Ax = b where A = mxn, x = 1xm, b = 1xn. x_i and b_i are defined to be in
the ith row of their corresponding vectors.

If C is an invertible mxm matrix, CAx = Cb is equivalent to Ax = b.
(1)
    If x_0 satisfies Ax_0 = b, CAx_0 = Cb is satisfied.
(2)
    If x_1 satisfies CAx_0 = Cb, C^{-1} CAx_0 = C^{-1} Cb \to Ax_0 = b
    is satisfied.

(I believe this proof requires associativity)

    LU Decomposition/Factorization

Goal: Factorize A into a lower triangular and upper triangular matrices
L and U respectively. (These are very non-unique)

A = LU. Conventionally, L is mxm, and U is mxn.
Ax = b ---> LUx = b ---> Y = Ux, LY = b.

If we assert that L is an invertible matrix (e.g. ones down the
diagonal), this is trivial to solve
  1    0  0           y1       b1
( c21  1  0 ... ) * ( y2 ) = ( b2 )
  c31 c32 1           y3       b3
      ...             ...      ...
Y has a unique solution by construction. ( ex: y1 = b1, c21*y1 + y2 = b2 ...)

U is upper triangular, meaning that it is strictly zero below the
diagonal started at the top-left. (However, we have no control over the
values of the diagonal or the values above the diagonal)

[Note] How do we ensure that L and U retain the properties of the
original matrix's solution? (i.e. not just matrices of zero)

Using a series of row operations (written as invertible matrices C_1,
C_2, ..., C_k -- that is, adding a multiple to a lower row) will turn A
into echelon (upper triangular), and C_1^{-1}C_2^{-1}*...*C_k^{-1} = L
(lower triangular). U is not reduced-echelon, but it is echelon/upper
triangular.