Chapter 7.4, Problem 74E

To determine

To Find: Value of $x,y,z$

Answer to Problem 74E

Solutions of the system of equation is $x=-2a,y=a,z=a$

Explanation of Solution

Given information: System of equations are

  $\begin{array}{l}x+y+z=0\\ 2x+3y+z=0 \\ 3x+5y+z=0 \end{array}$

Calculation:

System of equations are

  $\begin{array}{l}x+y+z=0\\ 2x+3y+z=0 \\ 3x+5y+z=0 \end{array}$

Write the augmented matrix:

  $\left[\begin{array}{ccccc}1& 1& 1& :& 0\\ 2& 3& 1& :& 0\\ 3& 5& 1& :& 0\end{array}\right]$

Now, apply elementary row operations until obtain zeros above each of the leading 1’s, as follows:

  $\begin{array}{l}\left[\begin{array}{ccccc}1& 1& 1& :& 0\\ 2& 3& 1& :& 0\\ 3& 5& 1& :& 0\end{array}\right]\\ \left[\begin{array}{ccccc}1& 1& 1& :& 0\\ 0& 1& -1& :& 0\\ 3& 5& 1& :& 0\end{array}\right]R_2\to R_2-2R_1 \\ \left[\begin{array}{ccccc}1& 0& 2& :& 0\\ 0& 1& -1& :& 0\\ 3& 5& 1& :& 0\end{array}\right]R_2\to R_1-R_2\\ \left[\begin{array}{ccccc}1& 0& 2& :& 0\\ 0& 1& -1& :& 0\\ 0& 5& -5& :& 0\end{array}\right]R_3\to R_3 -3R_1\\ \left[\begin{array}{ccccc}1& 0& 2& :& 0\\ 0& 1& -1& :& 0\\ 0& 1& -1& :& 0\end{array}\right]R_3\to \frac{1}{5}{R}_3 \\ \left[\begin{array}{ccccc}1& 0& 2& :& 0\\ 0& 1& -1& :& 0\\ 0& 0& 0& :& 0\end{array}\right]R_3\to R_3-R_2 \end{array}$

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

Finally, letting $z=a$

Where $\text{'}a\text{'}$ is a real numbers

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

  $x=-2a,y=a$

The solution to the given system is all of the form

  $x=-2a,y=a,z=a$