Table of Contents
1. Overview
In this article, we’ll describe how to find the determinant of the matrix using Kraut method, which works in $O(N^3)$.
The Kraut algorithm finds decomposition of matrix $A$ as $A = L U$ where $L$ is lower triangular and $U$ is upper triangular matrix. Without loss of generality, we can assume that all the diagonal elements of $L$ are equal to 1. Once we know these matrices, it is easy to calculate the determinant of $A$: it is equal to the product of all the elements on the main diagonal of the matrix $U$.
There is a theorem stating that any invertible matrix has a LU-decomposition, and it is unique, if and only if all its principle minors are non-zero. We consider only such decomposition in which the diagonal of matrix $L$ consists of ones.
Let $A$ be the matrix and $N$ – its size. We will find the elements of the matrices $L$ and $U$ using the following steps:
- Let $L_{i i} = 1$ for $i = 1, 2, …, N$.
- For each $j = 1, 2, …, N$ perform:
- For $i = 1, 2, …, j$ find values $$U_{ij} = A_{ij} – \sum_{k=1}^{i-1} L_{ik} \cdot U_{kj}$$
- Next, for $i = j+1, j+2, …, N$ find values $$L_{ij} = \frac{1}{U_{jj}} \left(A_{ij} – \sum_{k=1}^{j-1} L_{ik} \cdot U_{kj} \right)$$.
2. Implementation
static BigInteger det (BigDecimal a [][], int n) { try { for (int i=0; i<n; i++) { boolean nonzero = false; for (int j=0; j<n; j++) if (a[i][j].compareTo (new BigDecimal (BigInteger.ZERO)) > 0) nonzero = true; if (!nonzero) return BigInteger.ZERO; } BigDecimal scaling [] = new BigDecimal [n]; for (int i=0; i<n; i++) { BigDecimal big = new BigDecimal (BigInteger.ZERO); for (int j=0; j<n; j++) if (a[i][j].abs().compareTo (big) > 0) big = a[i][j].abs(); scaling[i] = (new BigDecimal (BigInteger.ONE)) .divide (big, 100, BigDecimal.ROUND_HALF_EVEN); } int sign = 1; for (int j=0; j<n; j++) { for (int i=0; i<j; i++) { BigDecimal sum = a[i][j]; for (int k=0; k<i; k++) sum = sum.subtract (a[i][k].multiply (a[k][j])); a[i][j] = sum; } BigDecimal big = new BigDecimal (BigInteger.ZERO); int imax = -1; for (int i=j; i<n; i++) { BigDecimal sum = a[i][j]; for (int k=0; k<j; k++) sum = sum.subtract (a[i][k].multiply (a[k][j])); a[i][j] = sum; BigDecimal cur = sum.abs(); cur = cur.multiply (scaling[i]); if (cur.compareTo (big) >= 0) { big = cur; imax = i; } } if (j != imax) { for (int k=0; k<n; k++) { BigDecimal t = a[j][k]; a[j][k] = a[imax][k]; a[imax][k] = t; } BigDecimal t = scaling[imax]; scaling[imax] = scaling[j]; scaling[j] = t; sign = -sign; } if (j != n-1) for (int i=j+1; i<n; i++) a[i][j] = a[i][j].divide (a[j][j], 100, BigDecimal.ROUND_HALF_EVEN); } BigDecimal result = new BigDecimal (1); if (sign == -1) result = result.negate(); for (int i=0; i<n; i++) result = result.multiply (a[i][i]); return result.divide (BigDecimal.valueOf(1), 0, BigDecimal.ROUND_HALF_EVEN).toBigInteger(); } catch (Exception e) { return BigInteger.ZERO; } }