Chapter 10, Problem 9RCC

To determine

To explain: whether a matrix is in reduced row-echelon form or not

Answer to Problem 9RCC

A matrix must meet all four of these rules in ordered to be in reduced row-echelon form.

1. The first nonzero number in each row (going from left to right) is 1.

2. The leading entry (the first 1) in each row is to the right of the leading entry in the row immediately above it.

3. All rows that consist of only zeroes are at the bottom of the matrix.

4. Every number above and below each leading entry is a 0.

Explanation of Solution

In order for a matrix to be in reduced row-echelon form, it must also be in row-echelon form. There are three requirements for a matrix to be in row-echelon form.

1. The first nonzero number in each row (going from left to right) is 1. Here is an example of a matrix breaking this rule.

  $\left[\begin{array}{ccccc}0& 1& 2& 1& 3\\ 0& 0& 5& 1& 2\\ 1& 2& 3& 4& 5\\ 7& 1& 2& 4& 5\end{array}\right]$

You can see that in rows 2 and 4, the first nonzero number in the row is not 1. In row 2, it is 5, and in row 4 it is 7. Here is a slightly different matrix that fixes the problems of the previous one.

  $\left[\begin{array}{ccccc}0& 1& 2& 1& 3\\ 0& 0& 1& 1& 2\\ 1& 2& 3& 4& 5\\ 1& 1& 2& 4& 5\end{array}\right]$

Now every row has a 1 as its first nonzero number.

2. The leading entry (the first 1) in each row is to the right of the leading entry in the row immediately above it.

Here is an example of a matrix breaking this rule.

  $\left[\begin{array}{ccccc}0& 1& 2& 1& 3\\ 1& 0& 0& 1& 2\\ 0& 0& 0& 1& 5\\ 0& 1& 2& 4& 5\end{array}\right]$

This follow Rule 1, but rows 2 and 4 again break a rule, since their 1's are to the left of the 1 in the row above them. A different, valid matrix is

  $\left[\begin{array}{ccccc}0& 1& 2& 1& 3\\ 0& 0& 1& 1& 2\\ 0& 0& 0& 1& 5\\ 0& 0& 0& 0& 1\end{array}\right]$

Now, each leading entry is to the right of the one directly above it.

3. All rows that consist of only zeroes are at the bottom of the matrix.

Here is a matrix that does not follow this rule.

  $\left[\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& 1& 2& 1& 3\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 1& 5\end{array}\right]$

All rows of zeroes must be at the bottom. A different, valid matrix is

  $\left[\begin{array}{ccccc}0& 1& 2& 1& 3\\ 0& 0& 0& 1& 5\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right]$

If a matrix meets all of the above criteria, then it is in row-echelon form. There is one extra rule that designates reduced row-echelon form.

4. Every number above and below each leading entry is a 0. Here is a matrix in row-echelon form, but is not reduced.

  $\left[\begin{array}{ccccc}1& 1& 2& 1& 3\\ 0& 0& 1& 2& 5\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$

The terms above and below the leading entries (the 1's) are not zero. Here is a different matrix that fulfills all the criteria of reduced row-echelon form.

  $\left[\begin{array}{ccccc}1& 4& 0& 0& 3\\ 0& 0& 1& 0& 5\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$

A matrix must meet all four of these rules in ordered to be in reduced row-echelon form.