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Math Help - [SOLVED] Elementary Linear Algebra, Systems of Linear Equations

  1. #1
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    [SOLVED] Elementary Linear Algebra, Systems of Linear Equations

    Hello, I am currently stuck on one problem...and am not sure about my solution on another. I believe I am restricted to elementary operations as well. Those are:
    1. Interchange two equations
    2. Multiply an equation by a nonzero constant
    3. Add a multiple of an equation (equation 1) to another equation (equation 2) and replace the second equation (say equation 2) with the result.

    Now that I have listed what I must work off of, the first problem I am having is with this system...

    (\cos\theta)x + (\sin\theta)y = 1
    (-\sin\theta)x + (\cos\theta)y = 1

    I need to solve this system for x and y. I do know that my answer will more than likely be in terms of sines and cosines.

    The second problem I am uncertain about is this:

    kx + y = 4
    2x - 3y = -12

    I need to find the value(s) of k such that the system has an infinite number of solutions. I believe my answer of k = -2/3 is accurate and as far as I can tell with k being -2/3 the system has an infinite number of solutions. However, I am uncertain if there are other values of k which would give me the same answer. I apologize if this should be in a different forum, but as my class is Elementary Linear Algebra I figured this location was appropriate. Thanks for all your help.
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  2. #2
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    Ok, I think I found a solution to the first problem...I get the following:

    x = \cos\theta - \sin\theta
    y =  \cos\theta + \sin\theta

    Now, I wound up doing it the long way by adding -1 times the first equation to the second equation. Finding an expression for y in terms of x and plugging back in. I did it this way, as opposed to simply setting the two equations equal to each other because I wasn't really certain how my teacher wants a problem like this done, guess that's a question for him. Can anyone verify my solution?

    Thanks!
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  3. #3
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    Quote Originally Posted by Alterah View Post
    Hello, I am currently stuck on one problem...and am not sure about my solution on another. I believe I am restricted to elementary operations as well. Those are:
    1. Interchange two equations
    2. Multiply an equation by a nonzero constant
    3. Add a multiple of an equation (equation 1) to another equation (equation 2) and replace the second equation (say equation 2) with the result.

    Now that I have listed what I must work off of, the first problem I am having is with this system...

    (\cos\theta)x + (\sin\theta)y = 1 .... (1)
    (-\sin\theta)x + (\cos\theta)y = 1 .... (2)

    I need to solve this system for x and y. I do know that my answer will more than likely be in terms of sines and cosines.
    [snip]
    Multiply equation (1) by \sin \theta and equation (2) by \cos \theta and then add the two resulting equations together.

    Quote Originally Posted by Alterah View Post
    [snip]
    The second problem I am uncertain about is this:

    kx + y = 4
    2x - 3y = -12

    I need to find the value(s) of k such that the system has an infinite number of solutions. I believe my answer of k = -2/3 is accurate and as far as I can tell with k being -2/3 the system has an infinite number of solutions. However, I am uncertain if there are other values of k which would give me the same answer. I apologize if this should be in a different forum, but as my class is Elementary Linear Algebra I figured this location was appropriate. Thanks for all your help.
    Your answer of k = -2/3 is correct. I cannot explain to you why it's the only value without seeing your working (although I'm sure if yu look at your working the reason will be obvious).
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  4. #4
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    For this system:
    x = \cos\theta - \sin\theta
    y =  \cos\theta + \sin\theta

    I don't think I can multiply through by \cos\theta\ or \sin\theta because those aren't necessarily constants. And therefore that goes against the elementary rules I need to follow for this assignment at least. So, in the end I would up subtracting the first equation from the second one (add -1 * first equation to second one) and then solved for y in the second one. Then proceeded to plug the answer for y into the first one. Did you get the solution I got for that system?

    Also, my work for the second set:
    1. kx + y = 4
    2. 2x - 3y = -12

    Went like this:
    Multiply equation 1 by 3 and replace equation 1 with the result (3*eq1 ->eq1) . Then add equation 1 to equation 2 and replace equation 2 with the result. So the system will then look like this:
    1. 3kx + 3y = 12
    2. (3k + 2)x = 0

    Equation 2 is true when either x = 0 or (3k + 2) = 0. Since x = 0 is trivial I solved 3k + 2 = 0 for k and got k = -2/3.

    With that we can put in any value for x and there will be a corresponding value for y that will make the system true. For example (3,6) or (6, 8). Since we know a system can have only 1 unique solution, no solution, or infinitely many solutions, the fact that we can find two points which satisfy the system when k = -2/3 means it has an infinite number of solutions. I guess that ought to be good enough, There wouldn't be another value for k to make it true.
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  5. #5
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    Quote Originally Posted by Alterah View Post
    For this system:
    x = \cos\theta - \sin\theta
    y = \cos\theta + \sin\theta

    I don't think I can multiply through by \cos\theta\ or \sin\theta because those aren't necessarily constants.

    Mr F says: The unknowns you're solving for are x and y. So yes, you can.

    And therefore that goes against the elementary rules I need to follow for this assignment at least. So, in the end I would up subtracting the first equation from the second one (add -1 * first equation to second one) and then solved for y in the second one. Then proceeded to plug the answer for y into the first one. Did you get the solution I got for that system?

    Also, my work for the second set:
    1. kx + y = 4
    2. 2x - 3y = -12

    Went like this:
    Multiply equation 1 by 3 and replace equation 1 with the result (3*eq1 ->eq1) . Then add equation 1 to equation 2 and replace equation 2 with the result. So the system will then look like this:
    1. 3kx + 3y = 12
    2. (3k + 2)x = 0

    Equation 2 is true when either x = 0 or (3k + 2) = 0. Since x = 0 is trivial I solved 3k + 2 = 0 for k and got k = -2/3.

    With that we can put in any value for x and there will be a corresponding value for y that will make the system true. For example (3,6) or (6, 8). Since we know a system can have only 1 unique solution, no solution, or infinitely many solutions, the fact that we can find two points which satisfy the system when k = -2/3 means it has an infinite number of solutions. I guess that ought to be good enough, There wouldn't be another value for k to make it true. Mr F says: I guess so.
    ..
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  6. #6
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    Thank you for your help. So do you agree with my solution for my first problem or are you getting a different answer?
    The system is actually:
    (\cos\theta)x + (\sin\theta)y = 1
    (-\sin\theta)x + (\cos\theta)y = 1

    The solution I got was:
    x = \cos\theta - \sin\theta
    y = \cos\theta + \sin\theta
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  7. #7
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    Quote Originally Posted by Alterah View Post
    Thank you for your help. So do you agree with my solution for my first problem or are you getting a different answer?
    The system is actually:
    (\cos\theta)x + (\sin\theta)y = 1
    (-\sin\theta)x + (\cos\theta)y = 1

    The solution I got was:
    x = \cos\theta - \sin\theta
    y = \cos\theta + \sin\theta
    Correct. (But why trust me ..... Verification by substitution should be simple enough).
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  8. #8
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    True enough. Thanks for your assistance.
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