Chapter 7.3, Problem 58E

Solution

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

  $\begin{array}{c}x-y+w=-4\\ -2x+y+z=8\\ 2x-2y-z=-10\\ -2x+z+w=5\end{array}$

The required solution is $x=-1,y=2,z=4,\text{and}w=-1$ .

Given information:

The given system is:

  $\begin{array}{c}x-y+w=-4\\ -2x+y+z=8\\ 2x-2y-z=-10\\ -2x+z+w=5\end{array}$

Explanation:

Consider the given system:

  $\begin{array}{c}x-y+w=-4\\ -2x+y+z=8\\ 2x-2y-z=-10\\ -2x+z+w=5\end{array}$

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

  $\left[\begin{array}{cccc}1& -1& 0& 1\\ -2& 1& 1& 0\\ 2& -2& -1& 0\\ -2& 0& 1& 1\end{array}\begin{array}{c}-4\\ 8\\ -10\\ 5\end{array}\right]$

Apply the operations $R_2\to 2R_1+R_2,R_3\to -2R_1+R_3,R_4\to 2R_1+R_4$ ,

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

Apply the operations $R_2\to -R_2,R_3\to -R_3$ ,

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

Apply the operations $R_4\to 2R_2+R_4$ ,

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

Apply the operation $R_4\to R_3+R_4$ ,

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

Apply the operations $R_1\to -R_4+R_1,R_3\to -2R_4+R_3,R_2\to 2R_4+R_2$ ,

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

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

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

Apply the operations $R_1\to R_2+R_1$ ,

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

Therefore, the resultant solution for the system of equation is:

  $x=-1,y=2,z=4,\text{and}w=-1$