Chapter 8.1, Problem 82E

To determine

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

Answer to Problem 82E

The solutions to the given system of equation are $\boxed{\left(-6,8,2\right)}$

Explanation of Solution

Given information: The system of equation is $\{\begin{array}{l}2x+2y-z=2\\ x-3y+z=-28\\ -x+y=14\end{array}$

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 $\{\begin{array}{l}2x+2y-z=2\\ x-3y+z=-28\\ -x+y=14\end{array}$ in augmented form

  $\left[\begin{array}{cccc}2& 2& -1& ⋮2\\ 1& -3& 1& ⋮-28\\ -1& 1& 0& ⋮14\end{array}\right]$

Multiply ½ with the first Row

  $\begin{array}{c}1/2R_1\to \\ \\ \end{array}\left[\begin{array}{cccc}1& 1& -1/2& ⋮1\\ 1& -3& 1& ⋮-28\\ -1& 1& 0& ⋮14\end{array}\right]$

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

  $\begin{array}{c}\\ -1R_1+R_2\to \\ \end{array}\left[\begin{array}{cccc}1& 1& -1/2& ⋮1\\ 0& -4& 3/2& ⋮-29\\ -1& 1& 0& ⋮14\end{array}\right]$

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

  $\begin{array}{c}\\ \\ R_1+R_2\to \end{array}\left[\begin{array}{cccc}1& 1& -1/2& ⋮1\\ 0& -4& 3/2& ⋮-29\\ 0& 2& -1/2& ⋮15\end{array}\right]$

Multiply the 2nd row by -1/4

  $\begin{array}{c}\\ -1/4R_2\to \\ \end{array}\left[\begin{array}{cccc}1& 1& -1/2& ⋮1\\ 0& 1& -3/8& ⋮29/4\\ 0& 2& -1/2& ⋮15\end{array}\right]$

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

  $\begin{array}{c}\\ \\ -2R_2+R_3\to \end{array}\left[\begin{array}{cccc}1& 1& -1/2& ⋮1\\ 0& 1& -3/8& ⋮29/4\\ 0& 0& 1/4& ⋮1/2\end{array}\right]$

Multiply the 3rd row by 4

  $\begin{array}{c}\\ \\ 4R_3\to \end{array}\left[\begin{array}{cccc}1& 1& -1/2& ⋮1\\ 0& 1& -3/8& ⋮29/4\\ 0& 0& 1& ⋮2\end{array}\right]$

Add 3/8 times the 3rd row to the 2nd row

  $\begin{array}{c}\\ 3/8R_3+R_2\to \\ \end{array}\left[\begin{array}{cccc}1& 1& -1/2& ⋮1\\ 0& 1& 0& ⋮8\\ 0& 0& 1& ⋮2\end{array}\right]$

Add 1/2 times the 3rd row to the 1st row

  $\begin{array}{c}1/2R_3+R_1\to \\ \\ \end{array}\left[\begin{array}{cccc}1& 1& 0& ⋮2\\ 0& 1& 0& ⋮8\\ 0& 0& 1& ⋮2\end{array}\right]$

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

  $\begin{array}{c}-1R_2+R_1\to \\ \\ \end{array}\left[\begin{array}{cccc}1& 0& 0& ⋮-6\\ 0& 1& 0& ⋮8\\ 0& 0& 1& ⋮2\end{array}\right]$

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

  $\{\begin{array}{l}x=-6\\ y=8\\ z=2\end{array}$

Conclusion:

So, the solution set can be written as an ordered triple of the form $\left(-6,8,2\right)$ .