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We received many correct solutions describing each of the transformations. Well done Rahel from Dartford Grammar School for Boys, Harry from Culford, Dominic, Ellie, Finn and Max from Brenchley andMatfield Primary School, Jannis from Long Bay, Ben from the Perse School and Emilio from St Peter's College.
However, the second part caught almost everyone out: since it looked like the first four transformations were "undone" by their inverses, most people thought that the string of transformations returned the original shape back to where it started.
Only Emilio managed to work out that this was not the case:
$R$ reflects the figure vertically.
$S$ rotates it 90 degrees clockwise.
This leaves the figure like this:
$T$ translates the figure 1 right:
$I$ will do nothing.
$R^{-1}$ reflects the figure again because inverse reflections are the same as the original reflection.
$S^{-1}$rotates the figure $90^\circ$ anticlockwise:
$T^{-1}$ will translate the figure left one square.
$I$ does nothing, so $I^{-1}$ will do nothing.
Therefore a rotation of $180^\circ$ round (-0.5, -0.5) will have the same result as $R S T I R^{-1}S^{-1}T^{-1}I^{-1}$
Well done Emilio.
Editor's note: