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  1. #1
    Member mybrohshi5's Avatar
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    Matrix

    Determine the value of for which the system



    has no solutions.


    How do i do this? im confused because how can you make a system have no solutions when it really does. I know it does because i did row operations until i got it down to the 1's in the diagonal and the rest of the entries were 0's

    any help will be much appreciated. thank you
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  2. #2
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    Have you considered a Determinant?

    2k + 18 - 8 - 12 + 24 - k = 0

    I am really concerned with your opinion that this systems "really does" have a solution. You cannot know that until you determine the value of k. When you did those row operations, there was only one way to get rid of the k. That one way is to make an error. It's the only 'k' in the problem. Parity would suggest that you can't get rid of it. Whatever you do, there will always be a 'k'.

    Do you REALLY believe a system would be entirely independent of one of it's coefficients? No kidding, think about the implications for a minute. That '6y' is bothering me. Since the solution to the system doesn't care about that 6, let's just make it a 4. Awesome! That's much better. It is a mass of confusion.
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  3. #3
    Member mybrohshi5's Avatar
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    You are right. you cant find a solution until you find k. that was my mistake.

    I didnt even think about finding a determinant. But what i am wondering is how do you find the determinate of this matrix because it has the column of solutions and is not just a 3x3 matrix.

    i havent learned about determinates yet in class but i know how to find it im just not sure how you got the 2k + 18 - 8 -12 + 24 - k = 0 because the matrix has the column of the 3 solutions
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  4. #4
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    If you use proper row operations, you will find the same result. The determinant is not important. That's the thing with unique solutions, they don't care how you find them.

    As far as EXISTENCE, you do not need to be concerned with the constant side of things. Just the coefficients on the variables is enough. Determinant are not just parlor games. They are useful for things. This is one of those useful things. It's a rule. If the determinant is zero, there is no solution. Write it down.

    You do not have to see a 3x3 determinant out of the air. Feel free to use Expansion by Minors or however else you may have learned to do it.
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  5. #5
    Member mybrohshi5's Avatar
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    I am stuck on my row operations

    I am supposed to be trying to get the matrix to reduced row echelon form correct?

    I am stuck at this

    1 0 10 | 6
    0 1 -7 | -4
    0 0 k+22 |1
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  6. #6
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    No, you are not stuck. You are DONE!

    k+22 = 0 -- Solve for k.

    Compare this to my determinant equation.

    Good work.
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  7. #7
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    Quote Originally Posted by mybrohshi5 View Post
    I am stuck on my row operations

    I am supposed to be trying to get the matrix to reduced row echelon form correct?

    I am stuck at this

    1 0 10 | 6
    0 1 -7 | -4
    0 0 k+22 |1
    A little more- if k+ 22 is NOT 0, then you can divide the last row by that and continue your row reduction to get a unique solution to the equation. If k+22= 0, you cannot do that and, since 0\ne 1 there is no solution. If k+ 22= 0 and the last number in that row were also 0, then there would be an infinite number of solutions.
    Last edited by HallsofIvy; February 22nd 2010 at 03:46 AM.
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  8. #8
    Member mybrohshi5's Avatar
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    Quote Originally Posted by TKHunny View Post
    No, you are not stuck. You are DONE!

    k+22 = 0 -- Solve for k.

    Compare this to my determinant equation.

    Good work.
    My k+22 doesnt = 0 though i have it equal to 1 so k would actually be -21 and thats wrong cause the answer is -22 as your determinate stated.

    I have checked my operations multiple times so i know my matrix where i am stuck at is correct i just dont know what to do now
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  9. #9
    Member mybrohshi5's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    A little more- if k+2 is NOT 0, then you can divide the last row by that and continue your row reduction to get a unique solution to the equation. If k+22= 0, you cannot do that and, since 0\ne 1 there is no solution. If k+ 22= 0 and the last number in that row were also 0, then there would be an infinite number of solutions.
    I am a little confused on what you are trying to say. sorry

    What do yo mean by "if k+2 is NOT 0, then you can divide the last row by that "

    is that supposed to be k+22? and then what can i divide by?

    Thank you
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  10. #10
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    0 = 0 is fine. This normally would mean the system is under-defined and there are infinitely many solutions.
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  11. #11
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    Quote Originally Posted by mybrohshi5 View Post
    My k+22 doesnt = 0 though i have it equal to 1 so k would actually be -21 and thats wrong cause the answer is -22 as your determinate stated.

    I have checked my operations multiple times so i know my matrix where i am stuck at is correct i just dont know what to do now
    No, no, no! You do NOT have k+22= 1! The last row of your reduced matrix corresponds to the equation (k+22)z= 1. That is not solvable if and only if k+ 22= 0. If k+22 is NOT 0, you can divide the equation by it: z= 1/(k+22) and complete the solution. In terms of the augmented matrix, to complete the row reduction, you would divide the last row by k+ 22- that is not possible if k+22= 0.

    (Yes, I meant k+ 22, not "k+2". I've gone back and edited so no one will ever know I made that mistake!)
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