Since, A=LU
$\begin{bmatrix} 1 &2 \\ 3&8 \end{bmatrix}$ = $\begin{bmatrix} l1 &0 \\ l2&l3 \end{bmatrix}$ $\begin{bmatrix} 1 &u1 \\ 0&1 \end{bmatrix}$
= $\begin{bmatrix} l1 &l1 u1 \\l2&l2u1+l3 \end{bmatrix}$
Solving this,
$l1$ =1
$l3$ =2
So, trace(L)= $l1$ + $l3$ = 3