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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
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