Chapter 7, Problem 16RE

Solution

To calculate: The reduced row echelon form of the matrix.

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

The obtained matrix is $\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& 2\\ 0& 0& 1& -3\end{array}\right]$ .

Given Information:

The matrix is defined as,

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

Calculation:

Consider the given metrics,

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

The reduced row echelon form of the matrix is obtained when every column has a leading 1 and has 0's elsewhere by using elementary row operations. Unlike a row echelon form of the matrix, the reduced row echelon form of the matrix is unique so there is only one answer.

Starting from the given, apply the elementary row operations as shown:

  $\left[\begin{array}{cccc}2& 1& 1& 1\\ -3& -1& -2& 1\\ 5& 2& 2& 3\end{array}\right]\underrightarrow{\frac{1}{2}{R}_1}\left[\begin{array}{cccc}1& 1/2& 1/2& 1/2\\ -3& -1& -2& 1\\ 5& 2& 2& 3\end{array}\right]\underrightarrow{3R_1+R_2}\left[\begin{array}{cccc}1& 1/2& 1/2& 1/2\\ 0& 1/2& -1/2& 5/2\\ 5& 2& 2& 3\end{array}\right]$

  $\left[\begin{array}{cccc}1& 1/2& 1/2& 1/2\\ 0& 1/2& -1/2& 5/2\\ 5& 2& 2& 3\end{array}\right]\underrightarrow{2R_2}\left[\begin{array}{cccc}1& 1/2& 1/2& 1/2\\ 0& 1& -1& 5\\ 5& 2& 2& 3\end{array}\right]\underrightarrow{5R_1-R_3}\left[\begin{array}{cccc}1& 1/2& 1/2& 1/2\\ 0& 1& -1& 5\\ 0& 1/2& 1/2& -1/2\end{array}\right]$

  $\left[\begin{array}{cccc}1& 1/2& 1/2& 1/2\\ 0& 1& -1& 5\\ 0& 1/2& 1/2& -1/2\end{array}\right]\underrightarrow{2R_3}\left[\begin{array}{cccc}1& 1/2& 1/2& 1/2\\ 0& 1& -1& 5\\ 0& 1& 1& -1\end{array}\right]\underrightarrow{-\frac{1}{2}{R}_3}\left[\begin{array}{cccc}1& 1/2& 1/2& 1/2\\ 0& 1& -1& 5\\ 0& 0& 1& -3\end{array}\right]$

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

  $\underrightarrow{-\frac{1}{2}{R}_2+R_1}\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& 2\\ 0& 0& 1& -3\end{array}\right]$

Hence, the obtained reduce form is $\left[\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& 2\\ 0& 0& 1& -3\end{array}\right]$ .