Thread: Hard linear transformation question (corrected)

1. Hard linear transformation question (corrected) The diagram above shows a tessellation of a plane of equilateral triangular tiles. A right-angled triangle is of such a shape that it can cover exactly half of any tile. Its initial position is marked t in the diagram.

The operation of reflecting t in its shortest side is written A, and starting with the given position t the result is A(t), which is marked with the figure 1.

Reflection in the second side is denoted by B, so that B(t) is 2.

Reflection in the hypotenuse is C so that C(t) is 3.

Position 4 is B(3) = B(t), so that the operation carrying t to 4 is BC.

Questions...

(i) Show that operations AB and BA are the same

(ii) Describe the combined operation BC geometrically.

(iii)Simplify C(CB), (CC)B and BC(CB)...using I to denote the identity operation if necessary.

(iv) Obtain expressions, as combinations of A, B and C for operations which carry

(a) t to x
(b) t to y
(c) x to t

If you can answer some or all of this it is much appreciated.

2. Re: Hard linear transformation question (corrected)

(ii) if initial triangle is $mnp$ then

$C(m)=m,C(p)=p,C(n)=q$

and

$B(m)=m,B(p)=r,B(q)=q$

therefore

$BC(m)=m,BC(p)=r,BC(n)=q$

that's a rotation

3. Re: Hard linear transformation question (corrected)

My guess at (iV) is that

(a) BC

(b) ABC

(c) CB