# Tricky Matrices Problem

• Mar 16th 2010, 11:38 AM
craig
Tricky Matrices Problem
Got a bit of a confusing problem involving matrices.

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

For P,

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

Show that:

$\displaystyle P \left( \begin{array}{c} -p \\ q \\ r \\ \end{array} \right)$, $\displaystyle P \left( \begin{array}{c} p \\ -q \\ r \\ \end{array} \right)$ and $\displaystyle P \left( \begin{array}{c} -p \\ -q \\ r \\ \end{array} \right)$ 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 :)

• Mar 16th 2010, 02:41 PM
tonio
Quote:

Originally Posted by craig
Got a bit of a confusing problem involving matrices.

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

For P,

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

Show that:

$\displaystyle P \left( \begin{array}{c} -p \\ q \\ r \\ \end{array} \right)$, $\displaystyle P \left( \begin{array}{c} p \\ -q \\ r \\ \end{array} \right)$ and $\displaystyle P \left( \begin{array}{c} -p \\ -q \\ r \\ \end{array} \right)$ 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 :)

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

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

Tonio
• Mar 16th 2010, 05:25 PM
craig
Quote:

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

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

Tonio

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

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

and

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

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?

• Mar 16th 2010, 07:28 PM
tonio
Quote:

Originally Posted by craig

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

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

and

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

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:

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

Tonio