Chapter 7.4, Problem 68E

To determine

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

Answer to Problem 68E

The solution to the given system is all of the form

  $x=\frac{3-3a}{2},y=\frac{1+a}{3},z=a$

Explanation of Solution

Given information: System of equations are

  $\begin{array}{l}2x+3z=3\\ 4x-3y+7z=5\\ 8x-9y+15z=9 \end{array}$

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

  $\begin{array}{l}2x+3z=3\\ 4x-3y+7z=5\\ 8x-9y+15z=9 \end{array}$

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

Write the augmented matrix:

  $\left[\begin{array}{ccccc}2& 0& 3& :& 3\\ 4& -3& 7& :& 5\\ 8& -9& 15& :& 9\end{array}\right]$

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

  $\left[\begin{array}{ccccc}2& 0& 3& :& 3\\ 4& -3& 7& :& 5\\ 8& -9& 15& :& 9\end{array}\right]$

  $R_2\to R_2-2R_1$

  $\left[\begin{array}{ccccc}2& 0& 3& :& 3\\ 0& -3& 1& :& -1\\ 8& -9& 15& :& 9\end{array}\right]$

  $R_1\to \frac{1}{2}{R}_1$

  $\left[\begin{array}{ccccc}1& 0& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& :& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& -3& 1& :& -1\\ 8& -9& 15& :& 9\end{array}\right]$

  $R_2\to \frac{-1}{3}{R}_2$

  $\left[\begin{array}{ccccc}1& 0& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& :& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& 1& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& :& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 8& -9& 15& :& 9\end{array}\right]$

  $R_3\to R_3-8R_1$

  $\left[\begin{array}{ccccc}1& 0& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& :& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& 1& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& :& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 0& -9& 3& :& -3\end{array}\right]$

  $R_3\to R_3+9R_2$

  $\left[\begin{array}{ccccc}1& 0& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.& :& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0& 1& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.& :& \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\\ 0& 0& 0& :& 0\end{array}\right]$

The corresponding system of equation is

  $x+\frac{3}{2}z=\frac{3}{2}$ equation 1, $y-\frac{1}{3}z=\frac{1}{3}$ equation 2, $0=0$ equation 3

The result means that equation 3 depends on equation 1 and equation2

Finally, letting $z=a$

Where $\text{'}a\text{'}$ is a realnumber

Now substitute $\text{'}a\text{'}$ for z in the equations forx and y ,

  $y=\frac{1+a}{3}$

  $x=\frac{3-3a}{2}$

The solution to the given system is all of the form

  $x=\frac{3-3a}{2},y=\frac{1+a}{3},z=a$

So, the solution set can be written as an ordered triple with the form $\left(\frac{3-3a}{2},\frac{1+a}{3},a\right)$