Skip to main content

Section 1.2 Row Reduction of Matrices (E2)

Definition 1.2.1.

Two systems of linear equations (and their corresponding augmented matrices) are said to be equivalent if they have the same solution set.

For example, both of these systems share the same solution set \(\setList{ \left[\begin{array}{c} 1 \\ 1\end{array}\right] }\text{.}\)

\begin{align*} 3x_1 &\,-\,& 2x_2 &\,=\,& 1 \\ x_1 &\,+\,& 4x_2 &\,=\,& 5 \end{align*}
\begin{align*} 3x_1 &\,-\,& 2x_2 &\,=\,& 1 \\ 4x_1 &\,+\,& 2x_2 &\,=\,& 6 \end{align*}

Therefore these augmented matrices are equivalent, which we denote with \(\sim\text{:}\)

\begin{equation*} \left[\begin{array}{cc|c} 3 & -2 & 1\\ 1 & 4 & 5\\ \end{array}\right] \sim \left[\begin{array}{cc|c} 3 & -2 & 1\\ 4 & 2 & 6\\ \end{array}\right] \end{equation*}
Activity 1.2.1.

Following are seven procedures used to manipulate an augmented matrix. Label the procedures that would result in an equivalent augmented matrix as valid, and label the procedures that might change the solution set of the corresponding linear system as invalid.

  1. Swap two rows.
  2. Swap two columns.
  3. Add a constant to every term in a row.
  4. Multiply a row by a nonzero constant.
  5. Add a constant multiple of one row to another row.
  6. Replace a column with zeros.
  7. Replace a row with zeros.
Definition 1.2.2.

The following row operations produce equivalent augmented matrices:

  1. Swap two rows, for example, \(R_1\leftrightarrow R_2\text{:}\)

    \begin{equation*} \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right] \sim \left[\begin{array}{cc|c} 4 & 5 & 6 \\ 1 & 2 & 3 \end{array}\right] \end{equation*}

  2. Multiply a row by a nonzero constant, for example, \(2R_1\rightarrow R_1\text{:}\)

    \begin{equation*} \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right] \sim \left[\begin{array}{cc|c} 2(1) & 2(2) & 2(3) \\ 4 & 5 & 6 \end{array}\right] \end{equation*}

  3. Add a constant multiple of one row to another row, for example, \(R_2-4R_1\rightarrow R_2\text{:}\)

    \begin{equation*} \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & 2 & 3 \\ 4-4(1) & 5-4(2) & 6-4(3) \end{array}\right] \end{equation*}

Whenever two matrices \(A,B\) are equivalent (so whenever we do any of these operations), we write \(A\sim B\text{.}\)

Activity 1.2.2.

Consider the following (equivalent) linear systems.

  1. \begin{align*} x &\,+\,& 2y &\,+\,& z &\,=\,& 3 \\ -x &\,-\,& y &\,+\,& z &\,=\,& 1 \\ 2x &\,+\,& 5y &\,+\,& 3z &\,=\,& 7 \end{align*}
  2. \begin{align*} 2x &\,+\,& 5y &\,+\,& 3z &\,=\,& 7 \\ -x &\,-\,& y &\,+\,& z &\,=\,& 1 \\ x &\,+\,& 2y &\,+\,& z &\,=\,& 3 \end{align*}
  3. \begin{align*} x & & &\,-\,& z &\,=\,& 1 \\ & & y &\,+\,& z &\,=\,& 1 \\ & & y &\,+\,& 2z &\,=\,& 4 \end{align*}
  4. \begin{align*} x &\,+\,& 2y &\,+\,& z &\,=\,& 3 \\ & & y &\,+\,& z &\,=\,& 1 \\ 2x &\,+\,& 5y &\,+\,& 3z &\,=\,& 7 \end{align*}
  5. \begin{align*} x & & &\,-\,& z &\,=\,& 1 \\ & & y &\,+\,& z &\,=\,& 1 \\ & & & & z &\,=\,& 3 \end{align*}
  6. \begin{align*} x &\,+\,& 2y &\,+\,& z &\,=\,& 3 \\ & & y &\,+\,& z &\,=\,& 1 \\ & & y &\,+\,& 2z &\,=\,& 4 \end{align*}

Rank the six linear systems from most complicated to simplest.

