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  • 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.

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  1. phi
    • one year ago
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    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
    • one year ago
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    Thanks. I really appreciate your helpful and detailed responses!

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