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Math Help - General Solution

  1. #1
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    General Solution

    Hello,

    Can someone please help me figure out the general solution for this matrix?

    x+y+z = 0
    2x+2y+2z = 0
    3x+3y+3z = 0

    Thanks!
    l Flipboi l
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  2. #2
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    Set it up as

    Ax = b \implies x = A^{-1}b

    where A is the coefficient matrix of x,y,z
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Set it up as

    Ax = b \implies x = A^{-1}b

    where A is the coefficient matrix of x,y,z
    so take the inverse of A and multiple by b?

    [ 1 1 1 ]^-1 [ 0 ]
    [ 2 2 2 ] [ 0 ]
    [ 3 3 3 ] [ 0 ]

    the books says the answer is:

    x = -s-t
    y = s
    z = t

    what i did earlier was reduced it to row echelon form and obtained the matrix

    [ 1 1 1 ]
    [ 0 0 0 ]
    [ 0 0 0 ]

    but i don't get how y=s and z=t
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  4. #4
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    Pickslides, the fact that the problem mentions a "general" solution should tell you that this problem does NOT have one unique solution and so the matrix A is NOT invertible!

    |flipboi| you did not write the original equations as a matrix and I see no reason to do so. You are making this much more difficult that it should be.

    x+y+z = 0
    2x+2y+2z = 0
    3x+3y+3z= 0

    are all the same equation! So you really just have z= -x- y. That is the "general solution". x and y can be any numbers at all and z= -x- y.

    If you really intended
    \begin{bmatrix}1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}0 \\0  \\ 0\end{bmatrix} then the vector solution is
    \begin{bmatrix} x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ -x-y\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ -1\end{bmatrix}

    Row reducing to
    \begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
    just tells you, again, that all three equations reduce to x+ y+ z= 0 so that z= -x -y.

    "but i don't get how y=s and z= t". I solved for z= -x- y so I could use x and y as parameters or just rename them "s" and "t" and write the solution as x= s, y= t, z= -s- t. Apparently in your text, they solve for x= -y- z. You don't "get" y= s and z= t, you assign those names. Just rename "y" as "s" and "z" as "t". Then x= -s- t, y= s, z= t. There are an infinite number of ways to write parametric equations for a "general" solution.
    Last edited by HallsofIvy; October 27th 2010 at 05:34 AM.
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