Chapter 7.3, Problem 108E

Solution

  

To determine: A row echelon from of matrix not unique ad show that the matrix can have two unequal row echelon forms, explain using an example.

The echelon form is not unique for any matrix.

Given Information:

The general matrix is defined as,

  $\left[\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& b_1\\ a_{21}& a_{22}& a_{23}& b_2\\ a_{31}& a_{32}& a_{33}& b_3\end{array}\right]$

Explanation:

Consider the given information,

A row echelon form of a matrix is not unique because the elementary row operations can be in any order. For example, we have the system:

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

One way to write its row-echelon form is shown below:

  $\begin{array}{l}\left[\begin{array}{cccc}1& 2& 2& 5\\ 2& 3& -2& 3\\ -1& -1& 1& -1\end{array}\right]\underrightarrow{2R_1-R_2}\left[\begin{array}{cccc}1& 2& 2& 5\\ 0& 1& 6& 7\\ -1& -1& 1& -1\end{array}\right]\underrightarrow{2R_1-R_2}\left[\begin{array}{cccc}1& 2& 2& 5\\ 0& 1& 6& 7\\ 0& 1& 3& 4\end{array}\right]\\ \\ \left[\begin{array}{cccc}1& 2& 2& 5\\ 0& 1& 6& 7\\ 0& 1& 3& 4\end{array}\right]\underrightarrow{R_2-R_3}\left[\begin{array}{cccc}1& 2& 2& 5\\ 0& 1& 6& 7\\ 0& 0& 3& 3\end{array}\right]\underrightarrow{\frac{R_3}{2}}\left[\begin{array}{cccc}1& 2& 2& 5\\ 0& 1& 6& 7\\ 0& 0& 1& 1\end{array}\right]\end{array}$

Now, use another way to express the form.

  $\begin{array}{l}\left[\begin{array}{cccc}1& 2& 2& 5\\ 2& 3& -2& 3\\ -1& -1& 1& -1\end{array}\right]\underrightarrow{R_2=R_3}\left[\begin{array}{cccc}1& 2& 2& 5\\ -1& -1& 1& -1\\ 2& 3& -2& 3\end{array}\right]\underrightarrow{R_1+R_2}\left[\begin{array}{cccc}1& 2& 2& 5\\ 0& 1& 3& 4\\ 2& 3& -2& 3\end{array}\right]\\ \\ \left[\begin{array}{cccc}1& 2& 2& 5\\ 0& 1& 3& 4\\ 2& 3& -2& 3\end{array}\right]\underrightarrow{2R_1-R_3}\left[\begin{array}{cccc}1& 2& 2& 5\\ 0& 1& 3& 4\\ 0& 1& 6& 7\end{array}\right]\underrightarrow{\frac{R_2-R_3}{2}}\left[\begin{array}{cccc}1& 2& 2& 5\\ 0& 1& 3& 4\\ 0& 0& -3& -3\end{array}\right]\underrightarrow{-\frac{1}{3}{R}_3}\left[\begin{array}{cccc}1& 2& 2& 5\\ 0& 1& 3& 4\\ 0& 0& 1& 1\end{array}\right]\end{array}$

A row echelon form of a matrix is not unique because the elementary row operations can be in any order.

Therefore, the echelon form is not unique for any matrix.