= Solution
{c}
Eliminate the first column with multipliers $\ell_{21}=2$ and $\ell_{31}=-1$. The new second and third rows are $(0,2,4)$ and $(0,-1,3)$. The next multiplier is $\ell_{32}=-1/2$, giving final row $(0,0,5)$. Recording these multipliers gives the <LU decomposition>
$$
\boxed{L=\begin{pmatrix}1&0&0\\2&1&0\\-1&-1/2&1\end{pmatrix},\qquad
U=\begin{pmatrix}1&2&2\\0&2&4\\0&0&5\end{pmatrix},\qquad A=LU.}
$$
For an upper-triangular <matrix> with nonzero diagonal, the <back substitution> algorithm works from bottom to top:
$$
x_i=\frac{b_i-\sum_{j=i+1}^n u_{ij}x_j}{u_{ii}},\qquad i=n,n-1,\ldots,1.
$$
In the separate triangular system printed in the question, $5x_3=10$, $2x_2+4x_3=6$, and $x_1+2x_2+2x_3=3$. Hence
$$
\boxed{x=(1,-1,2)^{\mathsf T}.}
$$
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