Chapter 7, Problem 82RE

To determine

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

Answer to Problem 82RE

The solutions to the given system of equation are $\boxed{\left(4,-3,2\right)}$

Explanation of Solution

Given information: The system of equation is $\{\begin{array}{l}x-3z=-2\\ 3x+y-2z=5\\ 2x+2y+z=4\end{array}$

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 $\{\begin{array}{l}x-3z=-2\\ 3x+y-2z=5\\ 2x+2y+z=4\end{array}$ in augmented form

  $\left[\begin{array}{cccc}1& 0& -3& ⋮-2\\ 3& 1& -2& ⋮5\\ 2& 2& 1& ⋮4\end{array}\right]$

Multiply -3 with the first Row and add it with the second Row

  $\begin{array}{c}\\ -3R_1+R_2\to \\ \end{array}\left[\begin{array}{cccc}1& 0& -3& ⋮-2\\ 0& 1& 7& ⋮11\\ 2& 2& 1& ⋮4\end{array}\right]$

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

  $\begin{array}{c}\\ \\ -2R_1+R_3\to \end{array}\left[\begin{array}{cccc}1& 0& -3& ⋮-2\\ 0& 1& 7& ⋮11\\ 0& 2& 7& ⋮8\end{array}\right]$

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

  $\begin{array}{c}\\ \\ -2R_2+R_3\to \end{array}\left[\begin{array}{cccc}1& 0& -3& ⋮-2\\ 0& 1& 7& ⋮11\\ 0& 0& -7& ⋮-14\end{array}\right]$

Multiply the 3rd row by -1/7

  $\begin{array}{c}\\ \\ -1/7R_3\to \end{array}\left[\begin{array}{cccc}1& 0& -3& ⋮-2\\ 0& 1& 7& ⋮11\\ 0& 0& 1& ⋮2\end{array}\right]$

Add -7 times the 3rd row to the 2nd row

  $\begin{array}{c}\\ -7R_3+R_2\to \\ \end{array}\left[\begin{array}{cccc}1& 0& -3& ⋮-2\\ 0& 1& 0& ⋮-3\\ 0& 0& 1& ⋮2\end{array}\right]$

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

  $\begin{array}{c}3R_3+R_1\to \\ \\ \end{array}\left[\begin{array}{cccc}1& 0& 0& ⋮4\\ 0& 1& 0& ⋮-3\\ 0& 0& 1& ⋮2\end{array}\right]$

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

  $\{\begin{array}{l}x=4\\ y=-3\\ z=2\end{array}$

Conclusion:

So, the solution set can be written as an ordered triple of the form $\left(4,-3,2\right)$ .