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Math Help - Matrices

  1. #1
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    Matrices

    I would really appreciate the help for the question below

    Solve the linear equations
    2x+3y+z=0
    x-y+2z = 0 ( THe first equation is at the top and the second is at the bottom row in matrix form)

    and hence solve
    2x+3y+z=0
    x-y+2z = 0
    ax+y-z=0 ( respectively from top to bottom row in the matrix) for all aER.

    Pleas ehelp me with this! I solved the first part with parametric equations, I got x=-7/3t and y =t and z=5/3t. However, how do I apply it to the second one? I just went about the sam way for the second one as I did the first one but I got stuck and they have to be related somehow. I only can figure out that both have the same two equations, but I'm a blockhead for anything more.

    2.)Consider the system of equations
    x+3y+kz=2
    kx-2y+3z=k
    4x-3y+10z= 5

    I reduced it to my very best, I had 4 times Row =2 minus k times Row 3 to change row 2 and then for channging row 3, I had -1 times Row 3 + Row 2.
    I got
    [tex]\begin{bmatrix} 1 & 3 & k \\0 & 3k-8 & 12-10k \\0 & 0 & 0 \end{bmatrix}\begin{bmatrix} x \\y \\z \end{bmatrix} = \begin{bmatrix} 2 \\-k \\0 \end{bmatrix}

    However, the answer had a very different form of doing the row reduction, it had subtracted the multiple of another row form some row. SOrry that I could not be more specific as the answer sheet is not with me now..please could you help me identify where I had gone wrong here?

    Appreciate it loads, thank you!
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  2. #2
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    Re: Matrices

    1) we know that any solution to the first two equations is of the form t(-7/3,1,5/3). if this is to be a solution to the third equation as well, we must have:

    a(-7/3)t + t - (5/3)t = 0 that is:

    t(-7a - 2)/2 = 0.

    since it makes no difference if we multiply this by -2 (the RHS is still 0), we can write this as:

    t(7a + 2) = 0.

    now taking t = 0 works, leading to the solution (0,0,0) which is always a solution of any system of homogeneous equations (ones where the RHS is equal to 0 in every equation).

    but if a = -2/7, then the third equation is 0 = 0 no matter what t is, so in this case we have the same solution as with just the first 2 equations: any vector of the form t(-7/3,1,5/3).

    (it is worth noting that this is the same set of vectors as the set of vectors of the form s(-7,3,5), since the two vectors in the parentheses only differ by a scalar multiple, so {(-7,3,5)} is a basis for the solution space).

    2) here is how i would do the second problem:

    i would form the augmented matrix

    \left[\begin{array}{ccc|c}1&3&k&2\\k&-2&3&k\\4&-3&10&5 \end{array}\right]

    R2-->-kR1+R2 (subtract k times the first row from the second row):

    \left[\begin{array}{ccc|c}1&3&k&2\\0&-2-3k&3-k^2&-k\\4&-3&10&5\end{array}\right]

    R3-->-4R1+R3:

    \left[\begin{array}{ccc|c}1&3&k&2\\0&-2-3k&3-k^2&-k\\0&-15&10-4k&-3\end{array}\right]

    R2-->15R2

    \left[\begin{array}{ccc|c}1&3&k&2\\0&-30-45k&45-15k^2&-15k\\0&-15&10-4k&-3\end{array}\right]

    R3-->(-2-3k)R3:

    \left[\begin{array}{ccc|c}1&3&k&2\\0&-30-45k&45-15k^2&-15k\\0&30+45k&-20-22k+12k^2&6+9k\end{array}\right]

    R3-->R2+R3:

    \left[\begin{array}{ccc|c}1&3&k&2\\0&-30-45k&45-15k^2&-15k\\0&0&25-22k-3k^2&6-6k\end{array}\right]

    just to clean things up, let's divide row 2 by -15:

    \left[\begin{array}{ccc|c}1&3&k&2\\0&2+3k&-3+k^2&k\\0&0&25-22k-3k^2&6-6k\end{array}\right]

    now let's see what we can do with this. with a little algebraic rearrangement, the last row becomes:

    (25+3k)(1-k)z = 6(1-k). now IF k ≠ 1, this means that:

    z = \frac{6}{25+3k} <--only defined when k ≠ -25/3.

    the second row is now:

    (2+3k)y + \frac{-18+6k^2}{25+3k} = k = \frac{25k+3k^2}{25+3k}

    (2+3k)y = \frac{18+25k-3k^2}{25+3k} = \frac{(9-k)(2+3k)}{25+3k} IF k ≠ -2/3, we can divide both sides by 2+3k, to get:

    y = \frac{9-k}{25+3k}<--again, only defined when k ≠ -25/3

    the top row now becomes:

    x + \frac{27-3k}{25+3k} + \frac{6k}{25+3k} = 2, so:

     x= \frac{50+6k}{25+3k} - \frac{27-3k}{25+3k} - \frac{6k}{25+3k} = \frac{23+3k}{25+3k}

    now, i've done the hard stuff, you can decide what happens when k = 1, -2/3 or -25/3.
    Thanks from HallsofIvy
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  3. #3
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    Re: Matrices

