# Calculating the determinant using Kraut method in $O(N^3)$

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:

1. Let $L_{i i} = 1$ for $i = 1, 2, ..., N$.
2. 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 $$L_{ij} = \frac{1}{U_{jj}} \left(A_{ij} - \sum_{k=1}^{j-1} L_{ik} \cdot U_{kj} \right)$$.

## 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;
}
}