Computed three-by-three LU decomposition ($A=LU$)

Here is an original example of Gaussian elimination that can be checked entirely with integers. Start withThe first pivot element is $2$. The elimination multipliers are $l_{21}=4/2=2$ and $l_{31}=-2/2=-1$. Apply the elementary row operations $R_2\leftarrow R_2-2R_1$ and $R_3\leftarrow R_3+R_1$ to obtainNow the second pivot element is $3$, and $l_{32}=9/3=3$. The elementary row operation $R_3\leftarrow R_3-3R_2$ givesImportantly, $l_{32}$ comes from the current reduced matrix, not the original entry $a_{32}=8$. To reconstruct $A$ from $U$, its second row is twice the first row of $U$ plus the second; its third row is minus the first row of $U$ plus three times the second plus the third. These are exactly the coefficients in the lower triangular matrix $L$.

The unpivoted calculation here illustrates the structure of LU decomposition; applying partial pivoting to this particular $A$ would choose a different first row and different factors. Both procedures can represent the same linear system when their permutation matrices are handled consistently.