I tried this:
Clear[b1, b2, A, p]
b1 = {-1, 1};
b2 = {1, 1};
A = {{2, 1}, {-1, 1}};
p = Parallelogram[{0, 0}, {b1, b2}];
Then drew this:
Graphics[{
{Red, p},
Blue, Thick,
Arrow[{{0, 0}, b1}],
Arrow[{{0, 0}, b2}],
{Opacity[0.6], GeometricTransformation[p, A]},
Red,
Arrow[{{0, 0}, A.b1}],
Arrow[{{0, 0}, A.b2}]
}]
Then I tried:
Area[GeometricTransformation[p, A]]
But it didn't work, giving me:
Area::reg: GeometricTransformation[p,{{2,1},{-1,1}}] is not a correctly specified region. >>
Now, I know I can get the area of the transformed region with:
Abs[Det[A.Transpose[{b1, b2}]]]
Which is 6, but I am wondering if there is a simple way (for students just beginning with Mathematica) to use the Area command in this situation.
Update: Thanks to MichaelE2, J.M., and rcollyer, I was able to transform the region:
t = TransformedRegion[p, AffineTransform[A]]
Which returned Parallelogram[{0, 0}, {{-1, 2}, {3, 0}}]
, which is terrific information for students.
Then I was able to draw the same image like this:
Graphics[{
{Red, p},
Blue, Thick,
Arrow[{{0, 0}, b1}],
Arrow[{{0, 0}, b2}],
{Opacity[0.6], t},
Red,
Arrow[{{0, 0}, A.b1}],
Arrow[{{0, 0}, A.b2}]
}]
Then I was able to get the area.
Area[t]
Which returned a 6.
Area@DiscretizeGraphics@Graphics@GeometricTransformation[p, A]
fails miserably, andNormal@GeometricTransformation[p, A]
fails, too. :/ $\endgroup$TransformedRegion[]
instead. $\endgroup$