4 added 90 characters in body

I am hazarding a guess, I believe this should do the job

A = { {1, x, 0}, {0, -1, 0}, {0, y, 1} }

It really does not matter what x and y are, they can be chosen arbitrarily and can even be two formal symbols

B = { {0, 0, -$i$}, {$i$, 0, 0}, {0, 1, 0} }

Then A.A= B.B.B = Id

the order of A.B should $A.B$ would be infinite as when $x$ and $y$ are suitably chosen, for example one can choose $x$ and $y$ so that the coefficient of the matrices are $(A.B)^n$ unbounded ?

3 deleted 4 characters in body

I am hazarding a guess, I believe this should do the job

A = { {1, x, 0}, {0, -1, 0}, {0, y, 1} }

It really does not matter what x and y are, they can be chosen arbitrarily and can even be two formal symbols

B = { {0, 0, -\iota}, $i$}, {\iota, $i$, 0, 0}, {0, 1, 0} }

Then A.A= B.B.B = Id

the order of A.B should be infinite as the coefficient of the matrices are unbounded ?

I am hazarding a guess, I believe this should do the job

A = { {1, x, 0}, {0, -1, 0}, {0, 0y, 1} }

It really does not matter what x and y are, they can be chosen arbitrarily and can even be two formal symbols

B = { {0, 0, -\iota}, {\iota, 0, 0}, {0, 1, 0} }

Then A.A= B.B.B = Id

we can even choose x appropriately if needed

the order of A.B should be infinite as the coefficient of the matrices are unbounded ?

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