1. ## 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

2. 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

3. 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?

4. 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...

Tonio