## anonymous one year ago My question is with respect to problem set 1, 1c-7. I don't understand the solution. In particular, what is the connection between the area of the parallelogram and the max value of the determinant? Could someone please explain in detail.

1. phi

given two vectors <a,b> and <c,d> they can define a parallelogram |dw:1442232715212:dw| We know the magnitude of the cross product of vectors x and y has the definition $|x\times y| = |x| |y| \sin \theta$ where theta is the angle between the vectors x and y This expression also represents the area of a parallelogram defined by x and y Thus, given two vectors <a,b> and <c,d> if we extend these to vectors in 3D by setting the z component to 0 <a,b,0> and <c,d,0> we can form the cross product <0,0, ad-bc> and the magnitude is $$\sqrt{0^2+0^2 +(ad-bc)^2} = ad-bc$$ and ad-bc is the determinant of a matrix whose columns are <a,b> and <c,d> I don't know that this gives any intuition about why a determinant is linked to area, but it does show there is that condition.

2. anonymous