Activity 1.2.3.

We can rewrite the previous in terms of equivalences of augmented matrices

\begin{align*} \left[\begin{array}{ccc|c} 2 & 5 & 1 3 & 7 \\ -1 & -1 & 1 & 1 \\ 1 & 2 & 1 & 3 \end{array}\right] & \sim & \left[\begin{array}{ccc|c} \circledNumber{1} & 2 & 1 & 3 \\ -1 & -1 & 1 & 1 \\ 2 & 5 & 1 & 3 \end{array}\right] & \sim & \left[\begin{array}{ccc|c} \circledNumber{1} & 2 & 1 & 3 \\ 0 & 1 & 1 & 1 \\ 2 & 5 & 1 & 3 \end{array}\right] \sim \\ \left[\begin{array}{ccc|c} \circledNumber{1} & 2 & 1 & 3 \\ 0 & \circledNumber{1} & 1 & 1 \\ 0 & 1 & 2 & 4 \end{array}\right] & \sim & \left[\begin{array}{ccc|c} \circledNumber{1} & 0 & -1 & 1 \\ 0 & \circledNumber{1} & 1 & 1 \\ 0 & 1 & 2 & 4 \end{array}\right] & \sim & \left[\begin{array}{ccc|c} \circledNumber{1} & 0 & -1 & 1 \\ 0 & \circledNumber{1} & 1 & 1 \\ 0 & 0 & \circledNumber{1} & 3 \end{array}\right] \end{align*}

Determine the row operation(s) necessary in each step to transform the most complicated system's augmented matrix into the simplest.

Definition 1.2.3.

A matrix is in reduced row echelon form (RREF) if

  1. The leading term (first nonzero term) of each nonzero row is a 1. Call these terms pivots.
  2. Each pivot is to the right of every higher pivot.
  3. Each term above or below a pivot is zero.
  4. All rows of zeroes are at the bottom of the matrix.

Every matrix has a unique reduced row echelon form. If \(A\) is a matrix, we write \(\RREF(A)\) for the reduced row echelon form of that matrix.

Activity 1.2.4.

Recall that a matrix is in reduced row echelon form (RREF) if

  1. The leading term (first nonzero term) of each nonzero row is a 1. Call these terms pivots.
  2. Each pivot is to the right of every higher pivot.
  3. Each term above or below a pivot is zero.
  4. All rows of zeroes are at the bottom of the matrix.
  1. \begin{equation*} \left[\begin{array}{ccc|c} 1 & 0 & 0 & 3 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}
  2. \begin{equation*} \left[\begin{array}{ccc|c} 1 & 2 & 4 & 3 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}
  3. \begin{equation*} \left[\begin{array}{ccc|c} 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & -1 \end{array}\right] \end{equation*}
  4. \begin{equation*} \left[\begin{array}{ccc|c} 1 & 0 & 2 & -3 \\ 0 & 3 & 3 & -3 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}
  5. \begin{equation*} \left[\begin{array}{ccc|c} 0 & 1 & 0 & 7 \\ 1 & 0 & 0 & 4 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}
  6. \begin{equation*} \left[\begin{array}{ccc|c} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 0 \end{array}\right] \end{equation*}

For each matrix, circle the leading terms, and label it as RREF or not RREF. For the ones not in RREF, find their RREF.

Remark 1.2.4.

In practice, if we simply need to convert a matrix into reduced row echelon form, we use technology to do so.

However, it is also important to understand the Gauss-Jordan elimination algorithm that a computer or calculator uses to convert a matrix (augmented or not) into reduced row echelon form. Understanding this algorithm will help us better understand how to interpret the results in many applications we use it for in Module V.

Activity 1.2.5.

