Chapter 8.1, Problem 73E

To determine

To solve:The system of linear equations using Gaussian elimination method with back-substitution.

Answer to Problem 73E

The solution of the given system is $x=3,y=-2,z=5\text{and}w=0$ .

Explanation of Solution

Given information:

The system of equations is

  $\{\begin{array}{c}3x+2y-z+w=0\\ x-y+4z+2w=25\\ -2x+y+2z-w=2\\ x+y+z+w=6\end{array}$

Calculations:

Writing the given system of linear equations in the augmented matrix form, we have

  $\left[\begin{array}{cccccc}3& 2& -1& 1& :& 0\\ 1& -1& 4& 2& :& 25\\ -2& 1& 2& -1& :& 2\\ 1& 1& 1& 1& :& 6\end{array}\right]$

Performing the row operations to bring it in row-echelon form, we get

  $\left[\begin{array}{cccccc}1& 1& 1& 1& :& 6\\ 1& -1& 4& 2& :& 25\\ -2& 1& 2& -1& :& 2\\ 3& 2& -1& 1& :& 0\end{array}\right]\begin{array}{c}\leftarrow \\ \\ \\ \end{array}\begin{array}{c}{R}_1\leftrightarrow R_4\\ \\ \\ \end{array}$

  $\left[\begin{array}{cccccc}1& 1& 1& 1& :& 6\\ 0& -2& 3& 1& :& 19\\ 0& 3& 4& 1& :& 14\\ 0& -1& -4& -2& :& -18\end{array}\right]\begin{array}{c}\\ \leftarrow \\ \leftarrow \\ \leftarrow \end{array}\begin{array}{c}\\ R_2-R_1\\ R_3+2R_1\\ R_4-3R_1\end{array}$

  $\left[\begin{array}{cccccc}1& 1& 1& 1& :& 6\\ 0& 1& 4& 2& :& 18\\ 0& 3& 4& 1& :& 14\\ 0& -2& 3& 1& :& 19\end{array}\right]\begin{array}{c}\\ \leftarrow \\ \\ \end{array}\begin{array}{c}\\ R_2\leftrightarrow -R_4\\ \\ \end{array}$

  $\left[\begin{array}{cccccc}1& 1& 1& 1& :& 6\\ 0& 1& 4& 2& :& 18\\ 0& 0& -8& -5& :& -40\\ 0& 0& 11& 5& :& 55\end{array}\right]\begin{array}{c}\\ \\ \leftarrow \\ \leftarrow \end{array}\begin{array}{c}\\ \\ R_3-3R_2\\ R_4+2R_2\end{array}$

  $\left[\begin{array}{cccccc}1& 1& 1& 1& :& 6\\ 0& 1& 4& 2& :& 18\\ 0& 0& 1& 5/8& :& 5\\ 0& 0& 11& 5& :& 55\end{array}\right]\begin{array}{c}\\ \\ \leftarrow \\ \end{array}\begin{array}{c}\\ \\ R_3/-8\\ \end{array}$

  $\left[\begin{array}{cccccc}1& 1& 1& 1& :& 6\\ 0& 1& 4& 2& :& 18\\ 0& 0& 1& 5/8& :& 5\\ 0& 0& 0& -15/8& :& 0\end{array}\right]\begin{array}{c}\\ \\ \\ \leftarrow \end{array}\begin{array}{c}\\ \\ \\ R_4-11R_3\end{array}$

  $\left[\begin{array}{cccccc}1& 1& 1& 1& :& 6\\ 0& 1& 4& 2& :& 18\\ 0& 0& 1& 5/8& :& 5\\ 0& 0& 0& 1& :& 0\end{array}\right]\begin{array}{c}\\ \\ \\ \leftarrow \end{array}\begin{array}{c}\\ \\ \\ -8R_4/15\end{array}$

Now, the matrix is in row-echelon form, and writing it in the system form, we have

  $\{\begin{array}{c}x+y+z+w=6\\ y+4z+2w=18\\ z+\frac{5}{8}w=5\\ w=0\end{array}$

Substituting $w=0$ in the third equation, we get

  $\begin{array}{l}z+\frac{5}{8}(0)=5\\ \Rightarrow z=5\end{array}$

Now, the second equation becomes

  $\begin{array}{l}y+4(5)+2(0)=18\\ \Rightarrow y+20+0=18\\ \Rightarrow y=-2\end{array}$

Substituting above finding values in the first equation, we get

  $\begin{array}{l}x+(-2)+5+0=6\\ \Rightarrow x+3=6\\ \Rightarrow x=3\end{array}$

Conclusion:

Therefore, the solution of the given system is $x=3,y=-2,z=5\text{and}w=0$ .