 Solution

First add row 1 to rows 2 and 3. This is the product of two elementary matrices, written compactly asSwap rows 2 and 4 with $P$, then subtract half of the new row 2 from row 3 with $E_2$:This gives the upper triangular matrixBecause $P^{-1}=P$, rearranging yields the convention $A=PLU$, whereThe conjugation by $P$ moves the earlier elimination multipliers to the appropriate rows. Ignoring this permutation when forming $L$ would give an incorrect PLU factorization. Here the same permutation is its own inverse, so $PA=LU$ also holds.