Chapter 8.1, Problem 85E

To determine

Find the solution set that can be written as an order triple.

Answer to Problem 85E

The solution set can be written as an order triple with the form

  $\left(0,2–4a,a\right)$

  $(-2a,a,a,0)$

Explanation of Solution

Given:

The system of equations

  $\begin{array}{l}3x+3y+12z=6\hfill \\ x+y+4z=2\hfill \\ 2x+5y+20z=10\hfill \\ –x+2y+8z=4\hfill \end{array}$

Let consider the following system of equations

  $\begin{array}{l}3x+3y+12z=6\hfill \\ x+y+4z=2\hfill \\ 2x+5y+20z=10\hfill \\ –x+2y+8z=4\hfill \end{array}$

The augmented matrix for the above system of equations is

  $\left[\begin{array}{cccc}3& 3& 12& 6\\ 1& 1& 4& 2\\ 2& 5& 20& 10\\ -1& 2& 8& 4\end{array}\right]$

Let's use matrix capabilities of a graphing utility or graphing calculator to write the matrix in reduced row-echelon form and reduced row-echelon form as

  $\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 4& 2\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$

By converting back to the corresponding system of equations,

  $\begin{array}{l} x=0\hfill \\ y+4z=2\hfill \end{array}$

Solving for y in terms of z,

  $y=2–4z$

Here the variables y and z depend on each other, so infinite number of solutions.

To write a solution of the system that does not use any of the three variables of the system, let a is real number such that

  $z=a$

Now, by the above equations,

  $x=0$ and $y=2–4a$

So, the solution set can be written as an order triple with the form

  $\left(0,2–4a,a\right)$ Where, a is any real number.