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+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.