Pivots, Free Variables and All Solutions, linear algebra I lesson

Two equations in three unknowns, such as $x + z = 4$ and $y + z = 3$ , can never pin down a single point. The triples $(4, 3, 0)$ , $(3, 2, 1)$ and $(2, 1, 2)$ all satisfy them, and so do infinitely many others. Listing a few solutions is not the same as knowing all of them. This lesson shows how the pivots of an echelon form decide whether a system has any solution, which unknowns can be chosen freely, and how to write every solution in one formula. Throughout, a tuple such as $(4, 3, 0)$ stands for a column vector.

Key terms

Pivot

Definition Let $A$ be a matrix and let $U$ be a row echelon form of $A$ , that is, a matrix in row echelon form that is row equivalent to $A$ . The leading entries of the nonzero rows of $U$ are pivots, and their locations are pivot positions. Every echelon form of $A$ has its pivots in the same positions, so the pivot positions belong to $A$ itself. A pivot column of $A$ is a column that contains a pivot position. For a system with augmented matrix $[\,A \mid \mathbf{b}\,]$ , the unknowns whose columns are pivot columns of $A$ are the pivot variables, also called basic variables.

In plain words A pivot is the first nonzero entry of a row once the staircase is built. Each pivot column marks an unknown that one of the equations can be solved for.

Example The system $x + z = 4$ , $y + z = 3$ has the augmented matrix

\left[\begin{array}{ccc|c} 1 & 0 & 1 & 4 \\ 0 & 1 & 1 & 3 \end{array}\right],

which is already in row echelon form. Its pivots are the leading entries $1$ in columns 1 and 2. The pivot columns are the $x$ and $y$ columns, so $x$ and $y$ are pivot variables, and the $z$ column contains no pivot.

Free variable

Definition In a consistent linear system, a free variable is an unknown whose column in the coefficient matrix is not a pivot column. Each choice of real values for the free variables determines exactly one value for each pivot variable, found by back-substitution, and every solution arises from exactly one such choice. A consistent system with $n$ unknowns and $r$ pivots in its coefficient matrix therefore has $n - r$ free variables. It has exactly one solution when $n - r = 0$ and infinitely many when $n - r > 0$ .

In plain words A free variable is a setting you may choose at will. It does not drop out of the problem, because it is still an entry of every solution, and the pivot variables are calculated from the values you choose.

Example In the system $x + z = 4$ , $y + z = 3$ the unknown $z$ is free. Choosing $z = 1$ forces $x = 3$ and $y = 2$ , and choosing $z = 2$ forces $x = 2$ and $y = 1$ . Once $z$ is fixed, the pivot variables are determined and cannot be chosen separately.

Reduced row echelon form

Definition A matrix is in reduced row echelon form (RREF) if it is in row echelon form and, in addition,

every pivot equals $1$ , and
each pivot is the only nonzero entry in its column.

Every matrix is row equivalent to exactly one matrix in reduced row echelon form, so any two sequences of row operations that reach RREF end at the same matrix. Ordinary row echelon forms lack this uniqueness. Scaling a pivot row by a number other than $0$ and $1$ gives another one, so a nonzero matrix has infinitely many of them.

In plain words Scale each pivot to $1$ , then clear the entries above it as well as below it. In a consistent system each nonzero equation then contains exactly one pivot variable, and every other unknown in it is free.

Example Start from the echelon matrix

\left[\begin{array}{ccc|c} 1 & 1 & 2 & 7 \\ 0 & 2 & 2 & 6 \end{array}\right].

The scaling $R_2 \to \frac{1}{2}R_2$ and then the replacement $R_1 \to R_1 - R_2$ give

\begin{aligned} &\left[\begin{array}{ccc|c} 1 & 1 & 2 & 7 \\ 0 & 1 & 1 & 3 \end{array}\right] \\[0.4em] \to\ &\left[\begin{array}{ccc|c} 1 & 0 & 1 & 4 \\ 0 & 1 & 1 & 3 \end{array}\right]. \end{aligned}

The pivot columns now match the columns of $I_2$ , while the third column now holds the entries $1$ and $1$ . A column without a pivot does not have to be zero in RREF. The result is the augmented matrix of the system $x + z = 4$ , $y + z = 3$ .