Consider the matrix

\begin{equation*} \left[\begin{array}{cccc}2 & 6 & -1 & 6 \\ 1 & 3 & -1 & 2 \\ -1 & -3 & 2 & 0 \end{array}\right]. \end{equation*}

Which row operation is the best choice for the first move in converting to RREF?

  1. Add row 3 to row 2 (\(R_2+R_3 \rightarrow R_2\))
  2. Add row 2 to row 3 (\(R_3+R_2 \rightarrow R_3\))
  3. Swap row 1 to row 2 (\(R_1 \leftrightarrow R_2\))
  4. Add -2 row 2 to row 1 (\(R_1-2R_2 \rightarrow R_1\))
Activity 1.2.6.

Consider the matrix

\begin{equation*} \left[\begin{array}{cccc} \circledNumber{1} & 3 & -1 & 2 \\ 2 & 6 & -1 & 6 \\ -1 & -3 & 2 & 0 \end{array}\right]. \end{equation*}

Which row operation is the best choice for the next move in converting to RREF?

  1. Add row 1 to row 3 (\(R_3+R_1 \rightarrow R_3\))
  2. Add -2 row 1 to row 2 (\(R_2-2R_1 \rightarrow R_2\))
  3. Add 2 row 2 to row 3 (\(R_3+2R_2 \rightarrow R_3\))
  4. Add 2 row 3 to row 2 (\(R_2+2R_3 \rightarrow R_2\))
Activity 1.2.7.

Consider the matrix

\begin{equation*} \left[\begin{array}{cccc}\circledNumber{1} & 3 & -1 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right]. \end{equation*}

Which row operation is the best choice for the next move in converting to RREF?

  1. Add row 1 to row 2 (\(R_2+R_1 \rightarrow R_2\))
  2. Add -1 row 3 to row 2 (\(R_2-R_3 \rightarrow R_2\))
  3. Add -1 row 2 to row 3 (\(R_3-R_2 \rightarrow R_3\))
  4. Add row 2 to row 1 (\(R_1+R_2 \rightarrow R_1\))
Activity 1.2.8.

Consider the matrix

\begin{equation*} \left[\begin{array}{ccc}2 & 1 & 0 \\ 1 & 0 & 0 \\ 3 & -1 & 1 \end{array}\right]. \end{equation*}
(a)

Perform three row operations to produce a matrix closer to RREF.

(b)

Finish putting it in RREF.

Activity 1.2.9.

Consider the matrix

\begin{equation*} A=\left[\begin{array}{cccc}2 & 3 & 2 & 3 \\ -2 & 1 & 6 & 1 \\ -1 & -3 & -4 & 1 \end{array}\right]. \end{equation*}

Compute \(\RREF(A)\text{.}\)

Activity 1.2.10.

Consider the matrix

\begin{equation*} A=\left[\begin{array}{cccc} 2 & 4 & 2 & -4 \\ -2 & -4 & 1 & 1 \\ 3 & 6 & -1 & -4 \end{array}\right]. \end{equation*}

Compute \(\RREF(A)\text{.}\)

Remark 1.2.5.

A video example of how to perform the Gauss-Jordan Elimination algorithm by hand is available at https://youtu.be/Cq0Nxk2dhhU.

Practicing several exercises on your own using this method is strongly recommended.

