I was reading about determinants and what they actually represent. I’ve been working with solutions to the Ax = b problem where A has a determinant which is zero (i.e. it has no unique inverse). Determinants were always just a thing you did and there were magic properties about them. I suddenly realise why intuitively inv(A) exists iff det(A) != 0, simply from the geometric interpretation of the determinant.
The determinant of an n x n matrix represents the “signed volume”* of an n-dimensional cube after being multiplied by A (transformed by). So A is a transformation matrix. If we map to a zero volume, and thus the determinant is zero, we have zero in one of the dimensions.
Now zero is evil because multiplying everything by it yields 0. There is no information left regarding the original dimension as any value would have sufficed to obtain a zero.
Simply speaking, can you inverse 0a = 0? What is a? a is any value you like, so there is no inverse. Likewise for one of our dimensions.
So, when det(A) = 0 we have a matrix transformation that kills one of the dimensions (it maps all values onto a (hyper)plane) no matter what is transformed by it. When we write Ax = b, we know that one of b has a zero in some dimension and therefore we cannot determine exactly where we started to get a solution, since x could have any position in the deficient dimension, and this was mapped to zero in the transformation. Therefore you could potentially find infinitely many inverses since once or more of those dimensions has no constraint.
Of course when I say that it is in a dimension, we are strictly talking about any n-dimensional basis. So suppose you have a plane in R^3. You have a transformation matrix that projects all points in 3D onto that plane. So in the direction of normal to that plane, we have a scale of zero (eigen value as it happens), and thus the projection matrix is a matrix with no unique inverse.
If you’re interested and didn’t quite get what I was talking about, see http://www.youtube.com/watch?v=n-S63_goDFg