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Plotting the linear system

  1. Mar 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Solve the linear system of equations:
    [itex]ax+by+z=1[/itex]
    [itex]x+aby+z=b[/itex]
    [itex]x+by+az=1[/itex]
    for [itex]a,b\in\mathbb R[/itex]
    and plot equations and solutions in cases where the system is consistent.
    2. Relevant equations
    -Cramer's rule
    -Kronecker-Capelli's theorem

    3. The attempt at a solution
    Using Cramer's rule, we find the determinant of the system and determinant for each variable:
    [itex]D=b(a-1)^2(a+2)[/itex]
    [itex]D_x=b(a-b)(a-1)[/itex]
    [itex]D_y=(a-1)(ab+b-2)[/itex]
    [itex]D_z=b(a-1)(a-b)[/itex]
    [itex][/itex]
    For [itex]b\neq 0 \land a\neq 1\land a\neq -2\Rightarrow D\neq 0[/itex] system has unique solution:
    [itex](x,y,z)=\left(\frac{a-b}{(a-1)(a+2)},\frac{ab+b-2}{b(a-1)},\frac{a-b}{(a-1)(a+2)}\right)[/itex].

    How to plot the equations with intersection (point) in this case?

    Second case, [itex]a=1[/itex].
    Solvind the system using Kronecker-Capelli's theorem gives:
    [itex]b=1\Rightarrow[/itex] infinitely many solutions.
    [itex]b\neq 1\Rightarrow[/itex] the system is inconsistent.
    This gives [itex](x,y,z)=(1-y-z,y,z)[/itex].

    How to plot the equations with intersection (line) in this case?

    Third case, [itex]a=b=-2\Rightarrow[/itex] infinitely many solutions.
    [itex](x,y,z)=\left(z,\frac{-z-1}{2},z\right)[/itex].

    How to plot the equations with intersection (line) in this case?
     
  2. jcsd
  3. Mar 9, 2016 #2
    I didn't check the correctness of your result, but assuming all is right :

    If it has a unique solution, so it's a point.

    That means that the solution verifies ## x+ y + z - 1 = 0 ## when ##x,y,z## describe ##\mathbb{R}^3##.
    It's a plane passing through (1,0,0), (0,1,0), and (0,0,1).

    We already discussed that in length yesterday :mad:
     
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