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Math Help - Tricky Matrices Problem

  1. #1
    Super Member craig's Avatar
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    Tricky Matrices Problem

    Got a bit of a confusing problem involving matrices.

    We assume that (p,q,r)^T is a Pythagorean triple, so that p^2 + q^2 = r^2 where p,q,r are positive integers.

    For P,

    P = \left( \begin{array}{ccc} -1 & -2 & 2 \\ -2 & -1 & 2 \\ -2 & -2 & 3 \\ \end{array} \right).

    Show that:

    P \left( \begin{array}{c} -p \\ q \\ r \\ \end{array} \right), P \left( \begin{array}{c} p \\ -q \\ r \\ \end{array} \right) and P \left( \begin{array}{c} -p \\ -q \\ r \\ \end{array} \right) are also Pythagorean triples.

    Not sure where exactly to start here. Do I multiply them and then form some kind of equation, also a bit confused by the transposed bit at the start of the question?

    Any points would be greatly appreciated

    Thanks in advance
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  2. #2
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    Quote Originally Posted by craig View Post
    Got a bit of a confusing problem involving matrices.

    We assume that (p,q,r)^T is a Pythagorean triple, so that p^2 + q^2 = r^2 where p,q,r are positive integers.

    For P,

    P = \left( \begin{array}{ccc} -1 & -2 & 2 \\ -2 & -1 & 2 \\ -2 & -2 & 3 \\ \end{array} \right).

    Show that:

    P \left( \begin{array}{c} -p \\ q \\ r \\ \end{array} \right), P \left( \begin{array}{c} p \\ -q \\ r \\ \end{array} \right) and P \left( \begin{array}{c} -p \\ -q \\ r \\ \end{array} \right) are also Pythagorean triples.

    Not sure where exactly to start here. Do I multiply them and then form some kind of equation, also a bit confused by the transposed bit at the start of the question?

    Any points would be greatly appreciated

    Thanks in advance

    Take the first one, for example: \left( \begin{array}{ccc} -1 & -2 & 2 \\ -2 & -1 & 2 \\ -2 & -2 & 3 \\ \end{array} \right)\left( \begin{array}{c} \!\!-p \\ q \\ r \\ \end{array} \right) =\begin{pmatrix}p-2q+2r\\2p-q+2r\\2p-2q+3r\end{pmatrix} .

    Now just show that (p-2q+2r)^2+(2p-q+2r)^2=(2p-2q+3r)^2 (under the assumption, of course, that p^2+q^2=r^2 ).

    Tonio
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by tonio View Post
    Take the first one, for example: \left( \begin{array}{ccc} -1 & -2 & 2 \\ -2 & -1 & 2 \\ -2 & -2 & 3 \\ \end{array} \right)\left( \begin{array}{c} \!\!-p \\ q \\ r \\ \end{array} \right) =\begin{pmatrix}p-2q+2r\\2p-q+2r\\2p-2q+3r\end{pmatrix} .

    Now just show that (p-2q+2r)^2+(2p-q+2r)^2=(2p-2q+3r)^2 (under the assumption, of course, that p^2+q^2=r^2 ).

    Tonio
    Hi thanks for the reply.

    That was my initial thought, however after working, both by hand and then checking in maple, they are not equal.

    (p-2q+2r)^2+(2p-q+2r)^2=5p^2-8pq+12pr+5q^2-12qr+8r^2

    and

    (2p-2q+3r)^2=4p^2-8pq+12pr+4q^2-12qr+9r^2

    Have double check the question, made sure I didn't make a mistake with my matrices, but they are all correct.

    Is this something I'm doing wrong?

    Thanks again for the reply
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    Quote Originally Posted by craig View Post
    Hi thanks for the reply.

    That was my initial thought, however after working, both by hand and then checking in maple, they are not equal.

    (p-2q+2r)^2+(2p-q+2r)^2=5p^2-8pq+12pr+5q^2-12qr+8r^2

    and

    (2p-2q+3r)^2=4p^2-8pq+12pr+4q^2-12qr+9r^2

    Have double check the question, made sure I didn't make a mistake with my matrices, but they are all correct.

    Is this something I'm doing wrong?


    Yes, concluding they aren't equal, because they are:

    5p^2-8pq+12pr+5q^2-12qr+8r^2=4p^2-8pq+12pr+4q^2-12qr+9r^2\Longleftrightarrow p^2+q^2=r^2 , and this last is true by assumption...

    Tonio

    Thanks again for the reply
    .
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  5. #5
    Super Member craig's Avatar
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    Sorry brain just wasn't working then, thanks again for the help!
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