Reading Solutions From Reduced Forms

From a Reduced Matrix to a Solution Set

A reduced augmented matrix is more than a compact record of row reduction: each of its rows is an equation, and the locations of its pivots show which variables those equations determine. Reading the matrix carefully turns it into a description of every solution, not just one possible answer. The central task is to distinguish the variables determined by the equations from those that may be chosen freely, then express the determined variables in terms of those choices.

Throughout, the systems have real coefficients and variables $x_1,\dots,x_n$. The augmented matrix places the coefficients in the first $n$ columns and the right-hand sides in the last column. A row $[a_1\ \cdots\ a_n\mid b]$ represents the equation $a_1x_1+\cdots+a_nx_n=b$. The last column records constants; it is not a variable column. Row operations preserve the solution set, so a reduced augmented matrix can be used in place of the original system.

Definition: In a reduced augmented matrix, a pivot variable is the variable whose coefficient column contains a pivot, and a free variable is a variable whose coefficient column contains no pivot. To write a parametric solution, assign a parameter to each free variable and express each pivot variable in terms of those parameters.

For example, if columns 1 and 3 contain pivots, then $x_1$ and $x_3$ are pivot variables, while any other variable columns are free. The order of the columns matters: column 3 corresponds to $x_3$, regardless of which row contains its pivot. In reduced row echelon form (RREF), each pivot is $1$, and every other entry in its column is $0$. Thus a nonzero row with a pivot in column $j$ gives an equation in which $x_j$ is isolated on the left.

A Complete Parametrization Rule

The process works because each nonzero row of an RREF coefficient part determines one pivot variable, while no equation imposes an additional restriction on the free variables in a consistent system. The following result gives the rule precisely. A system is called consistent if it has at least one solution.

Theorem: Suppose the augmented matrix of a consistent system in $n$ real variables is in RREF. Let its pivot-variable columns be $p_1,\dots,p_r$, and let its free-variable columns be $f_1,\dots,f_q$, where $q=n-r$. If the right-hand side of row $i$ is $b_i$, and the entry in row $i$, column $f_j$ is $a_{ij}$, then every solution, and no other vector, is obtained by choosing arbitrary real numbers $t_1,\dots,t_q$, setting $x_{f_j}=t_j$, and setting $$ x_{p_i}=b_i-\sum_{j=1}^{q}a_{ij}t_j \quad\text{for }i=1,\dots,r. $$ If $q=0$, the sum is empty and equals $0$.

Proof. Because the system is consistent, it has no row whose coefficient entries are all zero and whose right-hand side is nonzero. Each nonzero row therefore has a pivot. In RREF, row $i$ has coefficient $1$ in pivot column $p_i$, and coefficient $0$ in every other pivot column. Its equation is consequently $x_{p_i}+\sum_{j=1}^{q}a_{ij}x_{f_j}=b_i$. Choose any real numbers $t_1,\dots,t_q$, assign $x_{f_j}=t_j$, and define the pivot variables by the displayed formula. Substitution into row $i$ gives $b_i-\sum_j a_{ij}t_j+\sum_j a_{ij}t_j=b_i$, so every nonzero row is satisfied. The zero rows are $0=0$, so they are satisfied as well. Conversely, any solution has some values $t_j$ in its free-variable coordinates. Its row $i$ must satisfy $x_{p_i}+\sum_j a_{ij}t_j=b_i$, which forces $x_{p_i}=b_i-\sum_j a_{ij}t_j$. Thus the formula includes every solution and only solutions. If $q=0$, there are no free variables to assign and the same equations determine all pivot variables; if $r=0$, there are no pivot equations and all variables are free. $\square$

In practice, it is often useful to write the result in vector form. The vector obtained by setting every parameter to zero is a particular solution. Each parameter contributes a direction vector that records how the solution changes when that parameter changes. This format makes the number of parameters and the dependence between coordinates visible.

Proposition: Under the theorem's hypotheses, the parameter tuple determines exactly one solution, and different parameter tuples give different solutions.

Proof. The theorem's formulas assign one value to each pivot variable after a parameter tuple has been chosen, so that tuple determines one solution. If two parameter tuples give the same solution vector, then their entries must agree in every free-variable coordinate. Since $x_{f_j}=t_j$, the two tuples agree in every position $j$. Therefore distinct tuples cannot give the same solution. When there are no free variables, there is exactly one empty parameter tuple, and the same formulas give exactly one solution vector. $\square$

Worked Examples

Worked Example: One Free Variable

Read the solution set from this reduced augmented matrix:

\begin{bmatrix} 1&0&2&\mid&5\\ 0&1&-1&\mid&-1\\ 0&0&0&\mid&0 \end{bmatrix}.

The first two coefficient columns contain pivots, so $x_1$ and $x_2$ are pivot variables. Column 3 has no pivot, so $x_3$ is free. Reading the first row gives $x_1+2x_3=5$, and the second gives $x_2-x_3=-1$. Set $x_3=t$, where $t\in\mathbb{R}$. Solving the two equations gives

x_1=5-2t,\qquad x_2=-1+t,\qquad x_3=t.

Thus the complete solution set is

\mathbf{x} = \begin{bmatrix}5\\-1\\0\end{bmatrix} +t\begin{bmatrix}-2\\1\\1\end{bmatrix}, \qquad t\in\mathbb{R}.

