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

Eliminate the first column using multipliers $3$ and $2$. The two residual rows have second entries zero and four, so interchange their order for the next pivot. One resulting Pivoted LU decomposition isThe factors below the first pivot must follow the permuted row order, which is why $L$ contains $2$ before $3$. Direct multiplication gives rows $(3,1,4)$, $(6,6,6)$ and $(9,3,1)$ in both $PA$ and $LU$. This checks the permutation convention rather than confusing $PA=LU$ with a different placement of $P$.

Apply the same permutation to the right-hand side and solve two triangular systems. Forward substitution in $Ly=Pb$ givesso $y=(3,-6,-11{)}^T$. Back substitution in $Ux=y$ gives $-11x_3=-11$, $4x_2-2x_3=-6$, and $3x_1+x_2+4x_3=3$. Therefore $x=(0,-1,1{)}^T$. Substitution in the original system returns $(3,-2,0{)}^T$, as required.

A triangular matrix has determinant equal to the product of its diagonal entries, so $\det L=1$ and $\det U=3\cdot 4\cdot (-11)=-132$. A permutation matrix exchanging two rows has determinant $-1$. Using multiplicativity of the determinant in $PA=LU$ yields $(-1)\det A=-132$. Hence $\det A=132$.