Math and science::Algebra

Determinant from pivots

Let $A$ be a matrix that is decomposed through elimination to $A = L U$ . The truth of the following statements allow us to read the determinant of $A$ from the pivots that are placed along the diagonal of $U$ :

If $B = C D$ then $det (B) = det (C) det (D)$ .
The determinant of a triangular matrix is the product of its diagonal elements.
All entries along the diagonal of $L$ are 1.

From these statements, we can see that the determinant of $A$ is the product of the pivots that are placed along the diagonal of $U$ .

Example

A = [\begin{matrix} 4 & 2 & 0 \\ 2 & 2 & 3 \\ 0 & 3 & 1 \end{matrix}]

Which is decomposed as $A = L U$ to:

A = L U = [\begin{matrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ 0 & 3 & 1 \end{matrix}] [\begin{matrix} 4 & 2 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & - 8 \end{matrix}]

The determinant of $A$ is the product of the pivots along the diagonal of $U$ , so $det (A) = (4) (1) (- 8) = - 32$ .