Uniqueness of LDU Factorisation [Strang P105 2.6.18] 
Let $L$ be a lower triangular matrix, $D$ diagonal, and $U$ upper triangular.
  If $A = LDU$ and also $A = L_1D_1U_1$ with all factors invertible, then $L = L_1$ and
  $D = D_1$ and $U = U_1$. 'The three factors are unique!'      
Hint:
  Derive the equation $L_1^{-1}LD = D_1 U_1 U^{-l}.$ Are the two sides triangular or diagonal?
  Deduce $L = L^{-1}$ and $U = U^{-1}$ (they all have diagonal $1$'s). Then $D = D_1$.
Lemma: The inverse of a (lower/upper) triangular matrix is a (lower/upper) triangular matrix.

Uniqueness Proof: Assume $LDU = L_1D_1U_1$. Objective: Prove $X = X_1$ for each $X = L, D, U$. 
In keeping with the hint, $\color{green}{L_1^{-1}}LDU\color{#D555D1}{U^{-1}} = \color{green}{L_1^{-1}}L_1D_1U_1\color{#D555D1}{U^{-1}} \iff \color{green}{L_1^{-1}}LD = D_1U_1\color{#D555D1}{U^{-1}}$ $\Longrightarrow (\text{Lower triangular})D = D_1(\text{Upper triangular}).$ 
$\color{red}{\bigstar} $ So both sides are diagonal. $ \; L,U,L_1,U_1$ have diagonal $1$’s so $D = D_1$. Then $\color{green}{L_1^{-1}}L = I$ and $U_1\color{#D555D1}{U^{-1}} = I. \qquad \blacksquare$
$\Large{1.}$ I don't apprehend the (gruff) sentences after $\color{red}{\bigstar}$. Would someone please enlarge upon them?
$\Large{2.}$ Without the hint, how would one divine/previse to work with $L_1^{-1}LD = D_1 U_1 U^{-l}$?
This equation looks like the critical one in this proof.
 A: Google this problem and found this page :( but finally I worked out this problem somehow. So I will try to answer this question.
First notice that $L_1^{-1}L$ is a lower triangular matrix and has diagonal 1's, since both $L_1^{-1}$ and $L$ are lower triangular and have diagonal 1's.
Nonzero entries below the diagonal of $L_1^{-1}L$ will cause nonzero entries below the diagonal of $L_1^{-1}LD$ (since $D$ is also diagonal). And if that happens, $L_1^{-1}LD$ cannot equal to $D_1U_1U^{-1}$, thus entries below the diagonal of $L_1^{-1}L$ are all 0's.
Now we can conclude $L_1^{-1}L=I$. Similarly, $U_1U^{-1}=I$.
As for the second question, I can not help. 
A: I wanted to add that the non-singularity of $A$ is crucial.  E.g:
$\pmatrix{1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 }$
$\pmatrix{0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 }$
$\pmatrix{1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 } = $
$\pmatrix{0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 } = $
$\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 }$
$\pmatrix{0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 }$
$\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 }$
The zeros on the diagonal allow for "junk" to appear in the non-selected rows or columns.  In this example, the zero in $D_{1,1}$ annihilates the first column from $L$, and the first row from $R$.  In fact, we have
$\pmatrix{* & 0 & 0 \\ * & 1 & 0 \\ * & 0 & 1 }$
$\pmatrix{0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 }$
$\pmatrix{* & * & * \\ 0 & 1 & 0 \\ 0 & 0 & 1 } = $
$\pmatrix{0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 }$
Where $*$ ranges over $\mathbb{R}$ (or anything that annihilates with 0).
As for question 2, the question is "how do I derive the hint from the problem?"    Here's a rough train of thought...


*

*Because $L$ and $R$ are invertible (being composed of row/ column operations), you can move them around and see what happens.

*Since the inverse of row (column) operations is more row (column) operations, and composing lower (upper) triangular matrices maintains the lower (upper) triangular structure, consolidating them seems natural.

*If a lower and upper triangular matrix are equal, then they are both diagonal.

*You probably had to be a little clever to get to this point, but noticing that the ones are on the diagonals of $L_1^{-1}L$ and $U_1U^{-1}$ is probably something hard to find.  Maybe you realize it will be hard to prove anything about the other matrices, so you start trying to prove $D = D_1$, and hopefully you'll see this can happen if the diagonals are all 1's, at which point you may have to do some calculations to verify.

*Since $D$ and $D_1$ are non-singular, you have that there can be no 0s on the diagonals, so any garbage below or above the diagonals in $L_1^{-1}L$ and $U_1U^{-1}$ will get captured by $D$ and $D_1$, but this can't happen because we made sure our left and right hand sides were both diagonal.

*Finally, it takes a second to realize that a lower (upper) triangular matrix with ones on the diagonal and nothing below (above) the diagonal is the identity, so you finally get $L_1^{-1}L = I$ ($U_1U^{-1} = I$)

