Messy Matrices Question

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  1. Perpetuallity's Avatar
    1 11-03-2012  11:25
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    Messy Matrices Question
    The matrix A is given by \begin{pmatrix} 1 & 2 \\3 & 4 \end{pmatrix}

    (a) Evaluate A^2 and show that A^2 = 5A + 2I.

    (b) Using the result shown in (a), show that A^3 = \lambda A + \mu I, where \lambda and \mu are constants to be determined.

    I've done part A, but I get in a mess with part B, and I find myself trying to 'guess' the constants, as I don't really have a plan of attack with this one. Some hints and pushes in the right direction would be appreciated.
  2. Slumpy's Avatar
    2 11-03-2012  11:26
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    Re: Messy Matrices Question
    A^3=A.A^2
    Does this help?
  3. Perpetuallity's Avatar
    3 11-03-2012  11:31
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    Re: Messy Matrices Question
    I've tried that but I can't figure out how to determine the constants.

    A^3 = A \cdot A^2 = \begin{pmatrix} 37 & 58 \\81 & 126 \end{pmatrix} = \lambda A +\mu I .

    I've substituted A and I in on the RHS, but I'm still uncertain on how to proceed.
    Last edited by Perpetuallity; 11-03-2012 at 11:33.
  4. Intriguing Alias's Avatar
    4 11-03-2012  11:37
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    Re: Messy Matrices Question
    (Original post by Perpetuallity)
    I've tried that but I can't figure out how to determine the constants.

    A^3 = A \cdot A^2 = \begin{pmatrix} 37 & 58 \\81 & 126 \end{pmatrix} = \lambda A +\mu I .

    I've substituted A and I in on the RHS, but I'm still uncertain on how to proceed.
    Well think about it. You need the top left to be half of the top right just by taking away several I matrices (as I doesn't affect top right) so how many is that?
    Last edited by Intriguing Alias; 11-03-2012 at 11:40.
  5. Slumpy's Avatar
    5 11-03-2012  11:39
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    Re: Messy Matrices Question
    (Original post by Perpetuallity)
    I've tried that but I can't figure out how to determine the constants.

    A^3 = A \cdot A^2 = \begin{pmatrix} 37 & 58 \\81 & 126 \end{pmatrix} = \lambda A +\mu I .

    I've substituted A and I in on the RHS, but I'm still uncertain on how to proceed.
    A^3=A.A^2, but you can substitute in an expression for A^2. You don't need to use matrices at all.

    Edit- also, I haven't really checked, but I think your matrix for A^3 may be wrong anyway.
    Last edited by Slumpy; 11-03-2012 at 11:40.
  6. Perpetuallity's Avatar
    6 11-03-2012  11:39
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    Re: Messy Matrices Question
    (Original post by hassi94)
    Well think about it. You need the top left to be half of the top right just by taking away several I matrices (as I doesn't affect top left) so how many is that?
    8 I suppose. So now we know mu. I guess now just find a common factor in lambda A. Thanks.
  7. Perpetuallity's Avatar
    7 11-03-2012  11:42
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    Re: Messy Matrices Question
    (Original post by Slumpy)
    A^3=A.A^2, but you can substitute in an expression for A^2. You don't need to use matrices at all.

    Edit- also, I haven't really checked, but I think your matrix for A^3 may be wrong anyway.
    Hmm. A^3 = A(A^2) = A(5A + 2I). So now just sub everything in, multiply out and go from there, I guess.

    Edit: A^3 was wrong. I have an idea- A(5A + 2I) = lambdaA + muI
    Last edited by Perpetuallity; 11-03-2012 at 11:48.
  8. Intriguing Alias's Avatar
    8 11-03-2012  11:43
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    Re: Messy Matrices Question
    (Original post by Perpetuallity)
    8 I suppose. So now we know mu. I guess now just find a common factor in lambda A. Thanks.
    Right and since A is \begin{pmatrix} 1 & 2 \\3 & 4 \end{pmatrix} You know that \lambda = 37-8
    Last edited by Intriguing Alias; 11-03-2012 at 11:44.
  9. Slumpy's Avatar
    9 11-03-2012  11:44
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    Re: Messy Matrices Question
    (Original post by Perpetuallity)
    Hmm. A^3 = A(A^2) = A(5A + 2I). So now just sub everything in, multiply out and go from there, I guess.

    Edit: Checked A^3, its correct.
    Yes. You will need to sub for A^2 again.

    Edit-No it's not.
  10. Perpetuallity's Avatar
    10 11-03-2012  11:49
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    Re: Messy Matrices Question
    (Original post by Slumpy)
    Yes. You will need to sub for A^2 again.

    Edit-No it's not.

    (Original post by hassi94)
    Right and since A is \begin{pmatrix} 1 & 2 \\3 & 4 \end{pmatrix} You know that \lambda = 37-8
    Thanks chaps. I have it now.
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  11. Jammy4410's Avatar
    11 11-03-2012  12:55
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    Re: Messy Matrices Question
    (Original post by Perpetuallity)
    Thanks chaps. I have it now.
    Was the answer:

    lambent = 27
    mu = 10

    Also, what unit is this?
  12. Perpetuallity's Avatar
    12 11-03-2012  14:16
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    Re: Messy Matrices Question
    (Original post by Jammy4410)


    Was the answer:

    lambent = 27
    mu = 10

    Also, what unit is this?
    Yes, that was the answer I reached. This was WJEC FP1, Summer 2011. The link to the paper is here: http://www.wjec.co.uk/uploads/papers/s11-0977-01.pdf.
  13. hamoura93's Avatar
    13 03-06-2012  15:10
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    Re: Messy Matrices Question
    does anyone know a good link to matrix transformations?
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