    Thank you! For the second problem, could you tell me why the way I did it is not preferred?
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  4. #4
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    Re: Matrices

    i don't know exactly what you did (your notation is hard to read), but in general, to change row k, you do one of 3 things:

    1) add m times another row to it
    2) multiply it by a number
    3) exchange it with another row

    it's not a good idea to add m times row k plus n times row j back to row k. that's too many steps at once, and it's easy to get confused. it's also not a good idea to work on two rows simultaneously. do one operation at a time. in what i did above, i was careful to list every step BEFORE i took it, and only ONE row changed at a time. this makes your work easier to check (but uses a LOT more paper). row-reduction is very labor-intensive, and an early mistake can ruin everything. and you must have made a mistake SOMEWHERE because you ended up with a 0 row at the bottom where i did not.
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  5. #5
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    Re: Matrices

    Thank you so so so much!

    I have another really short question, could you help me again?

    Make Y the subject in A(Y^(-1))=B
    I had
    A(Y^(-1))Y = BY
    AI = BY
    A = BY
    A(B^(-1))= = B(B^(-1))Y
    Therefore
    Y = A(B^(-1))
    However the answer is
    Y = (A^(-1))B..

    Thank you so much for helping me!
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  6. #6
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    Re: Matrices

    .
    Last edited by tmoria; January 24th 2013 at 04:16 AM.
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  7. #7
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    Re: Matrices

    it's not clear what the "sizes" of the matrices A,Y, and B are.

    there also appears to be a type somewhere. i will assume A,B and Y are all invertible nxn matrices.

    IF:

    AY^{-1} = B

    then:

    AY^{-1}Y = BY

    so:

    A(Y^{-1}Y) = AI = A = BY so far, you're good.

    your next step is wrong, you CANNOT "put a matrix in the middle" (AB)C is usually NOT equal to A(C)B.

    in this case, what you have to do is multiply on the left by B-1:

    B^{-1}A = B^{-1}(BY) = (B^{-1}B)Y = IY = Y

    from this, it follows that:

    Y^{-1} = (B^{-1}A)^{-1} = A^{-1}(B^{-1})^{-1} = A^{-1}B.

    ***********************

    IF, however, the original equation is:

    Ay = b, where y and b are vectors, then you would have to left-multiply by A-1, to get:

    A^{-1}(Ax) = A^{-1}b
    Iy = A^{-1}b
    y = A^{-1}b
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  8. #8
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    Re: Matrices

    Quote Originally Posted by tmoria View Post
    Y^{-1} = \dfrac {1}{Y}

    Should be easy enough from there

    btw, if you have given the correct AY^{-1}=B, the answer in your book is wrong
    this makes no sense, for vectors OR matrices, division is not defined.
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  9. #9
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    Re: Matrices

    I maybe reading worngly, but is it that you got Y^{-1}=A^{-1}B? The answer is different, states that it is Y=A^{-1}B

    Nontheless, thank you so so much as usual!
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  10. #10
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    Re: Matrices

    Quote Originally Posted by Deveno View Post
    this makes no sense, for vectors OR matrices, division is not defined.
    Yep, that's why I deleted the post. I had only read Tutu's last post and missed the matrices connection
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  11. #11
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    Re: Matrices

    yes, it appears that either:

    a) you stated the problem incorrectly, or

    b) there is a typo in the book

    it is impossible for me to say which.
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  12. #12
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    Re: Matrices

    The question:

    Make Y the subject of
    AY^{-1} = B

    Hopefully I have not typed wrongly!
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  13. #13
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    Re: Matrices

    It doesn't say anything about them being of the same order, that they are square matrices etc. But from what I know of these questions, just solve as if they are square and of the same order..
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  14. #14
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    Re: Matrices

    well, the trouble is, not all square matrices HAVE inverses. for instance, we might have:

    A = \begin{bmatrix}0&1\\0&0 \end{bmatrix};\ Y^{-1} = \begin{bmatrix}1&0\\0&1 \end{bmatrix};\ B = A

    then we certainly have:

    AY^{-1} = B

    but there is NO way to express Y as a product of A (to any exponent) and B (to any exponent).
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  15. #15
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    Re: Matrices

    an interesting point from my earlier error though...

    treating as ordinary algebra

    AY^{-1}=B

    Y^{-1}=\dfrac {B}{A}

    since \dfrac{1}{A}=A^{-1}

    we have Y^{-1}=A^{-1}B

    which agrees with your matrix solution.

    Is this true in all situations or just peculiar to this one ?
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