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Math Help - Matrix

  1. #1
    Suy
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    Matrix

    1)There are 4 switches numbered 1 through 4. Is it possible to set the switches such that:
    i) Switch 3 is off.
    ii) An odd number of switches 1, 2 and 3 are OFF.
    iii) An odd number of switches 2, 3 and 4 are ON.
    iv) An even number of switches 1 and 3 are ON.
    Switch 1: ?
    Switch 2: ?
    Switch 3: ?
    Switch 4:?

    2)Find a non-trivial solution to the following homogeneous system of equations:
    (4+i)x −1y = 0
    (−18+4i)x +(4−2i)y = 0



    3)A student finds $1.05 in dimes, nickels
    and pennies. If there are 17 coins in all, how many coins of each
    type can he have?
    This is what i got..
    0.10 x + 0.05 y + 0.01 z = 1.05
    x + y + z = 17 ,

    The first question, I have no idea how to do it....
    For second, I tired to use matrix, but the answer doesn't make sense to me. the answer have to be in a+bi form...
    The third , I don't know what to do next, because there is infinite solution... but how can I find all the solution?
    Please help me...
    ty!
    Last edited by Suy; October 11th 2010 at 12:21 PM.
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  2. #2
    A Plied Mathematician
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    Generally, you shouldn't post more than two problems in a thread, as per Rule 8.

    For # 1, I would just write out a truth table, and check to see which rows, if any, satisfy all the conditions.

    For # 2, can you post your work so far?

    For # 3, I would recommend working in units of cents, not dollars. Then impose the condition that x, y, and z must all be integers. That might narrow down the solution space a bit for you.
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  3. #3
    Suy
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    I got #3 now
    #1 ,
    switch 1 2 3 4
    ii)------1 1 0..=0
    iii)-----...1 0 0=1
    iv)-----1....0...=1
    so the answer is
    switch 1) on
    2)on
    3) off
    4)off
    is that right?
    for 2,
    i use matrix
    (4+i)x −1y = 0
    (−18+4i)x +(4−2i)y = 0

    4+i 1 0
    -9+2i 2-i 0

    after adding and subtracting
    I got
    4+i 1 0
    17 i 0
    and then i tried to plug it into the x+y+z form, and solve for both equation, i got -4iy+18y=0, which doesn't make sense to me....
    Last edited by Suy; October 11th 2010 at 04:27 PM.
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  4. #4
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    #1: Your proposed switch configuration does not satisfy condition iv).

    #2: You did not correctly produce the augmented matrix from the simultaneous equations. That is,

    4+i 1 0
    -9+2i 2-i 0

    is incorrect. Look more closely.
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  5. #5
    Suy
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    -9+2i 2-i 0
    yea, I divided the row by 2...I thought i could do that....?

    iv) An even number of switches 1 and 3 are ON.
    1+0=1

    why isn't that satisfy?
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  6. #6
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    # 1 You have on switch 1 on and switch 3 off. Therefore, one switch of switches 1 and 3 are on. One is an odd number. Therefore, condition iv is not satisfied.


    # 2 You can divide by 2. In fact, your second row of the augmented matrix contains no errors. The error is in the first row of the matrix. That is,

    (4+i)x −1y = 0 does not go into the augmented matrix as

    4+i 1 0. You've made an error there. Do you see it?
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  7. #7
    Suy
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    I am confused...
    I don't think i understand the question
    the even number, is it mean 1 plus 3 is on, or 1 and 3 is on , on? or they are saying that how many switches are there?, there is two switches , so the number is even?

    Yea , I see the mistake... I will redo it ,ty!
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  8. #8
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    They're requiring the following:

    Get yourself tunnel vision, and focus in on switches 1 and 3. That's all you're looking at. The requirement iv) is that an even number of those switches are on. There are only two ways to do this: Switches 1 and 3 are both off (zero of those switches are on, and zero is an even number), or Switches 1 and 3 are both on (two of those switches are on, and two is an even number). Does that clear things up for you?

