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LinAlg: Determine the value(s) of h such that the matrix...

  1. Jan 17, 2016 #1
    The problem statement, all variables and given/known data
    Determine the values of h such that the matrix is the augmented matrix of a consistent linear system.

    1 4 -2
    3 h -6

    The attempt at a solution
    The answer I got differs from the back of the book.

    I tried solving it by adding R1(4) to R2

    1 3 -2
    -4 h 8

    becomes

    1 3 -2
    0 h-12 0

    I thought that h had to equal 12 so that (0)x1 + (0)x2 = 0, but the book says the answer is all h. I don't know what I'm missing
     
  2. jcsd
  3. Jan 17, 2016 #2

    Mark44

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    I don't understand what you did to get this matrix. Also, I don't understand your notation -- R1(4) -- does that mean you are adding 4 times R1 to R2? If so, R1 shouldn't change. How did the row 1, column 2 entry change from 4 to 3?
    If h = 12 the two equations are dependent, meaning that they both represent the same line, so there are an infinite number of solutions. For this problem, an inconsistent system would be two parallel lines (no intersection).
     
  4. Jan 17, 2016 #3

    SammyS

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    ## \begin{bmatrix}
    1 & 3 & | & -2\\
    0 & (h-12) & | & 0
    \end{bmatrix} ##

    Depending upon the value of h, this system may be consistent and dependent, or it may be consistent and independent.

    (0)x1 + (h - 12)x2 = 0

    simplifies to: (h - 12)x2 = 0 .

    Either
    h - 12 = 0, leaving x2 arbitrary.

    or
    x2 = 0 .
     
  5. Jan 18, 2016 #4
    Oh man, sorry people. I was confusing two different problems as I typed this one out... been a long, fartbrain day.

    What I meant to say is:

    1 4 -2
    3 h -6

    R1(-3)+R2=

    1 4 -2
    0 h-12 0

    After seeing both responses, I think I might be missing something very fundamental and obvious about this problem. My thought process is that to make this system consistent, R2 is (0)x1+(0)x2=0 so (h-12) must equal 0 and the only way that can occur is if h = 12. If, for example, h = 1, then R2 implies that 11x2=0. Actually... I guess that makes sense because no matter what h is, when you solve for x2, it will equal 0/h which is always 0. Is this correct?
     
  6. Jan 18, 2016 #5

    SammyS

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    If h ≠ 12, then we must have x2 = 0. The solution for x1 may also be determined.

    If h = 12, then x2 can be any value. The value of x1 will the depend upon the value chosen for x2 .
     
  7. Jan 18, 2016 #6

    Mark44

    Staff: Mentor

    To expand on what SammyS said, your final augmented matrix represents this system of equations:
    1x + 4y = -2
    .... (h - 12)y = 0

    For simplicity I'm going to use x and y instead of subscripts on x.
    Obviously (??) you can solve for y in the second equation, which yields y = 0 if h ≠ 12, and y is arbitrary if h = 12. In the latter case (h = 12), x's value will depend on y.

    To summarize, if h ≠ 12, the single solution is x = 0 and y = 0. If h = 12 there are an infinite number of solutions, as both equations represent the same line. Every solution of the system is a point on the line x + 4y = -2.
     
  8. Jan 18, 2016 #7

    Ray Vickson

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    What is your answer to the original problem (which asked for values of h such that the system is consistent)? Is it consistent if ##h \neq 12##? Is it consistent if ##h = 12##?
     
  9. Jan 18, 2016 #8
    Got it! For some reason, I was under the impression that the bottom row could not be [0 * 0] (as * represents a non-zero number), but that is totally wrong because such a row would still make a system consistent and is quite common, actually since * would be the pivot position. And [0 0 0] is also consistent and essentially means that the row is 'free' as my textbook refers to it.

    Thanks everyone, for your patience and great explanations!
     
  10. Jan 18, 2016 #9

    SammyS

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    @Velouria555 ,

    We neglected welcoming you to PF ... so Welcome to PF !
     
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