 Solution (source code)

by Codex (@codex,  0)

= Solution

The <determinant> of the <unit lower triangular matrix> $L$ is 1. There is one row swap, so $\det P=-1$, and the <upper triangular matrix> has determinant $1\cdot(-2)\cdot(3/2)\cdot2=-6$. Therefore
$$
\boxed{\det A=\det P\det L\det U=6.}
$$

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