Inconsistent row

Definition An inconsistent row of an augmented matrix is a row of the form

\left[\begin{array}{ccc|c} 0 & \cdots & 0 & c \end{array}\right], \qquad c \ne 0,

which stands for the equation $0 = c$ . A linear system is inconsistent exactly when an echelon form of its augmented matrix has an inconsistent row, that is, when the last column of the augmented matrix is a pivot column. A row that is zero everywhere, constant included, stands for $0 = 0$ and imposes no condition.

In plain words Before looking for free variables, scan the echelon form for a row that says zero equals a nonzero number. One such row means there is no solution at all. A row that says zero equals zero is harmless and can be set aside.

Example For the system $x + y = 3$ , $2x + 2y = 7$ , the step $R_2 \to R_2 - 2R_1$ gives

\left[\begin{array}{cc|c} 1 & 1 & 3 \\ 2 & 2 & 7 \end{array}\right] \to \left[\begin{array}{cc|c} 1 & 1 & 3 \\ 0 & 0 & 1 \end{array}\right],

and the second row says $0 = 1$ , so the system is inconsistent. With $6$ in place of $7$ the same step produces a zero row, which says $0 = 0$ . That system is consistent, and its one remaining equation $x + y = 3$ leaves $y$ free. The coefficient matrix is the same in both cases, so here the right-hand side, not the coefficients, decides consistency.

Parametric solution

Definition A parametric solution of a consistent system describes its entire solution set in terms of parameters, one for each free variable. Set each free variable equal to its own parameter, which may be any real number, and solve the reduced equations for the pivot variables. The result can be written in vector form

\mathbf{x} = \mathbf{p} + t_1\mathbf{v}_1 + \cdots + t_k\mathbf{v}_k

with real parameters $t_1, \dots, t_k$ , where $k$ is the number of free variables and $\mathbf{p}, \mathbf{v}_1, \dots, \mathbf{v}_k$ are fixed vectors. Every choice of the parameters gives a solution, and the description is complete because every solution has some value of each free variable, after which the equations force the pivot variables.

In plain words With at least one free variable, one formula with $k$ adjustable numbers replaces an endless list of solutions. Every setting of those numbers gives a solution, and every solution comes from some setting.

Example The reduced system $x_1 + 2x_2 - x_4 = 3$ , $x_3 + 2x_4 = -2$ has pivot columns 1 and 3, so $x_2$ and $x_4$ are free. Set $x_2 = s$ and $x_4 = t$ . Then $x_1 = 3 - 2s + t$ and $x_3 = -2 - 2t$ , so

\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ -2 \\ 0 \end{bmatrix} + s\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t\begin{bmatrix} 1 \\ 0 \\ -2 \\ 1 \end{bmatrix}.

Substituting back gives $(3 - 2s + t) + 2s - t = 3$ and $(-2 - 2t) + 2t = -2$ for all real $s$ and $t$ . Once $s$ and $t$ are chosen, the pivot variables cannot be chosen too. With $s = t = 0$ , for example, the equations force $x_1 = 3$ and $x_3 = -2$ .

Particular and homogeneous solutions

Definition A particular solution of $A\mathbf{x} = \mathbf{b}$ is any one fixed solution $\mathbf{p}$ . The system $A\mathbf{x} = \mathbf{0}$ with the same coefficient matrix is the associated homogeneous system, and its solutions are the homogeneous solutions. If $\mathbf{p}$ is a particular solution of a consistent system $A\mathbf{x} = \mathbf{b}$ , then the solution set of that system is exactly

\{\, \mathbf{p} + \mathbf{h} \mid A\mathbf{h} = \mathbf{0} \,\}.

Both inclusions follow from linearity. If $A\mathbf{x} = \mathbf{b}$ , then $A(\mathbf{x} - \mathbf{p}) = \mathbf{b} - \mathbf{b} = \mathbf{0}$ , and if $A\mathbf{h} = \mathbf{0}$ , then $A(\mathbf{p} + \mathbf{h}) = \mathbf{b} + \mathbf{0} = \mathbf{b}$ . In a parametric solution $\mathbf{x} = \mathbf{p} + t_1\mathbf{v}_1 + \cdots + t_k\mathbf{v}_k$ , the vector $\mathbf{p}$ is a particular solution and the sums $t_1\mathbf{v}_1 + \cdots + t_k\mathbf{v}_k$ run through all homogeneous solutions.

In plain words A homogeneous solution is a change of input that the matrix does not notice. Start at any one solution and add every such change, and you reach every solution. An inconsistent system has no particular solution to start from.

