Chapter 8.1, Problem 74E

To determine

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

Answer to Problem 74E

Thesolution of the given system is $x=-1,y=0,z=6\text{and}w=4$ .

Explanation of Solution

Given information:

The system of equations is

  $\{\begin{array}{c}x-4y+3z-2w=9\\ 3x-2y+z-4w=-13\\ -4x+3y-2z+w=-4\\ -2x+y-4z+3w=-10\end{array}$

Calculations:

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

  $\left[\begin{array}{cccccc}1& -4& 3& -2& :& 9\\ 3& -2& 1& -4& :& -13\\ -4& 3& -2& 1& :& -4\\ -2& 1& -4& 3& :& -10\end{array}\right]$

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

  $\left[\begin{array}{cccccc}1& -4& 3& -2& :& 9\\ 0& 10& -8& 2& :& -40\\ 0& -13& 10& -7& :& 32\\ 0& -7& 2& -1& :& 8\end{array}\right]\begin{array}{c}\\ \leftarrow \\ \leftarrow \\ \leftarrow \end{array}\begin{array}{c}\\ R_2-3R_1\\ R_3+4R_1\\ R_4+2R_1\end{array}$

  $\left[\begin{array}{cccccc}1& -4& 3& -2& :& 9\\ 0& 1& -4/5& 1/5& :& -4\\ 0& -13& 10& -7& :& 32\\ 0& -7& 2& -1& :& 8\end{array}\right]\begin{array}{c}\\ \leftarrow \\ \\ \end{array}\begin{array}{c}\\ R_2/10\\ \\ \end{array}$

  $\left[\begin{array}{cccccc}1& -4& 3& -2& :& 9\\ 0& 1& -4/5& 1/5& :& -4\\ 0& 0& -2/5& -22/5& :& -20\\ 0& 0& -18/5& 2/5& :& -20\end{array}\right]\begin{array}{c}\\ \\ \leftarrow \\ \leftarrow \end{array}\begin{array}{c}\\ \\ R_3+13R_2\\ R_4+7R_2\end{array}$

  $\left[\begin{array}{cccccc}1& -4& 3& -2& :& 9\\ 0& 1& -4/5& 1/5& :& -4\\ 0& 0& 1& 11& :& 50\\ 0& 0& -18/5& 2/5& :& -20\end{array}\right]\begin{array}{c}\\ \\ \leftarrow \\ \end{array}\begin{array}{c}\\ \\ -5R_3/2\\ \end{array}$

  $\left[\begin{array}{cccccc}1& -4& 3& -2& :& 9\\ 0& 1& -4/5& 1/5& :& -4\\ 0& 0& 1& 11& :& 50\\ 0& 0& 0& 40& :& 160\end{array}\right]\begin{array}{c}\\ \\ \\ \leftarrow \end{array}\begin{array}{c}\\ \\ \\ R_4+18R_3/5\end{array}$

  $\left[\begin{array}{cccccc}1& -4& 3& -2& :& 9\\ 0& 1& -4/5& 1/5& :& -4\\ 0& 0& 1& 11& :& 50\\ 0& 0& 0& 1& :& 4\end{array}\right]\begin{array}{c}\\ \\ \\ \leftarrow \end{array}\begin{array}{c}\\ \\ \\ R_4/40\end{array}$

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

  $\{\begin{array}{c}x-4y+3z-2w=9\\ y-\frac{4}{5}z+\frac{1}{5}w=-4\\ z+11w=50\\ w=4\end{array}$

Substituting $w=4$ in the third equation, we get

  $\begin{array}{l}z+11(4)=50\\ \Rightarrow z+44=50\\ \Rightarrow z=6\end{array}$

Now, the second equation becomes

  $\begin{array}{l}y-\frac{4}{5}(6)+\frac{1}{5}(4)=-4\\ \Rightarrow y-4=-4\\ \Rightarrow y=0\end{array}$

Substituting above finding values in the first equation, we get

  $\begin{array}{l}x-4(0)+3(6)-2(4)=9\\ \Rightarrow x-0+18-8=9\\ \Rightarrow x+10=9\\ \Rightarrow x=-1\end{array}$

Conclusion:

Therefore, the solution of the given system is $x=-1,y=0,z=6\text{and}w=4$ .