Exercises 1.2.1 Exercises

1.
  1. Show that
    \begin{equation*} \operatorname{RREF} \left[\begin{array}{ccccc} 1 & -2 & -1 & 3 & 4 \\ 2 & -3 & -3 & 5 & 6 \\ 0 & -5 & 5 & 5 & 10 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & -3 & 1 & 0 \\ 0 & 1 & -1 & -1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] . \end{equation*}
  2. Explain why the matrix \(B= \left[\begin{array}{ccccc} 1 & 4 & 0 & -1 & 1 \\ -5 & -20 & 1 & 7 & -6 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \) is or is not in reduced row echelon form.
Answer
  1. \(\displaystyle \operatorname{RREF} \left[\begin{array}{ccccc} 1 & -2 & -1 & 3 & 4 \\ 2 & -3 & -3 & 5 & 6 \\ 0 & -5 & 5 & 5 & 10 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & -3 & 1 & 0 \\ 0 & 1 & -1 & -1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] .\)
  2. \(B\) is not in reduced row echelon form because not every entry above and below each pivot is zero.
2.
  1. Show that
    \begin{equation*} \operatorname{RREF} \left[\begin{array}{cccc} 0 & 0 & 1 & -2 \\ 1 & -3 & 5 & -12 \\ 0 & 0 & -4 & 8 \\ 1 & -3 & 3 & -8 \end{array}\right] = \left[\begin{array}{cccc} 1 & -3 & 0 & -2 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] . \end{equation*}
  2. Explain why the matrix \(B= \left[\begin{array}{cccc} 1 & 0 & -1 & -3 \\ -7 & 1 & 5 & 20 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \) is or is not in reduced row echelon form.
Answer
  1. \(\displaystyle \operatorname{RREF} \left[\begin{array}{cccc} 0 & 0 & 1 & -2 \\ 1 & -3 & 5 & -12 \\ 0 & 0 & -4 & 8 \\ 1 & -3 & 3 & -8 \end{array}\right] = \left[\begin{array}{cccc} 1 & -3 & 0 & -2 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] .\)
  2. \(B\) is not in reduced row echelon form because not every entry above and below each pivot is zero.
3.
  1. Show that
    \begin{equation*} \operatorname{RREF} \left[\begin{array}{ccc} 3 & -10 & 0 \\ -2 & 7 & 0 \\ 1 & 2 & 0 \\ 2 & -7 & 0 \\ 3 & -9 & 0 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] . \end{equation*}
  2. Explain why the matrix \(B= \left[\begin{array}{ccc} 1 & 0 & -2 \\ -2 & 1 & 4 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \) is or is not in reduced row echelon form.
Answer
  1. \(\displaystyle \operatorname{RREF} \left[\begin{array}{ccc} 3 & -10 & 0 \\ -2 & 7 & 0 \\ 1 & 2 & 0 \\ 2 & -7 & 0 \\ 3 & -9 & 0 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] .\)
  2. \(B\) is not in reduced row echelon form because not every entry above and below each pivot is zero.
4.
  1. Show that
    \begin{equation*} \operatorname{RREF} \left[\begin{array}{cccc} 1 & -1 & -1 & -8 \\ 2 & -1 & -5 & -12 \\ 0 & -3 & 9 & -11 \\ -1 & 1 & 1 & 7 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & -4 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] . \end{equation*}
  2. Explain why the matrix \(B= \left[\begin{array}{cccc} 1 & 0 & 0 & -3 \\ 0 & 0 & 1 & 3 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 0 & 0 \end{array}\right] \) is or is not in reduced row echelon form.
Answer
  1. \(\displaystyle \operatorname{RREF} \left[\begin{array}{cccc} 1 & -1 & -1 & -8 \\ 2 & -1 & -5 & -12 \\ 0 & -3 & 9 & -11 \\ -1 & 1 & 1 & 7 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & -4 & 0 \\ 0 & 1 & -3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] .\)
  2. \(B\) is not in reduced row echelon form because the pivots are not descending to the right.
5.
  1. Show that
    \begin{equation*} \operatorname{RREF} \left[\begin{array}{ccc} 2 & -2 & 0 \\ -5 & 1 & 4 \\ -2 & 3 & -1 \\ -5 & -2 & 7 \\ 1 & -2 & 1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] . \end{equation*}
  2. Explain why the matrix \(B= \left[\begin{array}{ccc} 0 & 1 & -2 \\ 1 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \) is or is not in reduced row echelon form.
Answer
  1. \(\displaystyle \operatorname{RREF} \left[\begin{array}{ccc} 2 & -2 & 0 \\ -5 & 1 & 4 \\ -2 & 3 & -1 \\ -5 & -2 & 7 \\ 1 & -2 & 1 \end{array}\right] = \left[\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] .\)
  2. \(B\) is not in reduced row echelon form because the pivots are not descending to the right.

Additional exercises available at checkit.clontz.org.