Use parametric equations to describe its solution set

[frctl]

The question from the textbook:
Use parametric equations to describe its solution set.
(Each linear system has infinitely many solutions)

x₁ + 3x₂ - x₃ = -4
3x₁ + 9x₂ - 3x₃ = -12
-x1 - 3x₂ + x₃ = 4

My work and why I am stuck:
In the previous examples, which only had two equations I proceeded to eliminate a variable from the second equation by adding -# times the first equation to the second. This yielded a simplified system

I'm not sure what to do in this case, since there are three equations.

If required, the augmented matrix for this is:
1 3 -1 -4
3 9 -3 -12
-1 -3 1 4

 

[Otis]

… I'm not sure what to do in this case, since there are three equations …

Hi frctl. Additional equations and variables just add more steps; the solution processes are the same.

If you're not told to use a specific method, then you're free to choose. You can reduce the augmented coefficient matrix (Gaussian elimination) OR you can apply substitution/elimination on the original equations. Either way, the steps are similar to what you've been doing in your other threads.

How would you like to proceed? Show us how far you get. (Please double-check your arithmetic, at each step.)

?

 

[Otis]

PS: Here's a tip, when you have a system of equations: For each equation, examine the numbers (coefficients) on each side, to see whether they share a common factor. If they do, then divide each side by that factor (or multiply by the reciprocal, same thing) -- before you write a matrix or start trying to eliminate some variables. Dividing out common factor(s) makes numbers in equations smaller (hopefully, some become 1), and smaller numbers make arithmetic easier.

EG: -21x + 3y - 6z = 9

Those numbers each have a factor of 3, so we multiply each side by the reciprocal 1/3.

-7x + y - 2z = 3

You can use the tip below.

3x₁ + 9x₂ - 3x₃ = -12

After you get a simpler form, write the augmented coefficient matrix and examine the numbers, from row to row. Do you notice two things?

\(\;\)

 

[Steven G]

Are you really sure that you have three different equationsS

Here are the steps if you have three equations in three unlnowns

1) Pick two of the equations and eliminate one variable---this wll give you one equation in two variable
2) Using one equation from step 1 and the 3rd unused equation eliminate the same variable as in step one---this wll give you amother equation in two variable
3) Using the two equations obtained in step 1 and step 2 solve for both unknowns.
4) Pick any of the original equations and solve for the 3rd unknown using the the results from step 3

Please note that it probably is not best to start using this method with the system you listed above as this is a slight problem with this system.

 

[frctl]

1 3 -1 -4
3 9 -3 -12
-1 -3 1 4
Multiply row2 by 1/3
1 3 -1 -4
1 3 -1 -4
-1 -3 1 4

Which operation would be best now?
Row 1 and row2 are the same now.

 

[ksdhart2]

Multiply the third row by -1. What do you notice then? Why does this observation mean you can conclude the system has infinitely many solutions?

 

[pka]

The question from the textbook:
Use parametric equations to describe its solution set.
(Each linear system has infinitely many solutions)
x₁ + 3x₂ - x₃ = -4
3x₁ + 9x₂ - 3x₃ = -12

frcti, can you not see that the above is but one equation written in two forms?
Hint: multiply by three.

 

[HallsofIvy]

I have never liked using matrix methods for such simple sets of equations!

As others have pointed out, the third equation is just the first equation multiplied by -1. That's why we don't just have a single solution but infinitely many solutions that can be given in terms of parametric equations.

From the first equation, \(\displaystyle x_1= -3x_2+ x_3\). Putting that into the second equation, \(\displaystyle 3x_1 + 9x_2 - 3x_3= 3(-4- 3x_2+ x_3)- 3x_2+ x_3= -12- 9x_2+ 3x_3- 3x_2+ x_3= -12-12x_2+ 4x_3 = -12\) or \(\displaystyle -12x_2+ 4x_3= 0\). So \(\displaystyle x_3= 3x_2\). And then \(\displaystyle x1= -3x_2+ x_3= -3x_2+ 3x_2= 0\). Write \(\displaystyle x_2\) and \(\displaystyle x_3\) using \(\displaystyle x_2= t\) as parameter.

 

[frctl]

Should I multiply row3 by 3 or by -1?

 

[frctl]

ah multiplying by -1 gives me
1 3 -1 -4
1 3 -1 -4
1 3 -1 -4

Three equations being the same reveals
the system has infinitely many solutions

 

[HallsofIvy]

Yes, that was the point of asking you to use parametric equations. \(\displaystyle (x_1, x_2, x_3)= (0, t, 3t)\).