Question

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.

Answer 1

given two vectors <a,b> and <c,d>
they can define a parallelogram


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.