$A = \begin{bmatrix} 2 &5 &9 \\ 4 &6 &5 \\ 8&2 &3 \end{bmatrix}$
-- Using Gaussian Elimination, we will change $A$ into upper triangular matrix $U$ and simultaneously find out elementary matrices $E_{21}$, $E_{31}$ and $E_{32}$.
-- To get the elementary matrices, we have to do the same operation on the Identity matrix as we do on the original matrix $A.$
On applying $R_2 \leftarrow R_{2} - 2R_1$
$A$ becomes $\begin{bmatrix} 2 &5 &9 \\ 0 &-4 &-13 \\ 8 &2 &3 \end{bmatrix}$
and $E_{21}$ becomes $\begin{bmatrix} 1 &0 &0 \\ -2 &1 &0 \\ 0 &0 &1 \end{bmatrix}$ $[ \because R_2 \leftarrow R_{2} - 2R_1 $ on identity matrix $I$ will give $E_{21}]$
We can write it as $\begin{bmatrix} 2 &5 &9 \\ 0 &-4 &-13 \\ 8 &2 &3 \end{bmatrix}=$ $\begin{bmatrix} 1 &0 &0 \\ -2 &1 &0 \\ 0 &0 &1 \end{bmatrix}$* $\begin{bmatrix} 2 &5 &9 \\ 4 &6 &5 \\ 8&2 &3 \end{bmatrix} $
Now, applying $R_3 \leftarrow R_{3} - 4R_1$
$A$ becomes $\begin{bmatrix} 2 &5 &9 \\ 0 &-4 &-13 \\ 0 &-18 &-33 \end{bmatrix}$
and $E_{31}$ becomes $\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ -4 &0 &1 \end{bmatrix}$ $[ \because R_3 \leftarrow R_{3} - 4R_1 $ on identity matrix $I$ will give $E_{31}]$
Now, applying $R_3 \leftarrow R_{3} - \frac{18}{4}R_2$
$A$ becomes $\begin{bmatrix} 2 &5 &9 \\ 0 &-4 &-13 \\ 0 &0 &\frac{51}{2} \end{bmatrix}$
and $E_{32}$ becomes $\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0&\frac{-18}{4} &1 \end{bmatrix}$ $[ \because R_3 \leftarrow R_{3} - \frac{18}{4}R_2$ on identity matrix $I$ will give $E_{32}]$
So, here,
$U= \begin{bmatrix} 2 &5 &9 \\ 0 &-4 &-13 \\ 0 &0 &\frac{51}{2} \end{bmatrix},\;\; E_{21} =\begin{bmatrix} 1 &0 &0 \\ -2 &1 &0 \\ 0 &0 &1 \end{bmatrix},\;\; E_{31} =\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ -4 &0 &1 \end{bmatrix},\;\; $$E_{32} =\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &\frac{-18}{4} &1 \end{bmatrix},\;\;$
Now, we can write all these things as :-
$\Rightarrow E_{32}E_{31}E_{21}(A) = U$
$\Rightarrow A = E_{21}^{-1}E_{31}^{-1}E_{32}^{-1} U$
Here, $E_{21}^{-1},\;E_{31}^{-1},\; E_{32}^{-1}$ are very simple to compute. Just do the reverse operation means in row operations where we have done subtraction , we have to do addition.
So, $\Rightarrow A = E_{21}^{-1}E_{31}^{-1}E_{32}^{-1} U$ becomes
$A= \begin{bmatrix} 1 &0 &0 \\ 2 &1 &0 \\ 0 &0 &1 \end{bmatrix}*\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 4 &0 &1 \end{bmatrix}*\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &\frac{18}{4} &1 \end{bmatrix}*\begin{bmatrix} 2 &5 &9 \\ 0 &-4 &-13 \\ 0 &0 &\frac{51}{2} \end{bmatrix}$
$\Rightarrow A= \begin{bmatrix} 1 &0 &0 \\ 2 &1 &0 \\ 4 &\frac{18}{4} &1 \end{bmatrix}*\begin{bmatrix} 2 &5 &9 \\ 0 &-4 &-13 \\ 0 &0 &\frac{51}{2} \end{bmatrix}$
It is the LU decomposition where $L= \begin{bmatrix} 1 &0 &0 \\ 2 &1 &0 \\ 4 &\frac{18}{4} &1 \end{bmatrix}\;\;and\; U=\begin{bmatrix} 2 &5 &9 \\ 0 &-4 &-13 \\ 0 &0 &\frac{51}{2} \end{bmatrix}$
Since, we have not done any row exchange, So, permutation matrix will not be changed. It will remain identity matrix $I.$
So, Answer is :-
$L= \begin{bmatrix} 1 &0 &0 \\ 2 &1 &0 \\ 4 &\frac{18}{4} &1 \end{bmatrix}\;\; U=\begin{bmatrix} 2 &5 &9 \\ 0 &-4 &-13 \\ 0 &0 &\frac{51}{2} \end{bmatrix}$ and $P= \begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{bmatrix}$