Chapter 7.3, Problem 29E

To determine

To Find: The solution of the system of linear equation and check the solution algebraically

Answer to Problem 29E

The solution is $x=1,y=\frac{-3}{2},z=\frac{1}{2}$ which can be written as the ordered triple $\left(1,\frac{-3}{2},\frac{1}{2}\right)$

Explanation of Solution

Given information: The given system of three linear equations in three variables are

  $\begin{array}{l}3x+3y+5z=1\text{ Equation }1\\ 3x+5y+9z=0\text{ Equation }2\\ 5x+9y+17z=0\text{ Equation }3\end{array}$

Concept used:Gaussian Elimination is the process of using valid row operations on a matrix until it is in reduced row echelon form

Calculation:

  $\begin{array}{l}3x+3y+5z=1\text{ Equation }1\\ 3x+5y+9z=0\text{ Equation }2\\ 5x+9y+17z=0\text{ Equation }3\end{array}$

Adding $-1$ times the first equation to the second equation

  $\begin{array}{l}3x+3y+5z=1\text{ Equation }1\\ 3x+5y+9z=0\text{ Equation }2\end{array}$

  ______________________.

  $2y+4z=-1$

Adding $-1$ times the first equation to the second equation produces a new second equation

  $\begin{array}{l}3x+3y+5z=1\text{ Equation }1\\ 2y+4z=-1\text{ Equation }2\\ 5x+9y+17z=0\text{ Equation }3\end{array}$

Adding $-5$ times the first equation to $\text{3 }$ times the third equation

  $\begin{array}{l}-15x-15y-25z=-5\\ 15x+27y+51z=0\end{array}$

  ______________________.

  $12y+26z=-5$

Adding $-5$ times the first equation to $\text{3 }$ times the third equation produces a new third equation

  $\begin{array}{l}3x+3y+5z=1\text{ Equation }1\\ 2y+4z=-1\text{ Equation }2\\ 12y+26z=-5\text{ Equation }3\end{array}$

Adding $-6$ times the second equation to the third equation

  $\begin{array}{l}-12y-24z=6\\ 12y+26z=-5\end{array}$

  ______________________.

  $2z=1$

Adding $-6$ times the second equation to the third equation produces a new third equation

  $\begin{array}{l}3x+3y+5z=1\text{ Equation }1\\ 2y+4z=-1\text{ Equation }2\\ 2z=1\text{ Equation }3\end{array}$

Finally, we need a coefficient of $1$ for $z$ in the third equation

Multiplying the third equation by $\frac{1}{2}$ produces a new third equation

  $\begin{array}{l}3x+3y+5z=1\text{ Equation }1\\ 2y+4z=-1\text{ Equation }2\\ z=\frac{1}{2}\text{ Equation }3\end{array}$

From $Equation\text{ 3 }$ we know the value of $z$ . To solve for $x$ , substitute $z=\frac{1}{2}$ into $\text{Equation 2}$ to obtain

  $y=\frac{-3}{2}$

Finally, substitute

  $y=\frac{-3}{2}$

And $z=\frac{1}{2}$ into equation 1 to obtain

  $x=1$

The solution is $x=1,y=\frac{-3}{2},z=\frac{1}{2}$ which can be written as the ordered triple $\left(1,\frac{-3}{2},\frac{1}{2}\right)$