Check: Substituting the coordinates into the first equation gives $(5-2t)+2t=5$. Substituting into the second gives $(-1+t)-t=-1$, and the last row gives $0=0$. Every real $t$ therefore produces a solution. Conversely, any solution must have $x_3=t$ for some real $t$, after which the first two equations force the displayed values of $x_1$ and $x_2$.

Worked Example: Two Free Variables

Write the complete solution set for the system represented by

\begin{bmatrix} 1&-2&0&3&\mid&4\\ 0&0&1&-1&\mid&2\\ 0&0&0&0&\mid&0 \end{bmatrix}.

There are pivots in columns 1 and 3, so $x_1$ and $x_3$ are pivot variables. Columns 2 and 4 have no pivots, so assign $x_2=s$ and $x_4=t$, with $s,t\in\mathbb{R}$. The first row gives $x_1-2x_2+3x_4=4$, so $x_1=4+2s-3t$. The second row gives $x_3-x_4=2$, so $x_3=2+t$. Therefore

\mathbf{x} = \begin{bmatrix}4\\0\\2\\0\end{bmatrix} +s\begin{bmatrix}2\\1\\0\\0\end{bmatrix} +t\begin{bmatrix}-3\\0\\1\\1\end{bmatrix}, \qquad s,t\in\mathbb{R}.

Check: For the first row, the left side is $(4+2s-3t)-2s+3t=4$. For the second row, it is $(2+t)-t=2$, and the last row is $0=0$. The free coordinates of the vector are $s$ and $t$, so no solution is omitted and no two different parameter pairs produce the same vector.

Worked Example: Reading a Contradictory Row

Determine whether the following reduced augmented matrix describes any solutions:

\begin{bmatrix} 1&0&-2&\mid&1\\ 0&1&3&\mid&4\\ 0&0&0&\mid&2 \end{bmatrix}.

The first row represents $x_1-2x_3=1$, and the second represents $x_2+3x_3=4$. The last row represents $0=2$, which is false for every choice of the variables. Hence this system has no solutions. There is no parameter choice that can satisfy all three rows, so the first two equations alone must not be reported as the complete solution set.

Check: The coefficient part of the last row gives $0x_1+0x_2+0x_3=0$, while its augmented entry requires that same left side to equal $2$. Since $0\ne2$, the contradiction cannot be removed by choosing values for $x_1,x_2,x_3$.

Common Pitfall: Letting a Pivot Variable Be Free

A variable is not free merely because it appears in a row with other variables. Its status comes from its column: a pivot in that column makes the variable a pivot variable. For instance, consider

\begin{bmatrix} 1&2&\mid&3\\ 0&1&\mid&1 \end{bmatrix}.

A mistaken reading might assign $x_1=s$ and then use the second row to set $x_2=1$, producing $\mathbf{x}=(s,1)^{\top}$. But this does not satisfy the first row for arbitrary $s$: it requires $s+2=3$, hence $s=1$. The correction is to notice that both columns are pivot columns. The second row gives $x_2=1$, and substituting into the first gives $x_1+2(1)=3$, so $x_1=1$. The only solution is $\mathbf{x}=(1,1)^{\top}$. Checking both equations gives $1+2(1)=3$ and $1=1$.

Another common source of errors is a sign change when solving for a pivot variable. From $x_1-2x_2+3x_4=4$, the correct rearrangement is $x_1=4+2x_2-3x_4$; moving terms across the equals sign changes their signs. Substitution into the original row is a quick way to check the signs.

A Reliable Reading Sequence

For a consistent reduced matrix, use the following order so that every variable receives exactly one role and every equation is used.

Match columns to variables.
Read the first $n$ columns as the coefficients of $x_1,\dots,x_n$. Treat the augmented column only as the right-hand side.

Mark pivot columns.
The variables in those columns are pivot variables. Variables in the other coefficient columns are free.

Check every row.
Translate each nonzero row into an equation. A row of zeros must have right-hand side zero for a solution to exist.

Assign parameters and solve.
Give each free variable its own real parameter, then rearrange the pivot equations to determine the pivot variables.

Verify the description.
Substitute the proposed coordinates into every reduced equation and confirm that the parameters occur in exactly the free-variable coordinates.

The last check matters even when the arithmetic looks familiar. A correct parametric expression must satisfy every row, including zero rows, and must describe all allowable choices of the free variables. If a row instead says $0=c$ for $c\ne0$, the system has no solution and no parametric solution set to write.

Check Your Understanding

For each reduced augmented matrix, identify the pivot and free variables before writing or interpreting its solution set.

For $\begin{bmatrix}1&0&-3&\mid&2\\0&1&4&\mid&-1\end{bmatrix}$, write the complete solution set using a parameter.
For $\begin{bmatrix}1&2&0&-1&\mid&3\\0&0&1&5&\mid&2\end{bmatrix}$, identify the free variables and write the solution in vector form.
Explain why $\begin{bmatrix}1&0&\mid&4\\0&0&\mid&-1\end{bmatrix}$ has no solutions.
In a consistent RREF system, why does assigning values to all free variables determine the pivot variables?
How can you check by substitution that a proposed parametric solution satisfies the reduced system?

Reading Solutions From Reduced Forms

What You'll Learn

From a Reduced Matrix to a Solution Set

A Complete Parametrization Rule