LU and determinant

When the LU factorization exists, we have

$$\det (A)=\prod _{i=1}^n{u}_{ii}$$

Proof: Use $A=LU$ and the property of the determinant:

$$\det (A)=\det (L)\det (U)$$

For a triangular matrix, the determinant is the product of the diagonal entries. $L$ has 1 along its diagonal.

Determinant, Triangular factorization