About

The cofactor expansion method (Laplace expansion) computes the determinant of an n × n matrix by reducing it to a weighted sum of (n − 1) × (n − 1) minor determinants. Each element a_ij is multiplied by its cofactor C_ij = (−1)^i+j ⋅ M_ij, where M_ij is the determinant of the submatrix formed by deleting row i and column j. Manual expansion on matrices larger than 3 × 3 is error-prone. A single sign mistake in the checkerboard pattern propagates through every subsequent minor. This tool traces every recursive step so you can verify intermediate results against your own work.

The calculator supports expansion along any row or column. Choosing a row or column with zeros reduces computation since terms with a_ij = 0 contribute nothing. The tool approximates results assuming exact rational input. Floating-point entries may introduce rounding in deeply nested expansions beyond 5 × 5.

Formulas

The Laplace expansion of the determinant along row i of an n × n matrix A:

det(A) = n∑j=1 a_ij ⋅ C_ij

where the cofactor C_ij is defined as:

C_ij = (−1)^{i + j} ⋅ M_ij

and M_ij is the minor determinant obtained by deleting row i and column j from A. The base case for a 2 × 2 matrix:

det abcd = ad − bc

For a 1 × 1 matrix, det([a]) = a.

Where: a_ij = element at row i, column j. C_ij = cofactor of element (i, j). M_ij = minor determinant (submatrix determinant). n = matrix dimension.

Reference Data

Matrix Size	Terms in Expansion	Total Multiplications	Minor Size	Recommended Strategy
2 × 2	2	2	1 × 1	Direct formula ad − bc
3 × 3	3	12	2 × 2	Expand along row/column with zeros
4 × 4	4	64	3 × 3	Row reduce first if possible
5 × 5	5	320	4 × 4	Combine with row operations
6 × 6	6	1920	5 × 5	Use LU decomposition for speed
Cofactor Sign Pattern (Checkerboard)
+	−	+	−	+
−	+	−	+	−
+	−	+	−	+
−	+	−	+	−
+	−	+	−	+
Key Properties of Determinants
Swapping two rows		Multiplies determinant by −1
Scaling a row by k		Multiplies determinant by k
Adding row multiple to another		Determinant unchanged
Two identical rows		det(A) = 0
Triangular matrix		Product of diagonal entries
det(A^T)		= det(A)
det(AB)		= det(A) ⋅ det(B)
det(A⁻¹)		= 1 ÷ det(A)
det(kA) for n×n		= kⁿ ⋅ det(A)

Frequently Asked Questions

Choose the row or column containing the most zeros. Each zero element contributes nothing to the sum, eliminating that entire branch of recursive minor computation. For a 4×4 matrix with two zeros in row 3, expanding along row 3 halves the work from four 3×3 determinants to two.

No. The Laplace expansion theorem guarantees the same determinant regardless of which row or column you expand along. If you get different values, there is an arithmetic or sign error in your cofactor computation. The checkerboard sign pattern (−1)^i+j is the most common source of mistakes.

Cofactor expansion has O(n!) time complexity. LU decomposition runs in O(n³). For a 10 × 10 matrix, cofactor expansion requires roughly 3,628,800 operations while LU needs about 1,000. Cofactor expansion is pedagogically valuable for matrices up to 5 × 5 but impractical for larger systems.

A singular matrix has det(A) = 0. The cofactor expansion will still compute correctly and return zero. This indicates the matrix has linearly dependent rows or columns, the system Ax = b has no unique solution, and the matrix is not invertible.

Yes. IEEE 754 double-precision arithmetic introduces rounding errors that compound through recursive minor computation. For a 5 × 5 matrix with entries like 0.1, accumulated error can reach 10⁻¹² or worse. Use integer or simple fractional entries when possible. This tool displays results rounded to 10 significant digits.

The inverse of a matrix A can be computed as A⁻¹ = (1 ÷ det(A)) ⋅ C^T, where C^T is the transpose of the cofactor matrix (the adjugate). This only exists when det(A) ≠ 0.