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Suppose an invertible matrix has two LU factorizations, $A=L_1{U}_1=L_2{U}_2$, with $L_1,L_2$ unit lower triangular matrices and $U_1,U_2$ upper triangular matrices. All four factors are invertible matrices, since their square products are invertible. Multiplication by the appropriate inverse matrices givesBy the supplied closure properties, the left side is a unit lower triangular matrix and the right side is an upper triangular matrix. A matrix of both types is diagonal; its diagonal entries are all one. Thus both sides are the identity matrix, whenceQualification of the printed wording: this establishes uniqueness if the specified LU factorization exists. Invertibility alone does not guarantee an unpivoted LU factorization. For example $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ is invertible, but in a product $\left(\begin{array}{cc}1& 0\\ \ell & 1\end{array}\right)\left(\begin{array}{cc}{u}_{11}& u_{12}\\ 0& u_{22}\end{array}\right)$, the zero top-left entry forces $u_{11}=0$, making the bottom-left entry $\ell u_{11}=0$ as well, contrary to the required value one. A permutation matrix is needed in general, as in PLU factorization.

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.