Finding the $LU$ factorization of the matrix 
Find the $LU$ factorization of the matrix:
  $$\begin{bmatrix} 1 & 1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7 \end{bmatrix}$$

I am aware that I need to find $L=\begin{bmatrix} 1 & 0 & 0 \\ * & 1 & 0 \\ * & * & 1 \end{bmatrix}$ and $U=\begin{bmatrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \end{bmatrix}$
I did row transformations and got $U=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}$ but I couldn't understand how to find $L=\begin{bmatrix} 1 & 0 & 0 \\ * & 1 & 0 \\ * & * & 1 \end{bmatrix}$
Can anyone explain how to find $L$
 A: For doing LU decomposition, you need to do Gaussian elimination. Here I'll just help you with the procedure, but if you want to understand why I recommend you to see this pdf http://www.math.iit.edu/~fass/477577_Chapter_7.pdf.  Lets apply Gaussian elimination to A
\begin{equation*}
A = 
\left[
\begin{matrix}
1 & 1 & 1\\
3 & 5 & 6 \\
-2 & 2 & 7
\end{matrix}
\right]
\end{equation*}
For eliminating $A_{12}$ and $A_{13}$, we need to multiply by $-3$ and by $2$ the first row and add this to the second and third row respectively, obtaining
\begin{equation*}
\left[
\begin{matrix}
1 & 1 & 1\\
0 & 2 & 3 \\
0 & 4 & 9
\end{matrix}
\right]
\end{equation*}
Now we eliminate $A_{32}$ multiplying by $-2$ the second row and adding it to the third one, obtaining
\begin{equation*}
\left[
\begin{matrix}
1 & 1 & 1\\
0 & 2 & 3 \\
0 & 0 & 3
\end{matrix}
\right]
\end{equation*}
Which is the $U$ matrix, for the $L$ matrix we use the factors by which we multiplied each row for obtaining the $U$ matrix, i.e. $3,-2,2$. We use this elements in the position of the elements they eliminated, then
\begin{equation*}
L = 
\left[
\begin{matrix}
1 & 0 & 0\\
3 & 1 & 0 \\
-2 & 2 & 1
\end{matrix}
\right]
\end{equation*}
A: I use to call $E_{ij}(d)$ the operation of summing to the $i$-th row the $j$-th row multiplied by $d$.
Thus the Gaussian elimination runs as
\begin{align}
\begin{bmatrix}
1 & 1 & 1 \\
3 & 5 & 6 \\
-2 & 2 & 7
\end{bmatrix}
&\xrightarrow{\begin{gathered} E_{31}(2) \\ E_{21}(-3) \end{gathered}}
\begin{bmatrix}
1 & 1 & 1 \\
0 & 2 & 3 \\
0 & 4 & 9
\end{bmatrix}
\\[6px]
&\xrightarrow{E_{32}(-2)}
\begin{bmatrix}
1 & 1 & 1 \\
0 & 2 & 3 \\
0 & 0 & 3
\end{bmatrix}=U
\end{align}
Now it's just a matter of replacing each transformation by its inverse:
$$
L=E_{21}(3)E_{31}(-2)E_{32}(2)=
\begin{bmatrix}
1 & 0 & 0 \\
3 & 1 & 0 \\
-2 & 2 & 1
\end{bmatrix}
$$
At place $(2,1)$ put $3$, and so on.
This is justified by the fact that, if you consider $E_{ij}(d)$ the matrix you obtain by applying the transformation to the identity, then performing the row operation is the same as multiplying by this matrix.
Thus we have written
$$
U=E_{32}(-2)E_{31}(2)E_{21}(-3)A
$$
(where $A$ is your original matrix) and so
$$
A=\underbrace{E_{21}(3)E_{31}(-2)E_{32}(2)}_{L}U
$$
If one follows a strict order in doing the Gaussian elimination (top down and left to right), filling the matrix $L$ is just putting the coefficients in the indicated place.
A: Going off of GBes's comment, you can find $L$ through some quick matrix multiplication. Let's call the original matrix $A$. We know that $A=LU$, and since you have $U$ (which is invertible), you can multiply by $U^{-1}$ on the right on both sides to get the following:
$$\begin{align}
A&=LU \\
AU^{-1}&=LUU^{-1} \\
AU^{-1}&=L\end{align}$$
A: Gaussian Elimination without Pivoting is as follows. The primary purpose of Gaussian elimination if you follow this is to find $\ell_{jk}$ which zeros out the row below. That is why it is the ratio of the two rows and then you subtract them. This continues on and on. 

Suppose that 
$$ A  = \begin{bmatrix} 1 &  1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7  \end{bmatrix} $$
$$ A = LU $$
$$ U =A, L=I$$
$$ k=1,m=3,j=2$$
$$\ell_{21} = \frac{u_{21}}{u_{11}}  = \frac{a_{21}}{a_{11}} = 3 $$
$$ u_{2,1:3} = u_{2,1:3} - 3 \cdot u_{1,1:3}  $$
Then we're going to subtract 3 times the 1st row from the 2nd row
$$ \begin{bmatrix} 3 & 5 & 6 \end{bmatrix} - 3 \cdot \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3\end{bmatrix}  $$
Updating each of them
$$U = \begin{bmatrix} 1 &  1  & 1 \\ 0 & 2 & 3 \\ -2 & 2 & 7 \end{bmatrix} $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
$$k=1,j=3,m=3 $$
$$\ell_{31} = \frac{u_{31}}{u_{11}} = \frac{-2}{1} = -2 $$
$$ u_{3,1:3} = u_{3,1:3} +2 \cdot u_{1,1:3} $$
Then we add two times the first row to the third row
$$ \begin{bmatrix} -2 & 2 & 7 \end{bmatrix} +  2 \cdot \begin{bmatrix} 1 & 1& 1  \end{bmatrix} = \begin{bmatrix}0 & 4 & 9  \end{bmatrix} $$
Updating 
$$ U = \begin{bmatrix} 1 &  1  & 1 \\ 0 & 2 & 3 \\ 0 & 4 & 9 \end{bmatrix}  $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} $$
$$ k=2, j=3,m=3 $$
$$ \ell_{32} = \frac{u_{32}}{u_{22}} = \frac{4}{2} = 2$$
We're subtracting out little blocks
$$ u_{3,2:3} = u_{3,2:3} - 2 \cdot u_{2,2:3} $$
$$ \begin{bmatrix} 4 & 9 \end{bmatrix} - 2 \cdot\begin{bmatrix} 2& 3 \end{bmatrix} = \begin{bmatrix} 0 & 3 \end{bmatrix} $$
Updating 
$$ U = \begin{bmatrix} 1 &  1  & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}  $$
$$ L =\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix} $$
It now terminates
$$ A  = LU $$
$$ \underbrace{\begin{bmatrix} 1 &  1 & 1 \\ 3 & 5 & 6 \\ -2 & 2 & 7  \end{bmatrix}}_{A} = \underbrace{\begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 2 & 1 \end{bmatrix}}_{L} \underbrace{\begin{bmatrix} 1 &  1  & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{bmatrix}}_{U}   $$
