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Matrices and linear systems

  1. Oct 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Find all the values of h and k such that the system:

    hx + 6y = 2
    x + (h+1)y = k

    has: (a) No solutions (b) A unique solution (c) Infinitely many solutions

    2. Relevant equations
    -


    3. The attempt at a solution
    I've tried putting the system into echelon form and got the following (with the augmented matrix):

    (1 h + 1 k)
    (0 h2+h-6 hk-2)

    However, how can i proceed from this step? Note that the only way i can solve this problem is using matrices, i've got the answer by comparing the linear systems but i need to do it with matrices.
     
  2. jcsd
  3. Oct 1, 2013 #2

    arildno

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    It is simpler to deduce the features from the determinant of the coefficient matrix.
     
  4. Oct 1, 2013 #3

    Ray Vickson

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    If you don't really understand what augmented systems are doing you should stay away from them and just write out the equations explicitly. So, for example, you could solve for x in terms of y from the second equation, then substitute that expression (in place of x) into the first equation, to get a new equation that involves y alone. Once you have done that it is easier to see what you need to do next.

    Of course, you could do all that using an augmented system, but only if you understand what is happening---otherwise the augmented system is just getting in your way.
     
  5. Oct 1, 2013 #4
    aridno:

    I have computed the value of the determinant. What can i work out from that?

    Ray Vickson:

    I could try doing that but the prompt was to solve them using matrices (or augmented systems). If you could explain me a bit what is going on in that situation that would be really helpful. I tried to transform the system into reduced row echelon form, but one of the terms was a mess and i couldn't work out anything from there. Any suggestions in what i can do?
     
  6. Oct 1, 2013 #5

    Ray Vickson

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    I have already suggested what you should do: write out the new linear equation that comes from your row-echelon form. Examine carefully what the new equation 'says'. That is my final hint.
     
  7. Oct 1, 2013 #6

    arildno

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    Well:
    You know that if you have a unique solution, then that is gained by multiplying the INVERSE of that matrix to the "vector" on RHS, leaving the (x,y)-vector on LHS.

    But:
    What is the requirement that the inverse matrix exists at all?
     
  8. Oct 1, 2013 #7
    If the coefficient matrix is M, then detA ≠ 0 (as then the expression would be over 0). So this leaves me with the following values for no solutions:

    h = -3 and all k except (-2/3)
    h = 2 and all k except (1)
     
  9. Oct 1, 2013 #8

    arildno

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    And how many solutions will you get for a choice precisely with those exceptional k-values (coupled to its respective h-value)?
     
  10. Oct 1, 2013 #9
    We would then get infinite solutions as we would have a matrix of rank 1, meaning we need to set up a parameter. Then for infinite solutions:

    h=-3 k=-2/3
    h=2 k=1

    How would we know the values of when the system has solution though? I was testing for an h=1 and i got the following matrix:

    (1 6 2)
    (1 2 k)

    Then we would have:

    (1 6 2)
    (0 -4 k-2)

    I was thinking of when if k = 2, then there would be infinite solutions. However, then x would be 1/2. However, don't we have to set up a parameter for y? For example, if we have the following:

    (a b c)
    (0 d 0)

    Can i simply assume that the system has infinite solutions by rewriting:

    (a b c)
    (0 0 0)

    Or that isn't permitted? Wouldn't the y variable have a parameter?
     
  11. Oct 2, 2013 #10

    arildno

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    "
    (1 6 2)
    (0 -4 k-2)

    I was thinking of when if k = 2, then there would be infinite solutions."

    Incorrect. With k=2, y=0 and x=2.

    With an invertible coefficient matrix, you ALWAYS have a unique solution for x and y, in terms of whichever constants appear elsewhere.
     
  12. Oct 2, 2013 #11

    Ray Vickson

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    When k = 2 the second row of the row-echelon form says that -4*y=0. Do you really think that has infinitely many no solutions or no solutions?
     
    Last edited: Oct 2, 2013
  13. Oct 2, 2013 #12
    I kind of get it. So when we have a row sort of (where n is a constant):

    (0 0 n 0)

    In a system AX = B, where the last 0 is part of the augmented system, we NEVER say that the row turns into:

    (0 0 0 0)

    Right? We just leave the row as it is (can we turn it into a leading 1 by dividing the row by "n"?? Our professor told us that we could do that but now i see that he may be wrong.
     
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