Chapter 8.1, Problem 75E

To determine

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

Answer to Problem 75E

The solutions to the given system of equation are $\boxed{\left(7,-3,4\right)}$

Explanation of Solution

Given information: The system of equation is $\{\begin{array}{l}-x+y-z=-14\\ 2x-y+z=21\\ 3x+2y+z=19\end{array}$

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.

Calculation:

Write the system of equation $\{\begin{array}{l}-x+y-z=-14\\ 2x-y+z=21\\ 3x+2y+z=19\end{array}$ in augmented form

  $\left[\begin{array}{cccc}-1& 1& -1& ⋮-14\\ 2& -1& 1& ⋮21\\ 3& 2& 1& ⋮19\end{array}\right]$

Multiply -1 with the first Row

  $\begin{array}{c}-1R_1\to \\ \\ \end{array}\left[\begin{array}{cccc}1& -1& 1& ⋮14\\ 2& -1& 1& ⋮21\\ 3& 2& 1& ⋮19\end{array}\right]$

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

  $\begin{array}{c}\\ -2R_1+R_2\to \\ \end{array}\left[\begin{array}{cccc}1& -1& 1& ⋮14\\ 0& 1& -1& ⋮-7\\ 3& 2& 1& ⋮19\end{array}\right]$

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

  $\begin{array}{c}\\ \\ -3R_1+R_3\to \end{array}\left[\begin{array}{cccc}1& -1& 1& ⋮14\\ 0& 1& -1& ⋮-7\\ 0& 5& -2& ⋮-23\end{array}\right]$

Add -5 times the 2nd row to the 3rd row

  $\begin{array}{c}\\ \\ -5R_2+R_3\to \end{array}\left[\begin{array}{cccc}1& -1& 1& ⋮14\\ 0& 1& -1& ⋮-7\\ 0& 0& 3& ⋮12\end{array}\right]$

Multiply the 3rd row by 1/3

  $\begin{array}{c}\\ \\ 1/3R_3\to \end{array}\left[\begin{array}{cccc}1& -1& 1& ⋮14\\ 0& 1& -1& ⋮-7\\ 0& 0& 1& ⋮4\end{array}\right]$

Add 1 times the 3rd row to the 2nd row

  $\begin{array}{c}\\ R_3+R_2\to \\ \end{array}\left[\begin{array}{cccc}1& -1& 1& ⋮14\\ 0& 1& 0& ⋮-3\\ 0& 0& 1& ⋮4\end{array}\right]$

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

  $\begin{array}{c}-1R_3+R_2\to \\ \\ \end{array}\left[\begin{array}{cccc}1& -1& 0& ⋮10\\ 0& 1& 0& ⋮-3\\ 0& 0& 1& ⋮4\end{array}\right]$

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

  $\begin{array}{c}{R}_2+R_1\to \\ \\ \end{array}\left[\begin{array}{cccc}1& 0& 0& ⋮7\\ 0& 1& 0& ⋮-3\\ 0& 0& 1& ⋮4\end{array}\right]$

The matrix is now in reduced row-echelon form. Converting back to a system of linear equations, you have

  $\{\begin{array}{l}x=7\\ y=-3\\ z=4\end{array}$

Conclusion:

So, the solution set can be written as an ordered triple of the form $\left(7,-3,4\right)$ .