Function solve_lu 

pub fn solve_lu<T: Scalar>(
    l: &Matrix<T>,
    u: &Matrix<T>,
    b: &Matrix<T>,
) -> Result<Matrix<T>>

Solve LU * x = b

Given factors L and U such that A = L * U, solves A * x = b via forward and back substitution.

Examples

use tensor4all_tcicore::{from_vec2d, matrixlu::{rrlu, solve_lu}};

let m = from_vec2d(vec![vec![2.0_f64, 1.0], vec![1.0, 3.0]]);
let lu = rrlu(&m, None).unwrap();
let l = lu.left(false);
let u = lu.right(false);

// b = [5, 10] => solve Ax = b in permuted coordinates
let b = from_vec2d(vec![
    vec![m[[lu.row_permutation()[0], 0]] * 1.0 + m[[lu.row_permutation()[0], 1]] * 2.0],
    vec![m[[lu.row_permutation()[1], 0]] * 1.0 + m[[lu.row_permutation()[1], 1]] * 2.0],
]);
let x = solve_lu(&l, &u, &b).unwrap();
// Solution in permuted col order
assert!((x[[lu.col_permutation().iter().position(|&c| c == 0).unwrap(), 0]] - 1.0).abs() < 1e-10);
assert!((x[[lu.col_permutation().iter().position(|&c| c == 1).unwrap(), 0]] - 2.0).abs() < 1e-10);