= Solution
The coefficient matrix and right-hand side are
$$
A=\begin{pmatrix}2&4&3\\2&-2&-3\\-4&-2&2\end{pmatrix},\qquad
\mathbf b=\begin{pmatrix}4\\0\\-6\end{pmatrix}.
$$
For an <LU decomposition> with unit diagonal in the lower factor, subtract row 1 from row 2, add twice row 1 to row 3, and then add the new row 2 to row 3. The elimination multipliers are $l_{21}=1$, $l_{31}=-2$, and $l_{32}=-1$. Thus
$$
\boxed{L=\begin{pmatrix}1&0&0\\1&1&0\\-2&-1&1\end{pmatrix},\qquad
U=\begin{pmatrix}2&4&3\\0&-6&-6\\0&0&2\end{pmatrix}.}
$$
Multiplication verifies $LU=A$. Forward substitution in $L\mathbf y=\mathbf b$ gives $y_1=4$, $y_2=-4$, and $y_3=-2$. Back substitution then gives $x_3=-1$, $x_2=5/3$, and $x_1=1/6$:
$$
\boxed{\mathbf y=(4,-4,-2)^T,\qquad
\mathbf x=(1/6,5/3,-1)^T.}
$$
Every pivot is nonzero, so no row permutation is needed and the <linear system> has a unique solution.
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