Chapter 7.4, Problem 47E

To determine

To write: the given matrix in reduced row echelon form by using graphing utility.

Answer to Problem 47E

Thereduced row echelon form of the given matrix is

  $\left[\begin{array}{cccc}1& 0& -\frac{3}{7}& -\frac{8}{7}\\ 0& 1& -\frac{12}{7}& \frac{10}{7}\end{array}\right]$ .

Explanation of Solution

Given information:

A matrix is given by

  $\left[\begin{array}{cccc}-4& 1& 0& 6\\ 1& -2& 3& -4\end{array}\right]$

Concept used:

A matrix is in row echelon form if it has the following properties:

1. Any rows consisting entirely of zeros at the bottom of the matrix.
2. For each row that does not consist entirely of zeros, the first nonzero entry is 1.
3. For two successive rows, the leading 1 in the higher row is further to the left than the leading 1 in the lower row.

And a matrix in row-echelon form is in reduced row echelon form if every column that has a leading 1 has zeros in every position above and below its leading 1.

Consider the given matrix.

  $\left[\begin{array}{cccc}-4& 1& 0& 6\\ 1& -2& 3& -4\end{array}\right]$

Now, apply graphing utility to perform the row operations as shown:

Use T1-83 calculator.

Click on 2^nd button and $x^{-1}$ button in T1-83 calculator and go to edit tab.

The below screen will be displayed.

  

Select 1 for matrix A then below screen will be displayed to enter the matrix.

  

Now, enter the number of rows then press ENTER then enter number of columns then ENTER.

  

Press 2^nd and QUIT. Now go to the MATRIX menu, and under MATH find rref. Press ENTER.

  

Click on 2^nd button and $x^{-1}$ button in T1-83 calculator and go to MATRIX.

  

The below screen will display.

  

Click on ENTER.

rref ([A])

  

So, the required row echelon formis $\left[\begin{array}{cccc}1& 0& -\frac{3}{7}& -\frac{8}{7}\\ 0& 1& -\frac{12}{7}& \frac{10}{7}\end{array}\right]$ .