Chapter 7.4, Problem 3E

To determine

To fill: the blank in the given statement..

Answer to Problem 3E

The process of using row operations to write a matrix in reduced row echelon form is called Gauss Jordan elimination.

Explanation of Solution

Given information:

Given statement is

The process of using row operations to write a matrix in reduced roe echelon form is called -------.

Calculation:

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.

The process of using row operations to write a matrix in reduced roe echelon form is calledGauss Jordan elimination.

Now, the augmented matrix is given by

  $\left[\begin{array}{cccc}-2& 3& ⋮& 5\\ 6& 7& ⋮& 4\end{array}\right]$

Now, perform row operations as shown:

Apply $R_2\to R_2+3R_1$ .

  $\left[\begin{array}{cccc}-2& 3& ⋮& 5\\ 0& 16& ⋮& 19\end{array}\right]$

Apply $R_1\to -\frac{1}{2}{R}_1,R_2\to \frac{1}{16}{R}_2$ .

  $\left[\begin{array}{cccc}1& -\frac{3}{2}& ⋮& -\frac{5}{2}\\ 0& 1& ⋮& \frac{19}{16}\end{array}\right]$

Apply $R_1\to R_1+\frac{3}{2}{R}_2$ .

  $\left[\begin{array}{cccc}1& 0& ⋮& -\frac{23}{32}\\ 0& 1& ⋮& \frac{19}{16}\end{array}\right]$

The above is the reduced row-echelon form of the given matrix.

Here, the process is called Gaussian Elimination.