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Math 211 Activities Workbook: Adapted from: Understanding Linear Algebra by David Austin

David Austin, Erica Bernstein

Section 5.3 Pivot positions, free and basic variables

Remember that a leading entry in a reduced row echelon matrix is the leftmost nonzero entry in a row of the matrix. As we’ll see, the positions of these leading entries encode a lot of information about the solution space of the corresponding linear system. For this reason, we make the following definition.

Definition 5.3.1.

A pivot position in a matrix \(A\) is the position of a leading entry in the reduced row echelon matrix of \(A\text{.}\)
For instance, in this reduced row echelon matrix, the pivot positions are indicated in bold:
\begin{equation*} \begin{bmatrix} {\mathbf 1} \amp \gray{0} \amp \gray{*} \amp \gray{0} \\ \gray{0} \amp {\mathbf 1} \amp \gray{*} \amp \gray{0} \\ \gray{0} \amp \gray{0} \amp \gray{0} \amp {\mathbf 1} \\ \gray{0} \amp \gray{0} \amp \gray{0} \amp \gray{0} \\ \end{bmatrix}. \end{equation*}
We can refer to pivot positions by their row and column number saying, for instance, that there is a pivot position in the second row and fourth column.

Example 5.3.2. Describing the solution space from a reduced row echelon matrix.

  1. Consider the reduced row echelon matrix
    \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 2 \amp -1 \\ 0 \amp 1 \amp 1 \amp 2 \\ \end{array} \right] \end{equation*}
    and its corresponding linear system as
    \begin{equation*} \begin{alignedat}{4} x \amp \amp \amp {}+{} \amp 2z \amp {}={} \amp -1 \\ \amp \amp y \amp {}+{} \amp z \amp {}={} \amp 2. \\ \end{alignedat} \end{equation*}
    Let’s rewrite the equations as
    \begin{equation*} \begin{alignedat}{2} x \amp {}={} \amp -1 -2z\\ y \amp {}={} \amp 2-z. \\ \end{alignedat} \end{equation*}
    From this description, it is clear that we obtain a solution for any value of the variable \(z\text{.}\) For instance, if \(z=2\text{,}\) then \(x = -5\) and \(y=0\) so that \((x,y,z) = (-5,0,2)\) is a solution. Similarly, if \(z=0\text{,}\) we see that \((x,y,z) = (-1,2,0)\) is also a solution.
    Because there is no restriction on the value of \(z\text{,}\) we call it a free variable, and note that the linear system has infinitely many solutions. The variables \(x\) and \(y\) are called basic variables as they are determined once we make a choice of the free variable.
    We will call this description of the solution space, in which the basic variables are written in terms of the free variables, a parametric description of the solution space.
  2. Consider the matrix
    \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 0 \amp 4 \\ 0 \amp 1 \amp 0 \amp -3 \\ 0 \amp 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right]. \end{equation*}
    The last equation gives
    \begin{equation*} 0x +0y+0z = 0\text{,} \end{equation*}
    which is true for any \((x,y,z)\text{.}\) We may safely ignore this equation since it does not impose a restriction on \((x,y,z)\text{.}\) We then see that there is a unique solution \((x,y,z) = (4,-3,1)\text{.}\)
  3. Consider the matrix
    \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 2 \amp 0 \\ 0 \amp 1 \amp -1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ \end{array} \right]. \end{equation*}
    Beginning with the last equation, we see that
    \begin{equation*} 0x +0y+0z = 0 = 1\text{,} \end{equation*}
    which is not true for any \((x,y,z)\text{.}\) There is no solution to this particular equation and therefore no solution to the system of equations.

Activity 5.3.1.

  1. Shown below are three augmented matrices in reduced row echelon form.
    \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 0 \amp 3 \\ 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}
    \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 2 \amp 3 \\ 0 \amp 1 \amp -1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}
    \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 2 \amp 0 \\ 0 \amp 1 \amp -1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}
    For each matrix, identify the pivot positions and determine if the corresponding linear system is consistent. Explain how the location of the pivots determines whether the system is consistent or inconsistent.
  2. Each of the augmented matrices above has a row in which each entry is zero. What, if anything, does the presence of such a row tell us about the consistency of the corresponding linear system?
  3. Give an example of a \(3\times5\) augmented matrix in reduced row echelon form that represents a consistent system. Indicate the pivot positions in your matrix and explain why these pivot positions guarantee a consistent system.
  4. Give an example of a \(3\times5\) augmented matrix in reduced row echelon form that represents an inconsistent system. Indicate the pivot positions in your matrix and explain why these pivot positions guarantee an inconsistent system.
  5. Suppose we have a linear system for which the coefficient matrix has the following reduced row echelon form.
    \begin{equation*} \left[ \begin{array}{rrrrr} 1 \amp 0 \amp 0 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \amp 0 \amp 2 \\ 0 \amp 0 \amp 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \amp -3 \\ \end{array} \right] \end{equation*}
    What can you say about the consistency of the linear system?
Solution.
  1. The pivot positions are indicated below.
    \begin{equation*} \left[ \begin{array}{rrr|r} {\mathbf 1} \amp 0 \amp 0 \amp 3 \\ 0 \amp {\mathbf 1} \amp 0 \amp 0 \\ 0 \amp 0 \amp {\mathbf 1} \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}
    \begin{equation*} \left[ \begin{array}{rrr|r} {\mathbf 1} \amp 0 \amp 2 \amp 3 \\ 0 \amp {\mathbf 1} \amp -1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}
    \begin{equation*} \left[ \begin{array}{rrr|r} {\mathbf 1} \amp 0 \amp 2 \amp 0 \\ 0 \amp {\mathbf 1} \amp -1 \amp 0 \\ 0 \amp 0 \amp 0 \amp {\mathbf 1} \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right] \end{equation*}
    The first two augmented matrices correspond to consistent linear systems. The third does not, however, since the third row corresponds to the equation \(0=1\text{.}\)
    In general, a linear system is inconsistent exactly when there is a pivot position in the rightmost column of the augmented matrix.
  2. A row in which every entry is zero corresponds to the equation \(0=0\text{,}\) which is always true. Such an equation has no bearing on the consistency of the linear system.
  3. \begin{equation*} \left[\begin{array}{rrrr|r} {\mathbf 1} \amp 0 \amp 0 \amp 2 \amp 4 \\ 0 \amp {\mathbf 1} \amp 0 \amp -2 \amp 1 \\ 0 \amp 0 \amp {\mathbf 1} \amp 0 \amp 3 \\ \end{array}\right] \end{equation*}
    This corresponds to a consistent system because there is not a pivot in the rightmost column.
  4. \begin{equation*} \left[\begin{array}{rrrr|r} {\mathbf 1} \amp 0 \amp 0 \amp 2 \amp 4 \\ 0 \amp {\mathbf 1} \amp 0 \amp -2 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \amp {\mathbf 1} \\ \end{array}\right] \end{equation*}
    This is an inconsistent system because the third row corresponds to the equation \(0=1\text{,}\) which is never satisfied.
  5. \begin{equation*} \left[\begin{array}{rrrr} {\mathbf 1} \amp 0 \amp 0 \amp 2 \\ 0 \amp {\mathbf 1} \amp 0 \amp -2 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array}\right] \end{equation*}
    In the coefficient matrix, there is a row without a pivot position so that each entry is \(0\text{.}\) This allows a pivot position to appear in the rightmost column of the augmented matrix.
  6. This linear system must be consistent because the augmented matrix cannot have a pivot position in the rightmost column.
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