A <triangular matrix> has <determinant> equal to the product of its diagonal entries, so $\det L=1$ and $\det U=3\cdot4\cdot(-11)=-132$. A <permutation matrix> exchanging two rows has <determinant> $-1$. Using multiplicativity of the <determinant> in $PA=LU$ yields $(-1)\det A=-132$. Hence \b[$\det A=132$].

Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Digital systems, linear algebra, introduction to formal proof / 4 / c / iii / Solution

 Solution (source code)