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Linear system of equations

  1. Mar 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Plot the solution set of linear equations
    [tex]x-y+2z-t=1[/tex]
    [tex]2x-3y-z+t=-1[/tex]
    [tex]x+7z=8[/tex]

    and check if the set is a vector space.

    2. The attempt at a solution

    Augmented matrix of the system:
    [tex]
    \begin{bmatrix}
    1 & -1 & 2 & -1 & 1 \\
    2 & -3 & -1 & 1 & -1 \\
    1 & 0 & 7 & 0 & 8 \\
    \end{bmatrix}\sim \begin{bmatrix}
    1 & -1 & 2 & -1 & 1 \\
    0 & -1 & -5 & 3 & -3 \\
    0 & 1 & 5 & 1 & 7 \\
    \end{bmatrix}\sim \begin{bmatrix}
    1 & -1 & 2 & -1 & 1 \\
    0 & 1 & 5 & 1 & 7 \\
    0 & -1 & 3 & 0 & -3 \\
    \end{bmatrix}\sim \begin{bmatrix}
    1 & -1 & 2 & -1 & 1 \\
    0 & 1 & 5 & 1 & 7 \\
    0 & 0 & 0 & 4 & 4 \\
    \end{bmatrix}
    [/tex]
    [itex]\Rightarrow t=1[/itex]

    [tex]x-y+2z=2[/tex]
    [tex]y+5z=6[/tex]

    [itex]x,y[/itex] are pivot variables [itex]\Rightarrow x=8-7z,y=6-5z[/itex].
    The solution set can be described as [itex]S=\{(8-7z,6-5z,z,1):z\in\mathbb R\}[/itex].

    Algebraically, this means that the system is inconsistent (infinitely many solutions).
    This should represent a line in [itex]xyz[/itex] plane, but it seems that it contradicts with algebraic solution.

    First equation is a plane with coordinates [itex](6,-\frac{3}{2},3)[/itex].
    Second equation is a plane with coordinates [itex](-1,\frac{2}{3},2)[/itex].
    Third equation is a line in [itex]xz[/itex] plane with coordinates [itex](8,\frac{8}{7})[/itex].

    When drawn altogether, there is no intersection between all three planes, and also no line as an intersection.

    Is there a mistake in algebraic, or geometric solution?

    Assuming that algebraic solution is correct, solution set is a vector space because it is an infinite line (closed under vector addition and scalar multiplication). Is this correct?
     
    Last edited: Mar 7, 2016
  2. jcsd
  3. Mar 7, 2016 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    A vector space must contain the zero vector. Does your solution set have that property?
     
  4. Mar 7, 2016 #3
    So, the algebraic solution is not a vector space.

    But why algebraic and geometric solution is not the same?
    Algebraic solution set gives infinite solutions, and geometric solution doesn't give
    neither point nor line as an intersection.
     
  5. Mar 7, 2016 #4
    Assuming your solution is correct, ##S## is a line of ##\mathbb{R}^4## passing through ##(8,6,0,1)## and directed by vector ##(-7,-5,1,0)##. It is not a vector space but an affine subspace ##\mathbb{R}^4##
     
  6. Mar 7, 2016 #5

    Mark44

    Staff: Mentor

    An inconsistent system has no solutions, and there is no such thing as an "xyz" plane.
    I think you have some mistakes in your work. In the 3rd matrix you show, how did you get the 2nd and 3rd rows?
    Geometrically, the solution set I get is a line in ##\mathbb{R}^4## that passes through the point (8, 6, 0, 1).

    Edit: Corrected an earlier mistake I made.
     
    Last edited: Mar 7, 2016
  7. Mar 8, 2016 #6
    How do you get the point [itex](8,6,0,1)[/itex] and vector [itex](-7,-5,1,0)[/itex]?
     
  8. Mar 8, 2016 #7
    Your solution set ##S## (I see it was confirmed correct by Mark44) is a parametrization of line, i.e it can be written ## A + z \vec v ## where ##A = (8,6,0,1)## and ##\vec v = (-7,-5,1,0)##
     
  9. Mar 8, 2016 #8
    Could you explain how to do the parametrization of a line?
     
  10. Mar 8, 2016 #9
    I don't want go off topic so I'll just explain why your solution set is a line.
    A line passing through point ##A## and directed by a vector ##\vec v## is the set of points ##M## such that vector ##\vec {AM}## and ##\vec v## are colinear.
    Your solution set can be written ##S = \{ A + z \vec v, \ z\in \mathbb{R} \} ##, therefore ##M\in S \iff \exists z\in\mathbb{R},\ \vec{AM} = z\vec v##. It's a line. That's all there is to it.
     
  11. Mar 8, 2016 #10

    Mark44

    Staff: Mentor

    I have copied the work you showed earlier. To find the parametric form of the line, the last matrix needs to be completely row-reduced, which means that each nonzero row starts with a 1 entry, and each leading 1 entry (a pivot) has 0 entries above and below it.
    When the last matrix is completely reduced (in RREF or reduced row-echelon form), you can read off the solution, as
    x = ...
    y = ...
    z = ...
    t = ...
    For this matrix that will turn out to be z times a vector plus a constant vector, where in this case, z is the parameter.
     
  12. Mar 8, 2016 #11
    And how to represent the line in explicit form, [itex]y=kx+n[/itex]?
     
  13. Mar 8, 2016 #12

    Mark44

    Staff: Mentor

    The line in question is a line in ##\mathbb{R}^4##, so you wouldn't write it as y = mx + b. Instead, you would write it something like ##\vec{r} = k\vec{A} + \vec{B}##. Here ##\vec{r}## means ##\begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix}##. ##\vec{A}## and ##\vec{B}## are constant vectors that come directly out of the completely reduced final matrix.
     
  14. Mar 8, 2016 #13
    Which vector is [itex]\overrightarrow{A}[/itex] and which vector is [itex]\overrightarrow{B}[/itex] from RREF?
    Also, is [itex]k[/itex] undetermined?
     
    Last edited: Mar 8, 2016
  15. Mar 8, 2016 #14

    Mark44

    Staff: Mentor

    One more step takes your final matrix to RREF (reduced row-echelon form)
    ##\begin{bmatrix}
    1 & -1 & 2 & -1 &| & 1 \\
    0 & 1 & 5 & 1 & | & 7 \\
    0 & 0 & 0 & 4 & | & 4 \\
    \end{bmatrix} \equiv
    \begin{bmatrix}
    1 & 0 & 7 & 0 & | & 8 \\
    0 & 1 & 5 & 0 & | & 6 \\
    0 & 0 & 0 & 1 & | & 1 \\
    \end{bmatrix}##
    Now, write the system of equations that this matrix represents.
    x = ...
    y = ...
    z = ...
    t = ...
    If you line things up you should see a column vector times a parameter z, plus another vector.
     
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