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Linear Equations

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    3. The attempt at a solution

    For the sixth question, I got:
    cos a = [tex]\frac{-3}{8}[/tex]
    cos b = [tex]\frac{-5}{4}[/tex]
    sin 2a = [tex]\frac{7}{8}[/tex]

    Whenever I use the answer I got, it says it's incorrect, but every time I do it I get that answer.


    [​IMG]

    I ended up with:

    a = [tex]\frac{-1}{5}[/tex]
    b = [tex]\frac{32}{15} - \frac{1}{3}d[/tex]
    c = 0

    I chose option C as it is the only one that made sense to me. I can choose more than one option though... anyone have any input? Maybe I'm not solving the system correctly?


    [​IMG]

    I ended up with:

    a = [tex]-3c + \frac{2}{7}d + \frac{46}{7}[/tex]
    b = [tex]\frac{8}{7}d + \frac{2}{7}d[/tex]
    d = [tex]\frac{13}{11}[/tex]

    I chose option A and option D as being true... but got it wrong. Could someone possibly take a look please?
     
    Last edited: Sep 15, 2007
  2. jcsd
  3. Sep 15, 2007 #2

    nicksauce

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    I get 3/14... maybe you could show some work?

    Edit: Nevermind, I get the same
     
    Last edited: Sep 15, 2007
  4. Sep 15, 2007 #3

    HallsofIvy

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    6) is a very strange question. If I write "cos a" as x, "cos b" as y, and "sin 2a" as z, it is easy to see that x= -3/8, y= -5/4, z= 7/8 satisfy all 3 equations and are the only solution. But how can cos b= -5/4?? Also, if cos a= -3/8, then sin a must satisfy [itex]\sqrt{1- 9/64}= \sqrt{55/64}= sqrt{55}/8 or -sqrt{55}/8[/itex]. But then sin(2a) must be equal to 2sin(a)cos(a) and 7/8 is definitely not [itex]3\sqrt{55}/22[/itex]!

    I have no idea why they put those trig functions in there! Frankly, I would be inclined to say that those equations, as given, have no solution!

    7) Have you actually tried to solve those equations? There are 3 equations with 4 variables so at first glance you would think there would be one parameter in the solution. But the equations are not independent!

    "Is determined uniquely" mean it has exactly one possible value.
     
  5. Sep 15, 2007 #4
    6. Unfortunately, it must have a solution. I am completely stumped... anyone know? :confused:

    7. What do you mean they aren't independent? What would the answer be?

    Thanks for the help so far! :)
     
  6. Sep 15, 2007 #5

    HallsofIvy

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    You don't know what "independent" or "dependent" equations means? Here the three equations are NOT independent (are dependent) because a combination of two of them gives the third- you don't really have "three" equations!

    As for 6, since your solutions to the 3 equations are correct- even if they make no sense as trig functions- and it only asks for "cos(a)" go ahead and answer "-3/8". I might also make a notation to the teacher that "cos(b)= -5/4" makes no sense but you may not want to be that bold!
     
  7. Sep 15, 2007 #6
    6. I tried answering -3/8 and it is wrong. We have an online hand-in thing so it tells you if the answer is correct or incorrect right away.

    7. When you say they're dependent, are you talking about the first equations or the ones I've solved? I still don't understand what the answer is... to any of these. :tongue:
     
  8. Sep 15, 2007 #7
  9. Sep 15, 2007 #8

    learningphysics

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    Your answer for 6 looks right to me... I don't understand why it won't accept. For 7 and 8, looks like you made mistakes solving the equations... check your work.
     
  10. Sep 15, 2007 #9
    Okay, I'll try re-doing 7 and 8. As for 6, I don't get it either. I'm going to have to wait to hear back from my professor about that one.

    I'll post my results for 7 and 8 once I've re-done them.

    Thanks for the help so far guys. :)
     
  11. Sep 15, 2007 #10
    For #7, I think a set of linear equations is independant if you they cannot be created from each other with a sequence of row operations. Note how 3R1=R3 right? So that means you will get a zero row in RREF and end up with a matrix of rank 2 or less and the number of free variables will be at least 2.

    I don't really have too much to add but I had to ask. Are you going to the U of Calgary? I see that your vector problem matches mine for my mechanics course (complete with problem number). :-p
     
    Last edited: Sep 15, 2007
  12. Sep 15, 2007 #11
    Finalled solved #7. The answer ends up being A,B, and E.

    Here's what I did...

    I first subtracted 3R1 from R3 to make it all zero's, and then...

    We have:[tex]
    \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;7 & 3 & 1 & 1 & 5 \;\\\;11 & 9 & 1 & 3 & 17\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}
    [/tex]
    [tex]\frac{R1}{7}
    \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;11 & 9 & 1 & 3 & 17\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}
    [/tex]
    [tex]R2 - 11\times R1
    \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;0 & \frac{30}{7} & \frac{-4}{7} & \frac{10}{7} & \frac{64}{7}\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}
    [/tex]
    [tex]R2\times\frac{7}{30}
    \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;0 & 1 & \frac{-2}{15} & \frac{1}{3} & \frac{32}{15}\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}
    [/tex]
    [tex]R1 - \frac{3}{7}R2
    \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & \frac{1}{5} & 0 & \frac{-1}{5} \;\\\;0 & 1 & \frac{-2}{15} & \frac{1}{3} & \frac{32}{15}\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}
    [/tex]

    Therefore:
    [tex]a + \frac{1}{5}c = \frac{-1}{5}[/tex]

    [tex]b - \frac{2}{15}c + \frac{1}{3}d = \frac{32}{15}[/tex]

    Then I can solve the first equation for c and substitute it into the second equation, which satisfies Option E. Option A is satisfied by looking at the matrix. Option B is also satisfied.

    dontdisturbmycircles - Yep. :smile: First year engineering, you?
     
  13. Sep 16, 2007 #12
    First year engineering. =-)
     
    Last edited: Sep 16, 2007
  14. Sep 16, 2007 #13
    What a small world!
     
  15. Sep 16, 2007 #14
    Haha, I think you are in a different linear algebra lecture though since we have not been assigned anything yet.
     
  16. Sep 16, 2007 #15
    Must be. Our assignment is due tomorrow.
     
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