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May17-05, 10:03 AM   #12
 
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RE; matt's post #7, if you draw those vertices, you can compute the area of the parallelogram by combining (adding and subtracting) areas of several triangles and rectangles. You should get 2 as the area, and since the determinant is supposedly negative, you should note that the image of the vector (0,1) is clockwise from, rather than counterclockwise from, the image of (1,0).