$\newcommand{\W}[1]{ \; #1 \; } \newcommand{\R}[1]{ {\rm #1} } \newcommand{\B}[1]{ {\bf #1} } \newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} } \newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} } \newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} } \newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }$
Determinant Using Expansion by Lu Factorization

Syntax
# include <cppad/speed/det_by_lu.hpp>  det_by_lu<Scalar> det(n)  d = det(a) 
Inclusion
The template class det_by_lu is defined in the CppAD namespace by including the file cppad/speed/det_by_lu.hpp (relative to the CppAD distribution directory).

Constructor
The syntax       det_by_lu<Scalar> det(n)  constructs the object det which can be used for evaluating the determinant of n by n matrices using LU factorization.

Scalar
The type Scalar can be any NumericType

n
The argument n has prototype       size_t n 
det
The syntax       d = det(a)  returns the determinant of the matrix $A$ using LU factorization.

a
The argument a has prototype       const Vector &a  It must be a Vector with length $n * n$ and with It must be a Vector with length $n * n$ and with elements of type Scalar . The elements of the $n \times n$ matrix $A$ are defined, for $i = 0 , \ldots , n-1$ and $j = 0 , \ldots , n-1$, by $$A_{i,j} = a[ i * m + j]$$

d
The return value d has prototype       Scalar d 
Vector
If y is a Vector object, it must support the syntax       y[i]  where i has type size_t with value less than $n * n$. This must return a Scalar value corresponding to the i-th element of the vector y . This is the only requirement of the type Vector .

Example
The file det_by_lu.cpp contains an example and test of det_by_lu.hpp. It returns true if it succeeds and false otherwise.

Source Code
The file det_by_lu.hpp contains the source for this template function.