Chapter 7.4, Problem 45E

To determine

To write: the given matrix in row echelon form.

Answer to Problem 45E

The row echelon form of the given matrix is

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

Explanation of Solution

Given information:

A matrix is given by

  $\left[\begin{array}{cccc}1& 2& -1& 3\\ 3& 7& -5& 14\\ -2& -1& -3& 8\end{array}\right]$

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.

  $\left[\begin{array}{cccc}1& 2& -1& 3\\ 3& 7& -5& 14\\ -2& -1& -3& 8\end{array}\right]$

Multiply row 1 of the matrix by $\left(-3\right)$ and 2, and add with row 2 and row 3 respectively.

Apply $R_2\to \left(-3\right)R_1+R_2$ , $R_3\to 2R_1+R_3$

  $\left[\begin{array}{cccc}1& 2& -1& 3\\ 0& 1& -2& 5\\ 0& 3& -5& 14\end{array}\right]$

Multiply row 2 of the matrix by $\left(-3\right)$ and add with row 3.

Apply $R_3\to \left(-3\right)R_2+R_3$

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

Multiply row 3 of the matrix by $2$ and add with row 2.

Apply $R_2\to 2R_3+R_2$

  $\left[\begin{array}{cccc}1& 2& -1& 3\\ 0& 1& 0& 7\\ 0& 0& 1& 1\end{array}\right]$

Multiply row 2 of the matrix by $\left(-2\right)$ and add with row 1.

Apply $R_1\to \left(-2\right)R_2+R_1$

  $\left[\begin{array}{cccc}1& 0& -1& -11\\ 0& 1& 0& 7\\ 0& 0& 1& 1\end{array}\right]$

Multiply row 3 of the matrix by $1$ and add with row 1.

Apply $R_1\to R_3+R_1$

  $\left[\begin{array}{cccc}1& 0& 0& -10\\ 0& 1& 0& 7\\ 0& 0& 1& 1\end{array}\right]$

So, the required row echelon form is $\left[\begin{array}{cccc}1& 0& 0& -10\\ 0& 1& 0& 7\\ 0& 0& 1& 1\end{array}\right]$ .