Chapter 7.4, Problem 66E

To determine

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

Answer to Problem 66E

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.

Two matrices are row-equivalent when one can be 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 $\{\begin{array}{l}2x-y+3z=24\\ 2y-z=14\\ 7x-5y=6\end{array}$ 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 $\left(8,10,6\right)$ .