Chapter 10.2, Problem 19E

To determine

To find: the complete solution of the linear system, or show that it is inconsistent

Answer to Problem 19E

The solution of the system is $x=1$ , $y=2$ and $z=1$ . The solution as the ordered triple is $\left(1,2,1\right)$ .

Explanation of Solution

Given:

Consider the following system of equations:

  $\{\begin{array}{l}x+y+z=4\\ x+3y+3z=10\\ 2x+y-z=3\end{array}$

Calculation:

To change the given system to triangular system, first eliminate the x variable from equation (2) and (3).

Step 1: Perform the following operation to eliminate x from equation (2)

  $\{\begin{array}{l}x+y+z=4\hfill \\ \begin{array}{l}x+3y+3z=10Equation\left(2\right)+\left(-1\right)\times Equation\left(1\right)=NewEquation\left(2\right)\\ 2x+y-z=3\end{array}\hfill \end{array}$

That is,

  $\begin{array}{l}\underset{\_}{\begin{array}{l}x+3y+3=10\\ -x-y-z=-4\end{array}}\hfill \\ 2y+2z=6\hfill \end{array}$

This gives new equivalent system as shown below:

  $\begin{array}{l}x+y+z=4\\ 2y+2z=6\\ 2x+y-z=3\end{array}$

Step 2: Perform the following operation to eliminate x from equation (3):

  $\{\begin{array}{l}x+y+z=4\\ 2y+2z=6\\ 2x+y+z=3Equation\left(3\right)+\left(-2\right)\times Equation\left(1\right)=NewEquation\left(3\right)\end{array}$

That is,

  $\begin{array}{l}\underset{\_}{\begin{array}{l}2x+y-z=3\\ -2x-2y–2z=-8\end{array}}\hfill \\ -y-3z=-5\hfill \end{array}$

This gives new equivalent system as shown below:

  $\begin{array}{l}x+y+z=4\hfill \\ \begin{array}{l}2y+2z=6\\ -y-3z=-5\end{array}\hfill \end{array}$

Step 3: Perform the following operation to eliminate y from equation (3):

  $\{\begin{array}{l}x+y+z=4\\ 2y+2z=6\\ -y-3z=-5Equation\left(3\right)+\left(\frac{1}{2}\right)\times Equation\left(2\right)=NewEquation\left(3\right)\end{array}$

That is,

  $\begin{array}{l}\underset{\_}{\begin{array}{l}-y-3z=-5\\ y+z=3\end{array}}\hfill \\ -2z=-2\hfill \end{array}$

Therefore, the new equivalent system which is in triangular form is shown below:

  $\begin{array}{l} x+y+z=4\mathrm{......}\left(4\right)\\ 2y+2z=6\mathrm{......}\left(5\right)\\ -2z=-2\mathrm{......}\left(6\right)\end{array}$

Solve this system by using back substitution method.

From the equation (6), $-2z=-2$

First solve for z:

  $-2z=-2$

  $z=1$ Divide both sides by -2

Now plug the $z=1$ value into the equation (5) and solve for y as shown below:

  $2y+2z=6$

  $2y+2\left(1\right)=6$ Back substitute $z=1$ into equation (5)

  $2y=4$ Subtract 2 from both sides

  $y=2$ Divide both sides by 2

Next substitute $y=2$ and $z=1$ values in the equation (4), and solve for x as shown below:

  $x+y+z=4$

  $x+2+1=4$ Back substitute $y=2$ and $z=1$ into equation (4)

  $x=1$ Subtract 3 from both sides

Therefore, the solution of the system is $x=1$ , $y=2$ and $z=1$ . The solution as the ordered triple is $\left(1,2,1\right)$ .

Conclusion:

Therefore, the solution of the system is $x=1$ , $y=2$ and $z=1$ . The solution as the ordered triple is $\left(1,2,1\right)$ .