LU decomposition

First step. $L_2 \leftarrow L_2 - 2L_1$, $L_3 \leftarrow L_3 - 4L_1$. Multipliers: $\ell_{21} = 2$, $\ell_{31} = 4$.
$$A \to \begin{pmatrix} 2 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 3 & 5 \end{pmatrix}.$$

Second step. $L_3 \leftarrow L_3 - 3L_2$. Multiplier: $\ell_{32} = 3$.
$$A \to \begin{pmatrix} 2 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{pmatrix}.$$

Conclusion.
$$L = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 4 & 3 & 1 \end{pmatrix}, \qquad U = \begin{pmatrix} 2 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{pmatrix}.$$

Verification. $LU = A$, as can be checked directly.