Linear algebra question

  1. i got a set of linear equations with x,y,z,t variables.

    x=4,y=-2,z=-2,t=4 is a solution
    x=-2 y=4 z=4 t=-2 is a solution


    x=2,y=2,z=2,t=2 is not a solution.

    prove that this set is not homogeneous ?
    prove that x=0,y=2,z=2,t=0 is a solution?

    ??
     
  2. jcsd
  3. Defennder

    Defennder 2,616
    Homework Helper

    I'm not clear as to what homogenous here means. On the other hand, if a set of linear equations has more than one solutions, then it has infinite number of solutions. That allows you to come up with a general solution given just two specific possible solutions. This should help you with part 2.
     
  4. HallsofIvy

    HallsofIvy 40,678
    Staff Emeritus
    Science Advisor

    I believe that by "homogeneous" here, you mean that the equations are of the form
    [itex]a_1x+ b_1y+ c_1z+ d_1t= 0[/itex]
    [itex]a_2x+ b_2y+ c_2z+ d_2t= 0[/itex]
    [itex]a_3x+ b_3y+ c_3z+ d_3t= 0[/itex]
    [itex]a_4x+ b_4y+ c_4z+ d_4t= 0[/itex]
    That is the same as the matrix equation
    [tex]\left[\begin{array}{cccc}a_1 & b_1 & c_1 & d_1 \\a_2 & b_2 & c_2 & d_2 \\a_3 & b_3 & c_3 & d_3 \\a_4 & b_4 & c_4 & d_4 \end{array}\right]\left[\begin{array}{c} x \\ y \\ z \\ t\end{array}\right]= \left[\begin{array}{c}0 \\ 0 \\ 0 \\ 0\end{array}\right][/tex]

    Such an equation will have a unique solution if and only if the coefficient matrix is invertible. Since this has at least two solutions, the coefficient matrix is not invertible. Since there are at least two solutions, the coefficient matrix is not invertible. If it were homogeneous, then any linear combination of solutions would also be a solution: if [itex]Ax_1= 0[/itex] and [itex]Ax_2= 0[/itex] then [itex]A(cx_1+ dx_2)= cAx_1+ dAx_2= 0[/itex]
     
  5. how to prove that
    x=0,y=2,z=2,t=0 is not a solution?
     
  6. Defennder

    Defennder 2,616
    Homework Helper

    You're supposed to prove it's a solution, you mean.
     
  7. i am supposed
    but i dont know how??
     
  8. Mark44

    Staff: Mentor

    You said you have a set of linear equations (which you didn't show). Replace x, y, z, and t with 0, 2, 2, and 0, respectively, on the expressions on the left side of the equations you have. You should get 0 on the right side of all of your equations.
     
  9. i got a crazy idea
    if i use this given sentences and transform them into equations:
    x=4,y=-2,z=-2,t=4 is a solution
    x=-2 y=4 z=4 t=-2 is a solution

    (4,-2,-2,4)+2*(-2,4,4,-2)=(0,6,6,0)=3*(0,2,2,0)

    so i got my solution that i suppose to prove from algebraic manipulation

    is that a proove??
    if it is what formal words do i need to say in order to confirm this method?
     
  10. Mark44

    Staff: Mentor

    Sort of.
    Here's the situation as I see it (necessariy sketchy, since you didn't provide many details):
    You have a system of equations which I will represent as a matrix equation:
    Ax = 0

    Any vector x in R^4 is a solution to the equation above if A times x equals the zero vector.

    You are given that for x_1 = (4, -2, -2, 4)^T and x_2 = (-2, 4, 4, -2),
    A*x_1 = 0 and A*x_2 = 0.

    For x_3 = (0, 2, 2, 0), you have shown that x_3 is a linear combination of x_1 and x_2, namely x_3 = 1/3*x_1 + 2/3*x_2.
    By the linearity of matrix multiplication A*x_3 = A*(1/3*x_1 + 2/3*x_2)
    You should be able to fill in what's missing at the end to show that x_3 is a solution to the homogeneous equation Ax = 0.
     
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