Chapter 7, Problem 36RE

Solution

To calculate: The solution of the system of equations by finding the reduced row echelon form of the augmented matrix.

  $\begin{array}{c}x+2z-2w=5\\ 2x+y+4z-3w=7\\ 4x+y+7z-6w=15\\ 2x+y+5z-6w=9\end{array}$

The obtained solution is $\left(1,-3,2,0\right)$ .

Given Information:

The system of equation is defined as,

  $\begin{array}{c}x+2z-2w=5\\ 2x+y+4z-3w=7\\ 4x+y+7z-6w=15\\ 2x+y+5z-6w=9\end{array}$

Calculation:

Consider the given equations,

  $\begin{array}{c}x+2z-2w=5\\ 2x+y+4z-3w=7\\ 4x+y+7z-6w=15\\ 2x+y+5z-6w=9\end{array}$

First, write the augmented matrix.

  $\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 2& 1& 4& -3& 7\\ 4& 1& 7& -6& 15\\ 2& 1& 5& -6& 9\end{array}\right]$

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.

Apply the row operation to find the echelon form.

  $\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 2& 1& 4& -3& 7\\ 4& 1& 7& -6& 15\\ 2& 1& 5& -6& 9\end{array}\right]\underrightarrow{2R_1-R_2}\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& -1& 0& 1& 3\\ 4& 1& 7& -6& 15\\ 2& 1& 5& -6& 9\end{array}\right]\underrightarrow{-R_2}\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 4& 1& 7& -6& 15\\ 2& 1& 5& -6& 9\end{array}\right]$

  $\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 4& 1& 7& -6& 15\\ 2& 1& 5& -6& 9\end{array}\right]\underrightarrow{4R_1-R_3}\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 0& -1& 1& -2& 5\\ 2& 1& 5& -6& 9\end{array}\right]\underrightarrow{2R_1-R_4}\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 0& -1& 1& -2& 5\\ 0& -1& -1& 0& 1\end{array}\right]$

  $\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 0& -1& 1& -2& 5\\ 0& -1& -1& 0& 1\end{array}\right]\underrightarrow{R_2+R_3}\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 0& 0& 1& -3& 2\\ 0& -1& -1& 0& 1\end{array}\right]\underrightarrow{R_3+R_4}\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 0& 0& 1& -3& 2\\ 0& 0& 0& -4& 0\end{array}\right]$

  $\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 0& 0& 1& -3& 2\\ 0& 0& 0& -4& 0\end{array}\right]\underrightarrow{-\frac{1}{4}{R}_4}\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 0& 0& 1& -3& 2\\ 0& 0& 0& 1& 0\end{array}\right]\underrightarrow{3R_4+R_3}\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 1& 0\end{array}\right]$

  $\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& -1& -3\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 1& 0\end{array}\right]\underrightarrow{R_4+R_2}\left[\begin{array}{ccccc}1& 0& 2& -2& 5\\ 0& 1& 0& 0& -3\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 1& 0\end{array}\right]\underrightarrow{2R_4+R_1}\left[\begin{array}{ccccc}1& 0& 2& 0& 5\\ 0& 1& 0& 0& -3\\ 0& 0& 1& 0& 2\\ 0& 0& 0& 1& 0\end{array}\right]$

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

The obtained system of equations is defined as,

  $\begin{array}{c}x=1\\ y=-3\\ z=2\\ w=0\end{array}$

Hence, the obtained solution is $\left(1,-3,2,0\right)$ .