Solve $A\cdot x = b$ by Naïve Gaussian elimination The following is a homework problem:

Let:
  $$A = \left[\matrix{4& 2& -5 &1\\ -8& 0& 9& 7\\ -32& -4& 43& 18\\ 24 &4 &-22 &-8\\} \right]$$
$$b = \left[\matrix{-1 \\ -1 \\ 3 \\ 14}\right]$$
Solve $A\cdot x = b$ by Naïve Gaussian elimination. Note the relationship
  between the subtractive row multipliers and the elements of the L
  matrix. How is the U matrix related to the reduced terms in the
  augmented matrix before back-substitution?

I have solved the problem and found $L$ and $U$ by Naïve Gaussian elimination.

$$U = \left[\matrix{4& 2& -5& 1\\ 0 &4 &-1& 9\\ 0& 0 &6 &-1\\ 0& 0& 0& 5\\}\right]$$
  $$L = \left[\matrix{1& 0& 0& 0\\ -2& 0& 0& 0\\ -8& 3& 1& 0\\ 6& -2& -6 &1 \\ }\right]$$

However, what I seem to be having trouble with is the second part of the question ('How is the $U$ matrix related to the reduced terms in the augmented matrix before back-substitution?')
I understand that $U$ is the first 4 columns of the augmented $Ax$ matrix after row reduction, but what would be a proper answer to this question?
 A: The matrix $A$ is lower triangular and invertible, because all coefficients on the diagonal are nonzero. So when you do the $LU$ decomposition, you have $L=A$ and $U$ is the identity matrix.
If you properly do Gaussian elimination, the first step (eliminating under the first pivot) will give the matrix
$$
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 3 & 1 & 0\\
0 &-2 & 1 & 1
\end{bmatrix}
$$
and so on for the other columns.

Since the original matrix has been changed, here's what can be said; after elimination according to the Gauss-Doolittle method (no reduction of pivots), we find, for the augmented matrix, the reduced form
$$
[U\mid c]=\left[\begin{array}{cccc|c}
4 & 2 & -5 & 1 & -1 \\
0 & 4 & -1 & 9 & -3 \\
0 & 0 & 6 & -1 & 4 \\
0 & 0 & 0 & 5 & 10
\end{array}\right]
$$
and the matrix $L$ such that $L[\,U\mid c\,]=[\,A\mid b\,]$ is
$$
L=\begin{bmatrix}
1 & 0 & 0 & 0 \\
-2 & 1 & 0 & 0 \\
-6 & -3 & 1 & 0 \\
8 & 2 & -1 & 1
\end{bmatrix}.
$$
The question ‘How is the $U$ matrix related to the reduced terms in the augmented matrix before back-substitution?’ is not really clear. What I can say is that the system
$$
Ux=c,
$$
where $c$ denotes the last column in the reduced augmented matrix, is equivalent to the original linear system $Ax=b$. In particular, the form of $U$ tells you that the system has a unique solution. The fact that $L[\,U\mid c\,]=[\,A\mid b\,]$ implies that $Lc=b$, so $c=L^{-1}b$.
What can be done now is to multiply the last row by $1/5$ and do “backwards elimination”, reducing the pivots: we find
$$\left[\begin{array}{cccc|c}
1 & 0 & 0 & 0 & 12 \\
0 & 1 & 0 & 0 & -5 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 2
\end{array}\right]
$$
which shows the unique solution.
