= Solution
The first elimination multipliers are $l_{21}=2$ and $l_{31}=-1$. Subtracting those multiples of row one gives lower rows $(0,2,4)$ and $(0,-1,3)$. The next multiplier is $l_{32}=-1/2$, and the final elimination produces $(0,0,5)$. Thus
$$
\boxed{L=\begin{pmatrix}1&0&0\\2&1&0\\-1&-1/2&1\end{pmatrix},
\qquad U=\begin{pmatrix}1&2&2\\0&2&4\\0&0&5\end{pmatrix},\qquad A=LU.}
$$
For an upper triangular system $Ux=b$ with nonzero diagonal, <back substitution> computes
$$
\boxed{x_i=\frac{b_i-\sum_{j=i+1}^nu_{ij}x_j}{u_{ii}},
\qquad i=n,n-1,\ldots,1.}
$$
For the supplied triangular system, the last row gives $x_3=10/5=2$; the second gives $x_2=(6-4\cdot2)/2=-1$; the first gives $x_1=3-2(-1)-2(2)=1$. Hence
$$
\boxed{x=(1,-1,2)^T.}
$$
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