You're welcome.

Just in case you haven't noticed , we now have an easy solution to an infinite amount of problems of this form.

For \(\displaystyle \ p \ \ne \ 0\)

IF

\(\displaystyle A^2 \ + \ B^2 \ = \ \[p \ q \\ q \ p \]\)

THEN a solution will be

\(\displaystyle A \ = \[ \sqrt{p} \ \ \frac{q}{ \sqrt{p}} \\ \ 0 \ \ \ \ 0 \]\)

\(\displaystyle B \ = \ \[ \ \ 0 \ \ \ \ 0 \\ \frac{q}{ \sqrt{p}} \ \ \sqrt{p} \]\)

We can also transpose both A , B and get another family of solutions. Then we would be working with columns.

An interesting question to ask is 'what if \(\displaystyle \ p \ = \ 0 \ \ \text{AND} \ \ q \ \ne \ 0 \\)?'

A place to start could be ,

Find a 2×2 matrix A such that

\(\displaystyle A^2 \ = \ \[ 0 \ 1 \\ 1 \ 0 \]\)

Can it be found ?