sides and add the right sides. The \(y\)-terms cancel, leaving \(3x+3z=9\). Dividing by \(3\) gives \(x+z=3\), so \(x=3-z\). Substitute this expression into the first equation: \((3-z)+y+z=6\), hence \(y=3\). No equation restricts \(z\), so choose \(z=t\), where \(t\) is any real number. In the original variable order, the candidate description is
Check every parameter value: in the first original equation, \((3-t)+3+t=6\). In the second, \(2(3-t)-3+2t=6-2t-3+2t=3\). Both equations hold for every real \(t\). For completeness, any solution must satisfy \(x+z=3\), which forces \(x=3-z\), and the first equation then forces \(y=3\). Setting \(t=z\) therefore recovers that solution. The formula describes all solutions, not merely some of them.
The order of the coordinates matters: the parameter was chosen as \(z\), but \(z\) remains in the third position of the solution column. A parameter is a label for the value we are free to choose; it is not an extra coordinate to append to the answer.
Worked Example: A Different Variable as the Parameter
Find and check a complete one-parameter description of the solutions of
Add the equations: \((2x+y-z)+(x-y+z)=5+1\), so \(3x=6\) and \(x=2\). Substitute \(x=2\) into the second equation: \(2-y+z=1\), which gives \(y=1+z\). Choose \(z=t\), with \(t\in\mathbb{R}\). The ordered solution column is
Check: the left side of the first original equation becomes \(2(2)+(1+t)-t=5\). The left side of the second becomes \(2-(1+t)+t=1\). Thus every real \(t\) gives a solution. To see that none are missed, any solution has \(x=2\) from adding the equations, and then its second equation requires \(y=1+z\). Taking \(t=z\) gives exactly the displayed column.
In both examples, the chosen variable takes every real value, and the equations uniquely determine the other variables from it. Those facts establish completeness. If we had only substituted the proposed column and found that it worked, we would have verified the first requirement but not yet ruled out other solutions.
Completeness and Uniqueness of the Parameter
Proof. First take any real \(t\). The formulas were obtained by solving the original equations, so substituting \(x_k=t\) and the corresponding determined values of the other variables satisfies those equations. Thus every listed column is a solution. Now take any solution of the system. Its \(k\)th coordinate has some real value; call it \(t\). Because all other variables are uniquely determined by \(x_k\), they must equal the formulas obtained by setting \(x_k=t\). Therefore this solution occurs in the displayed family. Finally, if two parameter values \(t_1,t_2\) gave the same solution column, their \(k\)th coordinates would agree. Those coordinates are \(t_1\) and \(t_2\), so \(t_1=t_2\). Hence every solution is represented, and no solution is repeated. \(\square\)
The condition that the selected variable itself equals \(t\) makes the parameter easy to recover from the solution: read its \(k\)th coordinate. The proposition relies on having exactly one freely chosen variable. If a system has more independent choices, fixing one variable does not in general determine the others; such a solution set needs additional parameters.
Worked Example: A Redundant Equation and a Complete Description
Describe all solutions of the system below, and check both that the formulas work and that they include every solution.
Divide the first equation by \(2\) to get \(x+y=3\). The second equation gives the same condition when divided by \(4\). Choose \(y=t\), for any \(t\in\mathbb{R}\); then \(x=3-t\). In the original order, the candidate solution is
Check: the first left side is \(2(3-t)+2t=6\), and the second is \(4(3-t)+4t=12\). Every real \(t\) works. Conversely, any solution of the first equation satisfies \(x+y=3\), so once its \(y\)-coordinate is called \(t\), its \(x\)-coordinate must be \(3-t\). The second equation imposes no further restriction. Thus the description includes every solution exactly once.
A Common Mistake: Losing a Parameter Term
Return to the first system, \(x+y+z=6\) and \(2x-y+2z=3\). A mistaken addition may produce \(3x+3z=9\), followed by the incorrect conclusion \(x=3\), as if the \(3z\) term were absent. This might lead to the proposed column \(\begin{bmatrix}3\\3-t\\t\end{bmatrix}\). It satisfies the first equation, since \(3+(3-t)+t=6\), but the second left side is \(2(3)-(3-t)+2t=3+3t\), which equals \(3\) only when \(t=0\). The proposed formula therefore fails for general \(t\).
The correction is to keep the parameter term when solving: \(3x+3z=9\) gives \(x+z=3\), hence \(x=3-z\). Substituting into the first equation gives \(y=3\). With \(z=t\), the corrected column is \(\begin{bmatrix}3-t\\3\\t\end{bmatrix}\), and direct substitution gives \(6\) and \(3\) on the left sides of the two equations for every real \(t\). Checking all original equations is an effective way to catch a lost term or a coordinate placed in the wrong position.
There are two further checks to keep separate. First, verify that every parameter value works by substituting into every original equation, not just an equation used during the solving. Second, verify completeness by explaining why an arbitrary solution must have the stated form. A system with no solutions has no such description; a system with a single solution does not need a free parameter. Systems with more than one free variable require more than one parameter, rather than forcing all variation into a single one.
Check Your Understanding
For each proposed description, consider both whether every parameter value works and whether all solutions are included.
- For \(x+y+z=4\) and \(x-y+z=2\), choose \(z=t\). Find the ordered column and verify it in both equations.
- Describe all solutions of \(2x-y=3\) and \(4x-2y=6\) using \(y=t\). Check both original equations.
- A proposed description is \(\begin{bmatrix}1+t\\2\\t\end{bmatrix}\) for a system in \(x,y,z\). Which coordinate identifies \(t\), and why must the displayed coordinate order be preserved?
- State what must be checked to show that a one-parameter description gives every solution, not just solutions.
- Explain why one parameter cannot describe all solutions of \(0x+0y=0\) in \(\mathbb{R}^2\).
