Chapter 7.4, Problem 70E

To determine

To Find: the value of $x,y,z$

Answer to Problem 70E

Solutions of the system of equation is $x=22,y=36,z=114$

Explanation of Solution

Given information: System of equations are

  $\begin{array}{l}2x+2y-z=2\\ x-3y+z=28\\ -x+y=14 \end{array}$

Concept Used: Gauss-Jordan Elimination is the process of performing row operations to transform any matrix into row echelon form. In reduced row echelon form, each successive row of the matrix has less dependencies than the previous

Calculation:

System of equations are:

  $\begin{array}{l}2x+2y-z=2\\ x-3y+z=28\\ -x+y=14 \end{array}$

Gauss-Jordan Elimination is used to obtain the row echelon form of the linear system above

Write the augmented matrix:

  $\left[\begin{array}{ccccc}2& 2& -1& :& 2\\ 1& -3& 1& :& 28\\ -1& 1& 0& :& 14\end{array}\right]$

Now, apply elementary row operations until obtain zeros above each of the leading $1\text{'}s$ , as follows:

  $\left[\begin{array}{ccccc}2& 2& -1& :& 2\\ 1& -3& 1& :& 28\\ -1& 1& 0& :& 14\end{array}\right]$

  $\left[\begin{array}{ccccc}1& 1& -0.5& :& 1\\ 1& -3& 1& :& 28\\ -1& 1& 0& :& 14\end{array}\right]R_1\to \frac{1}{2}{R}_1$

  $\left[\begin{array}{ccccc}1& 1& -0.5& :& 1\\ 0& -4& 1.5& :& 27\\ -1& 1& 0& :& 14\end{array}\right]R_2\to R_2-R_1$

  $\left[\begin{array}{ccccc}1& 1& -0.5& :& 1\\ 0& -4& 1.5& :& 27\\ 0& 2& -0.5& :& 15\end{array}\right]R_3\to R_3+R_1$

  $\left[\begin{array}{ccccc}1& 1& -0.5& :& 1\\ 0& 0& 0.5& :& 57\\ 0& 2& -0.5& :& 15\end{array}\right]R_2\to R_2+2R_3$

  $\left[\begin{array}{ccccc}1& 1& -0.5& :& 1\\ 0& 2& 0& :& 72\\ 0& 2& -0.5& :& 15\end{array}\right]R_2\to R_2+R_3$

  $\left[\begin{array}{ccccc}1& 1& -0.5& :& 1\\ 0& 1& 0& :& 36\\ 0& 2& -0.5& :& 15\end{array}\right]R_2\to \frac{1}{2}{R}_2$

  $\left[\begin{array}{ccccc}-1& 0& 0.5& :& 35\\ 0& 1& 0& :& 36\\ 0& 2& -0.5& :& 15\end{array}\right]R_1\to -R_1 +R_2$

  $\left[\begin{array}{ccccc}-1& 0& 0.5& :& 35\\ 0& 1& 0& :& 36\\ 0& 0& 1& :& 114\end{array}\right]R_3\to -R_3+2R_2$

  $\left[\begin{array}{ccccc}-1& 0& 0& :& -22\\ 0& 1& 0& :& 36\\ 0& 0& 1& :& 114\end{array}\right]R_1\to R_1 -\frac{1}{2}{R}_3$

  $\left[\begin{array}{ccccc}1& 0& 0& :& 22\\ 0& 1& 0& :& 36\\ 0& 0& 1& :& 114\end{array}\right]R_1\to -R_1$

The matrix is now in reduced row echelon form. Converting back to a system of linear equation

  $x=22,y=36,z=114$

This solution can be written as the ordered triple $\left(x,y,z\right)=(22,36,114)$