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Solve this linear system?

  1. Apr 26, 2015 #1
    \begin{cases}
    x+ 2y - z + w - t = 0 \\
    x - y + z + 3w - 2t = 0
    \end{cases}

    Add 1st to the 2nd:

    $$2x + y + w - t = 0 \\
    y = -2x -w + t = 0$$

    Substitute y in the 1st:

    ##x + 2x + w - t + 3w - 2t + z = 0 \\
    z = 3x - 4w + 3t##

    Both z and y in terms of x,w,t. Writing using matrix form:

    ##\left[\begin{matrix}
    x \\
    y \\
    z \\
    w \\
    t \\
    \end{matrix}\right] =
    x\left[\begin{matrix}
    1 \\
    -2 \\
    -3 \\
    0 \\
    0 \\
    \end{matrix}\right] +
    w\left[\begin{matrix}
    0 \\
    -1 \\
    -4 \\
    1 \\
    0 \\
    \end{matrix}\right] +
    t\left[\begin{matrix}
    0 \\
    1 \\
    3 \\
    0 \\
    1 \\
    \end{matrix}\right]
    ##

    Solution ##x(1,-2,-3,0,0) + w(0,-1,-4,1,0) + t(0,1,3,0,1)##. I can isolate any two variables, but I have no idea if different solutions are in fact, part of the same set of solutions.
     
  2. jcsd
  3. Apr 26, 2015 #2

    Simon Bridge

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    How many unknowns do you have? How many equations? What does that tell you about the solution?
     
  4. Apr 26, 2015 #3
    Two equations of five unknowns, that means that up to two unknowns can be calculated, while the other three remain unknown.

    the answer in the book is also a sum of three vectors, in terms of x,w and t. But my values aren't equal nor multiples.
     
    Last edited: Apr 26, 2015
  5. Apr 26, 2015 #4

    Mark44

    Staff: Mentor

    You can check to see if your solution is actually a solution this way.
    Write your system of equations in matrix form:
    $$\begin{bmatrix} 1 & 2 & -1 & 1 & -1 \\ 1 & -1 & 1 & 3 & -2\end{bmatrix}$$
    Multiply each of your three vectors on the left by this matrix. If you get the zero vector for each multiplication, your vectors are correct.

    You can check to see whether the vectors given as answers in your book are correct by doing the same thing.

    I haven't checked your work, but from what you have, the solution space is a three-dimensional subspace of R5. Your vectors, if correct, are a basis for that subspace. It's possible that the vectors given in your textbook are also correct. If so, they just happen to be a different set of basis vectors.

    BTW, I don't know if you're covered this yet, but the solution space you have found is called the kernel or nullspace of the tranformation represented by the matrix I showed. If we call this matrix A, the kernel is the set of vectors x such that Ax = 0.
     
  6. Apr 26, 2015 #5
    My solution is wrong. The first vector multiplied by the matrix results in 0, but not the other two. The answer in the book is correct because all three vectors multiplied by the matrix result in 0.

    There was a mistake in the first addition. Fixed it and now I've found the same answer as in the book. This means that there are other solutions, I can isolate any two variables and find different linear combinatinos of three vectors.
     
    Last edited: Apr 26, 2015
  7. Apr 26, 2015 #6

    Simon Bridge

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    Well done.
     
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