    How is # 3 going?
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  9. #9
    Suy
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    I think it's make sense to me... ,
    since switch 3 is off, then switch 1 must be off, two switch is off, two is even
    Yea i got it


    #3
    I multiplied 0.10 x + 0.05 y + 0.01 z = 1.05 by 100, then use the matrix, and backward substitution,
    then I got something like x=?+?z and y=?+?z, I notice that z can only be 0,5 or 10, 15...otherwise x and y will be a fraction, and z can only be 0 or 5, then i just solve for x and y
    it is much easier when i multiplied by 100, ty for for telling me
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  10. #10
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    So you've solved all the problems, then? What are your answers?
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  11. #11
    Suy
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    No... i still didn't get #2
    this is what i got..
    -9+2i 2-i 0
    4+i -1 0

    row 1 minus two time row 2
    -17 4-i 0
    4+i -1 0

    1 (4+i)/(-17) 0
    1 (-1)/(4+i) 0

    Then I tried to minus row 1from row 2
    then i got

    1 (4+i)/(-17) 0
    0 0 0

    so y=s , x=s(4-i)/17
    but i think this is wrong ,cause the answer have to in a+bi form

    #3
    x=4,y 13,z 0
    x 8,y 4,z 5
    #1
    1off
    2off
    3off
    4on
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  12. #12
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    I agree with your answer to # 1.

    With respect to # 3, I would say you can rule out the first solution, because of the wording of the problem. It says there are dimes, nickels, AND pennies.

    For # 2, I don't think you understand complex arithmetic correctly. Let me do one or two computations for you so you get the idea.

    \left[\begin{matrix}4+i &-1\\-9+2i &2-i\end{matrix}\Big|<br />
\begin{matrix}0\\0\end{matrix}\right]

    First step: need a 1 in the 1,1 position. Divide first row by 4+i to obtain

    \left[\begin{matrix}1 &\frac{-1}{4+i}\\-9+2i &2-i\end{matrix}\Big|<br />
\begin{matrix}0\\0\end{matrix}\right]\to\left[\begin{matrix}1 &\frac{-1}{4+i}\cdot\frac{4-i}{4-i}\\-9+2i &2-i\end{matrix}\Big|<br />
\begin{matrix}0\\0\end{matrix}\right]\to\left[\begin{matrix}1 &\frac{-4+i}{17}\\-9+2i &2-i\end{matrix}\Big|<br />
\begin{matrix}0\\0\end{matrix}\right].

    So here you can see that division by a complex number will require multiplication by a fraction that is equal to 1: a special fraction. In general,

    \dfrac{a+bi}{c+di}=\dfrac{a+bi}{c+di}\cdot\dfrac{c-di}{c-di}=\dfrac{ac-adi+bci+bd}{c^{2}+d^{2}}=\dfrac{ac+bd+(bc-ad)i}{c^{2}+d^{2}}.

    Don't memorize this formula. Instead, remember the procedure: multiply top and bottom by the complex conjugate, and then foil it out.

    Moving on, the next step will be to multiply row 1 by 9-2i, add it to row 2, and store the result in row 2. What do you get when you do that?
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  13. #13
    Suy
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    from
    multiply row 1 by , add it to row 2
    1 (-4+i)/17 0
    0 (2-i)+(-4+i)(9-2i)/17 0

    1 (-4+i)/17 0
    0 (2-i)+(-2+i) 0

    1 (-4+i)/17 0
    0 0 0

    I still got 0 in the second row??
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  14. #14
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    Excellent. This is what you should expect, if you were at all able to get a nontrivial solution of a homogeneous linear equation. Now what?
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  15. #15
    Suy
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    Do you mean y=s , x=s(4-i)/17 is the answer?
    but there is two unique answer y = -46/5-(6 i)/5 and x = -38/17+(22 i)/85
    and y = 67/10-(67 i)/10 and x = 201/170-(67 i)/34, from wolfram
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