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Homework Help: Linear Algebra Matrices Problem(s)

  1. Sep 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Here is the problem:

    http://img801.imageshack.us/img801/6770/oaza.png [Broken]

    2. Relevant equations

    None really, just gauss jordon elimination I assume unless I am missing out on something

    3. The attempt at a solution

    First I multiplied the first row by -5 then added it to the second row to obtain the following:

    1 1 5 3
    0 -1 -27 -8

    and then multiplied the second row by 1 and added the result to the first row to obtain:

    1 0 -22 -5
    0 -1 -27 -8

    I then tried to solve inputting the numbers into the boxes but apparently I can't put letters or x1 x2 x3 etc.. meaning that the solution is just numbers ie: 1 + 2 is the solution, which is not what I got. My question is, and im having a little trouble wondering what they want is why do they have 2 separate columns? Am I supposed to transpose afterwards or did I make a mistake doing the elimination progress, I more or less want to know the steps more than anything as I have more questions like this one and have no idea how to go about them.

    here are the other 2 :

    http://img694.imageshack.us/img694/5315/m0a5.png [Broken]
    http://img843.imageshack.us/img843/4208/2cu0.png [Broken]

    thanks in advance!
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 24, 2013 #2


    Staff: Mentor

    This looks find, but you need a bit more work. This table represents this system:
    x1 = 22x3 -5
    x2 = -27x3 8
    x3 = x3 + 0

    This represents an infinite number of solutions, one solution for each value of x3.
    One of the boxes they show has a parameter, s, which has the same role as x3.

    One way to think of this is that all solutions x have the form <-5, -8, 0> + s<22, -27, 1>.
    Last edited by a moderator: May 6, 2017
  4. Sep 24, 2013 #3
    I think you meant positive 8, and it works with +8, thanks! I will attempt the other 2 and revive this if help is needed
  5. Sep 24, 2013 #4
    Ok, I already have come across a small issue for this problem:

    http://img694.imageshack.us/img694/5315/m0a5.png [Broken]

    The issue with this problem is that in the solution they are asking for "s" and "t" which I am guessing are substitutes for either x,y or z. Now when I break down the equation I get this as the solution:

    x = (7y - 3z + 11)/2
    y = (-2x -3z + 11)/-7
    z = (-2x + 7y + 11)/3

    So the only problem I can seem to find as to why it wont accept my answers (i didn't write it like above) is that the values of x y and z are being used in the equation and so when I try to input my answers one of the boxes won't make sense since one value has to be s and the other t and so I can't input the last value...anyways thats how I see it unless anyone could shed some light on this? any help is highly appreciated, thank you.
    Last edited by a moderator: May 6, 2017
  6. Sep 24, 2013 #5


    Staff: Mentor

    The given "system" is 2x - 7y + 3z = 11, which means you have one equation in three unknowns. This leaves 2 variables free, and that corresponds to having two parameters.

    Instead of solving for x in terms of y and z, and then y in terms of x and z, and then z in terms of x and y, just solve for x in terms of y and z, and leave y and z and parameters.

    Code (Text):

    x = (7/2)y - (3/2)z + 11/2
    y = .....y
    z = ..............z
    y and z are free variables (parameters), so you could call them s and t.
    If you squint your eyes at the equations above, you might notice that the right side consists of three vectors: two of them with parameters and one that is a constant.
    Last edited by a moderator: May 6, 2017
  7. Sep 24, 2013 #6


    Staff: Mentor

    Yeah, I did. I fixed it in my post.
  8. Sep 25, 2013 #7


    User Avatar
    Science Advisor

    This is NOT a "solution" at all because in order to find "x" you have to know "y" and in order to find "y" you have to know "x"!

    Starting from the intial equations [itex]x_1+ x_2+ 5x_3= 3[/itex], [itex]5x_1+ 4x_2- 2x_3= 7[/itex], you should see that there are two equations with three unknows so the "solution space" will have dimension 3- 2= 1. (Strictly speaking, because these equation are not "= 0", the set of all solutions is a "linear manifold" rather than a "subspace".) Geometrically, each equation is that of a plane and they are both satisfied on the intersection of those planes, a line. In any case we expect the general sollution to have a single parameter.

    Seeing that, in the first equation, [itex]x_1[/itex] has coefficient "1", the simplest thing to do is to eliminate [itex]x_1[/itex] by multiplying the first equation by 5 and subracting that from the second equation: [itex](5- 5)x_1+ (4- 5)x_2- (2+ 25)x_3= -x_2+ 27x_3= -8[/itex].

    We now have a single equation in two unknowns so the best we can do is solve for [itex]x_2[/itex] "in terms of" [itex]x_3: x_2= 27x_3+ 8[/itex]. We can then go back to the original equations, replace [itex]x_2[/itex] by [itex]27x_3+ 8[/itex] and solve for [itex]x_1[/itex] in terms of [itex]x_3[/itex]. Then we could, say, choose [itex]x_3[/itex] itself as parameter.

    You understand, I hope, that, because there are an infinite number of parmeterizations for a straight line, there are an infinite number of "correct" answers for this problem.
    Last edited by a moderator: May 6, 2017
  9. Sep 25, 2013 #8
    thanks for the lengthy response, I understood that first question after mark explained it, but now having troubles with the second one:

    2x - 7y +3z = 11

    how would I find x or y if I am always going to have 2 variables that I won't be able to find?
  10. Sep 25, 2013 #9


    Staff: Mentor

    I talked about this in post #5.
  11. Sep 25, 2013 #10
    I am not sure i quite understand what you mean by when you say

    "Instead of solving for x in terms of y and z, and then y in terms of x and z, and then z in terms of x and y, just solve for x in terms of y and z, and leave y and z and parameters".

    I thought thats what i did above, but of course theres the extra variable case. I kind of understand what you mean by parameters, i guess assigning y and z a different letter or equation, but im having trouble trying to solve that. Im really just lost after I assign each variable a new letter, where do I go from there?
    Last edited: Sep 25, 2013
  12. Sep 25, 2013 #11
    any last minute help?! I need to know within the next 10 minutes, thanks!
  13. Sep 26, 2013 #12


    User Avatar
    Science Advisor

    You seem to be misunderstanding everything you have been told here. Once again, you cannot solve for specific values of x, y, and z because there is no single solution. Geometrically, the two equations you give represent two planes. (x, y, z) points that satisfy both equations must lie on both planes and so lie on the intersection of the two planes. That intersection is a line- a one dimensional subset so that all solutions, all points on that line, can be written in terms of a single parameter. What I[ would do, though there are many other ways to do this, is exactly what I did before: combine the two equations so as to eliminate one of the unknowns (I eliminated [itex]x_1[/itex] or "x") so that you have a single equation in terms of the two other variables. You can then solve that for one variable in terms of the other:
    [itex]x_2= 27x_3+ 8[/itex]. Use that to solve for [itex]x_1[/itex] also in terms of [itex]x_3[/itex] and write "parametric" equations for the line using [itex]x_3[/itex] as parameter.

    Since you seem to be having great difficulty understanding any of this, perhaps we need to "back up". Do you understand how a single equation in three variables "represents" a plane?
    Do you understand that the intersection of two planes is a line? Do you understand how to write parametric equations for a line? All of those are pretty much pre-requisites for a problem like this.
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