Chapter 7, Problem 94RE

To determine

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

Answer to Problem 94RE

The solutions to the given system of equation are $\boxed{\left(8,10,6\right)}$

Explanation of Solution

Given information:The system of equation is $\{\begin{array}{l}2x-y+3z=24\\ 2y-z=14\\ 7x-5y=6\end{array}$

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 $\{\begin{array}{l}2x-y+3z=24\\ 2y-z=14\\ 7x-5y=6\end{array}$ in augmented form

  $\left[\begin{array}{cccc}2& -1& 3& ⋮24\\ 0& 2& -1& ⋮14\\ 7& -5& 0& ⋮6\end{array}\right]$

Multiply ½ with the first Row

  $\begin{array}{c}1/2R_1\to \\ \\ \end{array}\left[\begin{array}{cccc}1& -1/2& 3/2& ⋮12\\ 0& 2& -1& ⋮14\\ 7& -5& 0& ⋮6\end{array}\right]$

Multiply ½ with the second Row

  $\begin{array}{c}\\ 1/2R_2\to \\ \end{array}\left[\begin{array}{cccc}1& -1/2& 3/2& ⋮12\\ 0& 1& -1/2& ⋮7\\ 7& -5& 0& ⋮6\end{array}\right]$

Multiply -7 with the first Row and add it with the third Row

  $\begin{array}{c}\\ \\ -7R_1+R_3\to \end{array}\left[\begin{array}{cccc}1& -1/2& 3/2& ⋮12\\ 0& 1& -1/2& ⋮7\\ 0& -3/2& -21/2& ⋮-78\end{array}\right]$

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

  $\begin{array}{c}\\ \\ 3/2R_2+R_3\to \end{array}\left[\begin{array}{cccc}1& -1/2& 3/2& ⋮12\\ 0& 1& -1/2& ⋮7\\ 0& 0& -45/4& ⋮-135/2\end{array}\right]$

Multiply the 3rd row by -4/45

  $\begin{array}{c}\\ \\ -4/45R_3\to \end{array}\left[\begin{array}{cccc}1& -1/2& 3/2& ⋮12\\ 0& 1& -1/2& ⋮7\\ 0& 0& 1& ⋮6\end{array}\right]$

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

  $\begin{array}{c}\\ 1/2R_3+R_2\to \\ \end{array}\left[\begin{array}{cccc}1& -1/2& 3/2& ⋮12\\ 0& 1& 0& ⋮10\\ 0& 0& 1& ⋮6\end{array}\right]$

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

  $\begin{array}{c}-3/2R_3+R_1\to \\ \\ \end{array}\left[\begin{array}{cccc}1& -1/2& 0& ⋮3\\ 0& 1& 0& ⋮10\\ 0& 0& 1& ⋮6\end{array}\right]$

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

  $\begin{array}{c}1/2R_2+R_1\to \\ \\ \end{array}\left[\begin{array}{cccc}1& 0& 0& ⋮8\\ 0& 1& 0& ⋮10\\ 0& 0& 1& ⋮6\end{array}\right]$

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

  $\{\begin{array}{l}x=8\\ y=10\\ z=6\end{array}$

Conclusion:

So, the solution set can be written as an ordered triple of the form $\left(8,10,6\right)$ .