Example For $x + z = 4$ , $y + z = 3$ every solution is $(4 - t, 3 - t, t) = (4, 3, 0) + t(-1, -1, 1)$ . Here $(4, 3, 0)$ is a particular solution, and

\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix},

so every multiple of $(-1, -1, 1)$ is a homogeneous solution. There are no others, because $x + z = 0$ and $y + z = 0$ force $x = y = -z$ . Changing the right-hand side to $(1, -2)$ changes only the particular solution, and the solutions become $(1, -2, 0) + t(-1, -1, 1)$ . The homogeneous solutions always include $\mathbf{0}$ , while the solution set of $A\mathbf{x} = \mathbf{b}$ with $\mathbf{b} \ne \mathbf{0}$ never does.

Check yourself

Each augmented matrix below is in row echelon form, for a system in the unknowns $x_1, x_2, x_3$ . Decide whether the system is consistent, and if it is, name the free variables.

\begin{gathered} \text{(a)}\ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 0 & 0 & 3 & 6 \end{array}\right] \\[0.4em] \text{(b)}\ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 0 & 0 & 0 & 6 \end{array}\right] \\[0.4em] \text{(c)}\ \left[\begin{array}{ccc|c} 1 & 2 & -1 & 4 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{gathered}

Show the answer

System (a) is consistent. Its pivots are in columns 1 and 3, so $x_2$ is the only free variable. The second row gives $x_3 = 2$ , and then $x_1 = 6 - 2x_2$ .

System (b) is inconsistent, because its second row says $0 = 6$ .

System (c) is consistent, because its zero row only says $0 = 0$ . The only pivot is in column 1, so $x_2$ and $x_3$ are both free, and $x_1 = 4 - 2x_2 + x_3$ . Systems (b) and (c) differ only in the last entry of the second row, and that entry alone decides consistency.

Reduce the augmented matrix below to RREF and write every solution in parametric vector form. Check your answer in the original equations.

\left[\begin{array}{cccc|c} 1 & -2 & 1 & 3 & 4 \\ 2 & -4 & 3 & 8 & 11 \end{array}\right]

Show the answer

Apply $R_2 \to R_2 - 2R_1$ , then $R_1 \to R_1 - R_2$ ,

\begin{aligned} &\left[\begin{array}{cccc|c} 1 & -2 & 1 & 3 & 4 \\ 0 & 0 & 1 & 2 & 3 \end{array}\right] \\[0.4em] \to\ &\left[\begin{array}{cccc|c} 1 & -2 & 0 & 1 & 1 \\ 0 & 0 & 1 & 2 & 3 \end{array}\right]. \end{aligned}

The pivot columns are 1 and 3, so $x_2$ and $x_4$ are free. Set $x_2 = s$ and $x_4 = t$ . Then $x_1 = 1 + 2s - t$ and $x_3 = 3 - 2t$ , and

\mathbf{x} = \begin{bmatrix} 1 \\ 0 \\ 3 \\ 0 \end{bmatrix} + s\begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + t\begin{bmatrix} -1 \\ 0 \\ -2 \\ 1 \end{bmatrix}.

In the original equations the first left side becomes $(1 + 2s - t) - 2s + (3 - 2t) + 3t$ , which equals $4$ , and the second becomes $2(1 + 2s - t) - 4s + 3(3 - 2t) + 8t$ , which equals $11$ , for all real $s$ and $t$ .

Two students solve $x + z = 4$ , $y + z = 3$ . One writes the solutions as $(4, 3, 0) + t(-1, -1, 1)$ , and the other writes them as $(2, 1, 2) + u(1, 1, -1)$ . Which student is right?

Show the answer

Both are right. The vector $(2, 1, 2)$ is a solution, since $2 + 2 = 4$ and $1 + 2 = 3$ , so it can serve as the particular solution. The vector $(1, 1, -1)$ is $-1$ times $(-1, -1, 1)$ , so its multiples are the same homogeneous solutions. The two formulas describe the same line, and the point with parameter $u$ in the second formula is the point with $t = 2 - u$ in the first. A particular solution is any one solution, not a special one.

Pivots, Free Variables and All Solutions

Key terms

Pivot

Free variable

Reduced row echelon form

Inconsistent row

Parametric solution

Particular and homogeneous solutions

Check yourself

How this lesson connects

Builds on

Used later in