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Thread: Determining that the system has no solutions

  1. #1
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    Question Determining that the system has no solutions

    Hi,

    I hope someone can help me with the question I have. The image I attached below is me trying to solve a systems of linear equations (the system is in the top left of the image). I struggled to find a solution to this problem... and it turns out that apparently there are no solutions. I would really appreciate if you could solve this and show me the steps that brought you to your solution.

    Determining that the system has no solutions-21985802_1536101269779295_1817577389_o.jpg

    - Olivia
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  2. #2
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    Re: Determining that the system has no solutions

    Quote Originally Posted by otownsend View Post
    Hi,

    I hope someone can help me with the question I have. The image I attached below is me trying to solve a systems of linear equations (the system is in the top left of the image). I struggled to find a solution to this problem... and it turns out that apparently there are no solutions. I would really appreciate if you could solve this and show me the steps that brought you to your solution.

    Click image for larger version. 

Name:	21985802_1536101269779295_1817577389_o.jpg 
Views:	7 
Size:	365.9 KB 
ID:	38121

    - Olivia
    LOOK HERE for the row reduction. The system appears to be iinconsistant,
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  3. #3
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    Re: Determining that the system has no solutions

    The matrix of coefficients is singular.

    You can check this ahead of time by computing the determinant if you want. A singular matrix will have determinant 0.

    But usually what happens is that during the course of your Gaussian reduction the bottom row becomes all 0's except for the rightmost element.

    Let's take a look

    Determining that the system has no solutions-clipboard01.jpg
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  4. #4
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    Re: Determining that the system has no solutions

    Okay I see what you mean! Thanks for showing me how to compute a matrix on Wolfram
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    Re: Determining that the system has no solutions

    The order in which you do these calculations seems rather random and error-prone to me. I would have just subtracted 3 times the first row from the second row and add four times the first row to the third row to get just " \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}" for the first column. Also, perhaps just because I don't like fractions, I would not have divide the second row by 2. In any case, you error occurs when you add the first row, \begin{bmatrix}1 & -3 & 4 & -4\end{bmatrix} to the third row ]\begin{bmatrix}0 & 3 & -\frac{15}{2} & \frac{9}{2}\end{bmatrix}. First, you did not add the "1"" in the first column. Second, you divide the third row by -7/2 and 1/2 divided by -7/2 is -1/7, not 1/7.

    Here's how I would have row reduced this:
    Starting from
    \begin{bmatrix}1 & -3 & 4 & -4 \\ 3 & -7 & 7 & -8 \\ -4 & 6 & -1 & 7 \end{bmatrix}
    I see that I can get " \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}" in the first column by subtracting three times the first row from the second and adding 4 times the first row to the third:
    \begin{bmatrix} 1 & -3 & 4 & -4 \\ 0 & 2 & -5 & 4 \\ 0 & -6 & -1 & -8 \end{bmatrix}.

    Notice that by getting "0" in the first column in both second and third rows, I can ignore the first column from now on. I could now get "1" in "second row, second column" by dividing the second row by 2 but that would lead to those awful fractions so I won't do that!

    We can get "0" in the second column of the third row by adding 3 times the second row to the third row (notice that I do not touch either the first row or the first column from now on):
    \begin{bmatrix} 1 & -3 & 4 & -4 \\  0 & 2 & -5 & 4 \\ 0 & 0 & 0 & 4 \end{bmatrix}.

    Aha! That final row is equivalent to the equation 0x+ 0y+ 0z= 4 of just "0= 4" which is not true. This system of equations has no solution!
    Last edited by HallsofIvy; Sep 23rd 2017 at 02:21 PM.
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    Re: Determining that the system has no solutions

    ok gotcha. Thank you very much
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