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Not sure how to solve matrix/equation system

  1. Jul 20, 2016 #1
    [mentor note: moved here from another forum hence no HW template]

    i have homework about linear algebra(just started learning it) and i got the following question
    "to the linear equation systerm :
    x+6y-18z=0
    -4x+5z=0
    -3x+6y-13z=0
    -2x-2y+z=0
    1. there is infinite solutions
    2. there is only trivial solution
    "
    and i need to check if two of the claims are true or wrong or if only one of them is true.

    if i remember correctly when there is a zero row there is also infinite solutions but what i should do now? i need to try to reduce it to row echelon form and see what i get at the end? do i need to try to "clear" the first colume from all zero numbers? i don't know even how to begin solving this. i am realy new to this and i will be realy thankful for anyone who will tell me what to do and give me some more information.
     
    Last edited by a moderator: Jul 20, 2016
  2. jcsd
  3. Jul 20, 2016 #2

    jedishrfu

    Staff: Mentor

  4. Jul 20, 2016 #3

    perplexabot

    User Avatar
    Gold Member

    First try and formulate your problem as a linear equation (such as Ax=b). Next step, as you said, reduce it so that it is in row echelon form (not necessarily rref) and figure it out from there.

    I would also say try and do a little research first before you post : p
     
  5. Jul 20, 2016 #4
    i try to read as much as i can and do a "research" but i always get stuck with things like "formulate your problem as a linear equation", i can read about linear equation but can't get what you wanted me to do.

    and for the question, can you give me some hint please? i am writing it as a matrix and trying to reduce it but it realy hard, i got a matrix with
    1 6 -18 0
    0 4 3 0
    4 0 -5 0
    2 2 -1 0

    and i think i done the math right but i feel like i will just "ruin" the zeros i made every time i am trying to make some value a zero.
    everything is so short and simple in the videos but i never saw something like this...
     
  6. Jul 20, 2016 #5

    perplexabot

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    Gold Member

    Your initial linear system is as follows:
    [itex]
    \ \ \ \ \ \ x+6y-18z=0\\
    -4x+0y+\ \ 5z=0\\
    -3x+6y-13z=0\\
    -2x-2y+\ \ \ \ z=0
    [/itex]
    Your linear equation [itex]Ax=b[/itex] would be:
    [itex]
    \begin{bmatrix}
    1 & 6 & -18\\
    -4 & 0 & 5\\
    -3&6 &-13\\
    -2&-2&1
    \end{bmatrix}
    \begin{bmatrix}
    x\\
    y\\
    z
    \end{bmatrix}
    =
    \begin{bmatrix}
    0\\
    0\\
    0\\
    0
    \end{bmatrix}
    [/itex]
    If you want to learn how to turn your matrix into row echelon form, you have to read to know how, if we simply give you the answer you will not learn. A very basic link would be this (or the links jedishrfu provided).

    If you don't want to learn how to get the answer yourself (which I think is totally ok!), try and find more efficient methods to get your answer, such as the following: https://www.wolframalpha.com/input/?i=row+reduce+{{1,2},{3,5}} or even better install matlab (or an equivalent) on your computer and use it!

    I put a random matrix in this link, you simply have to write your coefficient matrix (A) in and just hit "enter" : )

    Also, I just realized you have three variables and 4 equations, maybe that says something...
     
  7. Jul 20, 2016 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Forget about matrices, etc. because you do not need them for this problem. Just use grade-school methods to deal with the equations!

    From the second equation you get ##z## in terms of ##x##: ##z = (4/5)x##. Putting that into the other three equations, you will be left with three equations in the two unknowns ##x## and ##y##. Again, solve for one of these variables in terms of the other; for example, you could use the first equation (now in terms of ##x## and ##y## alone) to solve for ##y## in terms of ##x##. Put that expression into the remaining two equations, and you will have two equations in the single variable ##x##. Finally, you will be able to complete the solution by looking at what these two equations actually say about ##x##. However, I will not say any more now; you need to actually sit down and carry out all the steps I outlined, and see what you get.
     
  8. Jul 21, 2016 #7
    i used symbolab site to reduce the system of equation as a matrix into a reduced row echelon form like this
    1 0 0 0
    0 1 0 0
    0 0 1 0
    0 0 0 0
    and followed up after the steps and learned a thing or two about how to use fractions for reducing matrices.
    what does it mean about the solutions? in my book i readed that in that case when such a matrix can be reduced into
    rref form with a row of all zero its mean that the matrix have a non-trivial solution. does i got it right? or i got confused?
     
  9. Jul 21, 2016 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Never mind what the book says; you are making the mistake of trying to apply "rules" and "prescriptions", instead of understanding.

    The fundamental fact is that when you perform row operations on the matrix you are essentially doing the same operations on the systems of equations. So, if you perform something like "New row 2 = Old row 2 - (1/3)* Old row 1" on the matrix, you are really performing the operations "New equation 2 = Old equation 2 - (1/3)* Old equation 1" on the system of equations. The new matrix you get after the row operations just gives you the variables' coefficients in the new system of equations.

    Therefore, your rref above just says
    [tex] \begin{array}{l}
    1 x + 0 y + 0 z = 0\\
    0 x + 1 y + 0 z = 0 \\
    0 x + 0 y + 1 z = 0 \\
    0 x + 0 y + 0 z = 0
    \end{array} [/tex]
    In other words, you have the simple system of equations ##x = 0, y = 0, z = 0## and ##0 = 0##. The last equation is redundant---that is, has no effect.

    Note, however, that if you had non-zero right-hand sides (rather than the zeros you had in your problem) your final equations could take the form ##x = a##, ##y = b##, ##z = c## and ##0 x + 0 y + 0 z = d##. In that case if you happen to have ##d = 0## (after performing all the row operations) then you again would have a unique solution. If you happen to have ##d \neq 0## that would mean that the original system of equations is inconsistent---has no solution.
     
    Last edited: Jul 21, 2016
  10. Jul 21, 2016 #9

    perplexabot

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    Gold Member

    Nice! First, you may find it useful to know that the system of equations you have is a homogeneous system (Ax = b, where b = 0). Let me ask you a few questions. Bear with me.
    1. Do you know what a trivial solution is?
    2. Do you know what a non-trivial solution is?
    3. A homogeneous system always has at least one solution, can you see why? Do you know what type of solution this is (trivial or non trivial)?
    4. So we know there must be at least one solution, but are there more?
    One way to find out (the answer to question 4) is by knowing the rank of matrix A. Do you know what the rank of a matrix is? I will tell you : P it is the number of independent rows (or columns) of your matrix. You can see this by simply looking at the matrix you reduced! Once you know the rank, you can use this theorem. It basically says (copy pasted from wikipedia):
    • if n = rank(A), the solution is unique,
    • otherwise there is an infinite number of solutions.
    Here, n is the number of variables.
    Maybe you can figure it out now. If not, then read, or ask (or both, but in the order listed :p ).
     
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