Chapter 8.1, Problem 80E

To determine

To find: The solution to the given system of linear equations

Answer to Problem 80E

The solutions to the given system of equation are $\boxed{\left(-5a,a,3\right)}$

Explanation of Solution

Given information: The system of equation is $\{\begin{array}{l}2x+10y+2z=6\\ x+5y+2z=6\\ x+5y+z=3\\ -3x-15y-3z=-9\end{array}$

Concept Involved:

Solution of a system of equation is the point which makes both the equation TRUE.

Graphically the solution to the system of equation is the point where the two lines meet.

A matrix derived from a system of linear equations (each written in standard form with the constant term on the right) is the augmented matrix of the system.

Elementary Row Operation:

The three operations that can be used on a system of linear equations to produce an equivalent system.

Operation	Notation
1.Interchange two equations	$R_a\leftrightarrow R_b$
2. Multiply an equation by a nonzero constant	$c{R}_a\left(c\ne 0\right)$
3. Add a multiple of an equation to another equation.	$c{R}_a+R_b$

In matrix terminology, these three operations correspond to elementary row operations.

An elementary row operation on an augmented matrix of a given system of linear equations produces a new augmented matrix corresponding to a new (but equivalent) system of linear equations.

Two matrices are row-equivalent when one can be obtained from the other by a sequence of elementary row operations.

Calculation:

Write the system of equation $\{\begin{array}{l}2x+10y+2z=6\\ x+5y+2z=6\\ x+5y+z=3\\ -3x-15y-3z=-9\end{array}$ in augmented form

  $\left[\begin{array}{ccccc}2& 10& 2& ⋮& 6\\ 1& 5& 2& ⋮& 6\\ 1& 5& 1& ⋮& 3\\ -3& -15& -3& ⋮& -9\end{array}\right]$

Multiply ½ with the first Row

  $\begin{array}{c}1/2R_1\to \\ \\ \end{array}\left[\begin{array}{ccccc}1& 5& 1& ⋮& 3\\ 1& 5& 2& ⋮& 6\\ 1& 5& 1& ⋮& 3\\ -3& -15& -3& ⋮& -9\end{array}\right]$

Add -1 times the 1st row to the 2nd row

  $\begin{array}{c}\\ -1R_1+R_2\to \\ \\ \end{array}\left[\begin{array}{ccccc}1& 5& 1& ⋮& 3\\ 0& 0& 1& ⋮& 3\\ 1& 5& 1& ⋮& 3\\ -3& -15& -3& ⋮& -9\end{array}\right]$

Add -1 times the 1st row to the 3rd row

  $\begin{array}{c}\\ \\ -1R_1+R_3\to \\ \end{array}\left[\begin{array}{ccccc}1& 5& 1& ⋮& 3\\ 0& 0& 1& ⋮& 3\\ 0& 0& 0& ⋮& 0\\ -3& -15& -3& ⋮& -9\end{array}\right]$

Add -1 times the 1st row to the 4th row

  $\begin{array}{c}\\ \\ \\ 3R_1+R_4\to \end{array}\left[\begin{array}{ccccc}1& 5& 1& ⋮& 3\\ 0& 0& 1& ⋮& 3\\ 0& 0& 0& ⋮& 0\\ 0& 0& 0& ⋮& 0\end{array}\right]$

Add -1 times the 2nd row to the 1^st row

  $\begin{array}{c}-R_2+R_1\to \\ \\ \\ \end{array}\left[\begin{array}{ccccc}1& 5& 0& ⋮& 0\\ 0& 0& 1& ⋮& 3\\ 0& 0& 0& ⋮& 0\\ 0& 0& 0& ⋮& 0\end{array}\right]$

The corresponding system of equation is

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

Solving x in terms of y, you have

  $x=-5y$

To write a solution of the system that does not use any of the three variables of the system, let $a$ represent any real number and let $y=a$ . Substitute $a$ for yin the equations for x.

  $x=-5a$

Conclusion:

So, the solution set can be written as an ordered triple of the form $\left(-5a,a,3\right)$ where a is any real number. Remember that a solution set of this form represents aninfinite number of solutions.