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

David Austin, Erica Bernstein

Section 5.1 Reduced row echelon form

Definition 5.1.1.

We say that a matrix is in reduced row echelon form if the following properties are satisfied.
  • If the entries in a row are all zero, then the same is true of any row below it.
  • If we move across a row from left to right, the first nonzero entry we encounter is 1. We call this entry the leading entry in the row.
  • The leading entry in any row is to the right of the leading entries in all the rows above it.
  • A leading entry is the only nonzero entry in its column.
We call a matrix in reduced row echelon form a reduced row echelon matrix.

Example 5.1.2. 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 5.1.1. Identifying reduced row echelon matrices.

Consider each of the following augmented matrices. Determine if the matrix is in reduced row echelon form. If it is not, perform a sequence of scaling, interchange, and replacement operations to obtain a row equivalent matrix that is in reduced row echelon form. Then use the reduced row echelon matrix to describe the solution space.
  1. \(\displaystyle \left[ \begin{array}{rrr|r} 2 \amp 0 \amp 4 \amp -8 \\ 0 \amp 1 \amp 3 \amp 2 \\ \end{array} \right].\)
  2. \(\displaystyle \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \amp 3 \\ 0 \amp 0 \amp 1 \amp 1 \\ \end{array} \right].\)
  3. \(\displaystyle \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 4 \amp 2 \\ 0 \amp 1 \amp 3 \amp 2 \\ 0 \amp 0 \amp 0 \amp 1 \\ \end{array} \right].\)
  4. \(\displaystyle \left[ \begin{array}{rrr|r} 0 \amp 1 \amp 3 \amp 2 \\ 0 \amp 0 \amp 0 \amp 0 \\ 1 \amp 0 \amp 4 \amp 2 \\ \end{array} \right].\)
  5. \(\displaystyle \left[ \begin{array}{rrr|r} 1 \amp 2 \amp -1 \amp 2 \\ 0 \amp 1 \amp -2 \amp 0 \\ 0 \amp 0 \amp 1 \amp 1 \\ \end{array} \right].\)
Answer.
  1. The row equivalent reduced row echelon form is
    \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 2 \amp -4 \\ 0 \amp 1 \amp 3 \amp 2 \\ \end{array} \right], \end{equation*}
    and there are infinitely many solutions.
  2. This matrix is in reduced row echelon form. There is a single solution \((-1,3,1)\text{.}\)
  3. This matrix is also in reduced row echelon form. However, this are no solutions since the third equation is \(0 = 1\text{.}\)
  4. The row equivalent reduced row echelon form is
    \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 4 \amp 2 \\ 0 \amp 1 \amp 3 \amp 2 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right]. \end{equation*}
    There are infinitely many solutions.
  5. The row equivalent reduced row echelon form is
    \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \amp 2 \\ 0 \amp 0 \amp 1 \amp 1 \\ \end{array} \right]. \end{equation*}
    This system has the single solution \((-1,2,1)\text{.}\)
Solution.
  1. Because the leading entry in the first row is not a \(1\text{,}\) this is not in reduced row echelon form. If we scale the first row by \(1/2\text{,}\) however, we have a matrix in reduced row echelon form.
    \begin{equation*} \left[ \begin{array}{rrr|r} 2 \amp 0 \amp 4 \amp -8 \\ 0 \amp 1 \amp 3 \amp 2 \\ \end{array} \right] \sim \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 2 \amp -4 \\ 0 \amp 1 \amp 3 \amp 2 \\ \end{array} \right]. \end{equation*}
    We may write the corresponding linear system as
    \begin{equation*} \begin{alignedat}{4} x_1 \amp \amp \amp {}+{} \amp 2x_3 \amp {}={} \amp -4 \\ \amp \amp x_2 \amp {}+{} \amp 3x_3 \amp {}={} \amp 2, \\ \end{alignedat} \end{equation*}
    which may be rewritten as
    \begin{equation*} \begin{alignedat}{2} x_1 \amp {}={} -8 - 4x_3 \\ x_2 \amp {}={} 2 - 3x_3. \\ \end{alignedat} \end{equation*}
    Since \(x_3\) may take on any value, this shows that there are infinitely many solutions.
  2. This matrix is in reduced row echelon form. There is a single solution \((-1,3,1)\text{.}\)
  3. This matrix is also in reduced row echelon form. However, this are no solutions since the third equation is \(0 = 1\text{.}\)
  4. This is not in reduced row echelon form because the row of zeroes should be at the bottom of the matrix. We also need another interchange so that the leading entry in the second row is to the right of the leading entry in the first row.
    \begin{equation*} \left[ \begin{array}{rrr|r} 0 \amp 1 \amp 3 \amp 2 \\ 0 \amp 0 \amp 0 \amp 0 \\ 1 \amp 0 \amp 4 \amp 2 \\ \end{array} \right] \sim \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 4 \amp 2 \\ 0 \amp 1 \amp 3 \amp 2 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right]. \end{equation*}
    Once again, there are infinitely many solutions.
  5. This is not in reduced row echelon form because the leading entry in the second and third rows are not the only nonzero elements in their columns. We can use replacement operations to remedy this
    \begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 2 \amp -1 \amp 2 \\ 0 \amp 1 \amp -2 \amp 0 \\ 0 \amp 0 \amp 1 \amp 1 \\ \end{array} \right] \sim \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 0 \amp -1 \\ 0 \amp 1 \amp 0 \amp 2 \\ 0 \amp 0 \amp 1 \amp 1 \\ \end{array} \right] \end{equation*}
    and see that the system has the single solution \((-1,2,1)\text{.}\)
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