Chapter 7.3, Problem 65E

Solution

To state: The solution for the system of equation using any method:

  $\begin{array}{c}x+y-z+2w=0\\ y-z+2w=-1\\ x+y+3w=3\\ 2x+2y-z+5w=4\end{array}$

The system has no solution.

Given information:

The given system is:

  $\begin{array}{c}x+y-z+2w=0\\ y-z+2w=-1\\ x+y+3w=3\\ 2x+2y-z+5w=4\end{array}$

Explanation:

Consider the given system:

  $\begin{array}{c}x+y-z+2w=0\\ y-z+2w=-1\\ x+y+3w=3\\ 2x+2y-z+5w=4\end{array}$

Write the given system of equation in matrix and solve the system of equations using matrix:

  $\left[\begin{array}{cccc}1& 1& -1& 2\\ 0& 1& -1& 2\\ 1& 1& 0& 3\\ 2& 2& -1& 5\end{array}\begin{array}{c}0\\ -1\\ 3\\ 4\end{array}\right]$

Apply the operations $R_3\to -R_1+R_3,R_4\to -2R_1+R_4$ ,

  $\left[\begin{array}{cccc}1& 1& -1& 2\\ 0& 1& -1& 2\\ 0& 0& 1& 1\\ 0& 0& 1& 1\end{array}\begin{array}{c}0\\ -1\\ 3\\ 4\end{array}\right]$

Apply the operations $R_4\to \left(-1\right)R_3+R_4$ ,

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

Apply the operations $R_3\to \left(-3\right)R_4+R_3$ ,

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

Apply the operation $R_2\to R_3+R_2$ ,

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

Apply the operations $R_1\to \left(-1\right)R_2+R_1$ ,

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

Apply the operations $R_1\to R_3+R_1$ ,

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

Apply the operations $R_1\to \left(-1\right)R_4+R_1$ ,

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

The augmented matrix is in reduced row echelon form. Since

  $0=1$ is never being true. Hence it can be concluded that the given system has no solution.