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 $\boxed{\left(5a+4,-9a+4,a\right)}$

Explanation of Solution

Given information:The system of equation is $\{\begin{array}{l}x+y+4z=5\\ 2x+y-z=9\end{array}$

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 formwith 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 linearequations 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 rowoperations.

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

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

Row-Echelon Form and Reduced Row-Echelon Form:

A matrix in row-echelon form has the following properties.

1. Any rows consisting entirely of zeros occur at the bottom of the matrix.

2. For each row that does not consist entirely of zeros, the first nonzero entryis 1 (called a leading 1).

3. For two successive (nonzero) rows, the leading 1 in the higher row is fartherto the left than the leading 1 in the lower row.

A matrix in row-echelon form is in reduced row-echelon form when every columnthat has a leading 1 has zeros in every position above and below its leading 1.

Calculation:

Write the system of equation $\{\begin{array}{l}x+y+4z=5\\ 2x+y-z=9\end{array}$ in augmented form

  $\left[\begin{array}{l}\begin{array}{ccccc}1& 1& 4& ⋮& 5\end{array}\\ \begin{array}{ccccc}2& 1& -1& ⋮& 9\end{array}\end{array}\right]$

Multiply -2 with the first Row and add it with the second Row

  $\begin{array}{c}\\ -2R_1+R_2\to \end{array}\left[\begin{array}{l}\begin{array}{ccccc}1& 1& 4& ⋮& 5\end{array}\\ \begin{array}{ccccc}0& -1& -9& ⋮& -1\end{array}\end{array}\right]$

Multiply -1 to the second Row

  $\begin{array}{c}\\ -R_2\to \end{array}\left[\begin{array}{l}\begin{array}{ccccc}1& 1& 4& ⋮& 5\end{array}\\ \begin{array}{ccccc}0& 1& 9& ⋮& 1\end{array}\end{array}\right]$

Multiply -1 to the second Row and add it with the first Row

  $\begin{array}{c}-R_2+R_1\to \\ \end{array}\left[\begin{array}{l}\begin{array}{ccccc}1& 0& -5& ⋮& 4\end{array}\\ \begin{array}{ccccc}0& 1& 9& ⋮& 1\end{array}\end{array}\right]$

The corresponding system of equation is

  $\{\begin{array}{l}x-5z=4\\ y+9z=1\end{array}$

Solving for x and y in terms of z, you have

  $x=5z+4$ and $y=-9z+1$

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 $z=a$ . Substitute $a$ for $z$ in the equations

for xand y.

  $x=5a+4$ and $y=-9a+1$

Conclusion:

So, the solution set can be written as an ordered triple of the form $\left(5a+4,-9a+4,a\right)$ where a is any real number. Remember that a solution set of this form represents aninfinite number of solutions.