Chapter 8.1, Problem 83E

To determine

Using matrix capabilities of a graphing utility or graphing calculator to write the matrix in row-echelon form

Answer to Problem 83E

The solution set can be written as

  $(-2a,a,a,0)$

Explanation of Solution

Given:

The system of equations

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

Let consider the following system of equations

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

The augmented matrix for the above system of equations is

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

Use matrix capabilities of a graphing utility or graphing calculator to write the matrix in row-echelon form and get the row-echelon form as

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

By converting back to the corresponding system of equations,

  $\begin{array}{c}x+2z+\frac{5}{2}w=0\\ y-z-\frac{3}{2}w=0\\ w=0\end{array}$

Solving for x and y in terms of z,

  $x=-2z$

  $y=z$

Here the variables x,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 four variables of the system, let a is a real number such that

  $z=a$

Now, by the above equation,

  $x=–2a$ and $y=a$

So, the solution set can be written as

  $(-2a,a,a,0)$

Where, a is any real number.