Chapter 7.4, Problem 25E

To determine

To find: the elementary row operation to obtain the row equivalent matrix from the original matrix.

Answer to Problem 25E

The elementary row operation is $R_1\to \left(-\frac{1}{4}\right)R_1$ .

Explanation of Solution

Given information:

Original matrix is given by

  $\left[\begin{array}{ccc}-4& 8& -20\\ -2& 6& 7\end{array}\right]$

New row equivalent matrix is given by

  $\left[\begin{array}{ccc}1& -2& 5\\ -2& 6& 7\end{array}\right]$

Concept used:

Co-efficient matrix of a system of linear equation of the form $\{\begin{array}{l}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}$ is given by

  $\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]$

The augmented matrix of the system of linear equation is given by

  $\left[\begin{array}{cccc}{a}_1& b_1& ⋮& c_1\\ a_2& b_2& ⋮& c_2\end{array}\right]$

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.

  $\left[\begin{array}{ccc}-4& 8& -20\\ -2& 6& 7\end{array}\right]$

If row 1 of the matrix is multiplied by $\left(-\frac{1}{4}\right)$ then it will reduces to

  $\left[\begin{array}{ccc}1& -2& 5\\ -2& 6& 7\end{array}\right]$

Now, denote the row 1 by $R_1$ .

So, the performed row operation is $R_1\to \left(-\frac{1}{4}\right)R_1$ .