Matrix Determinant Calculator

Geometrically, the determinant is the signed volume-scaling factor of the linear map the matrix represents. A 2x2 determinant scales areas, a 3x3 scales volumes, and an n x n scales n-dimensional volumes. If det = 2, the map doubles every region's measure; if det = -1, it preserves measure but flips orientation (a reflection); if det = 0, the map collapses space into a lower dimension. The same number is also the invertibility test: a square matrix has an inverse if and only if its determinant is non-zero. So one scalar answers two different questions at once: how the matrix stretches volume, and whether the linear system it defines has a unique solution.

A determinant of 0 means the matrix collapses n-dimensional space into something lower-dimensional: it is rank-deficient, with rank strictly less than n. Concretely, its columns (or rows) are linearly dependent, so they span a flat plane, line, or point instead of all of R^n. No inverse can exist because an inverse would have to undo the collapse and recover the lost dimension, which is impossible once distinct input vectors get mapped to the same output. In linear-system terms, Ax = b either has no solution (b is outside the collapsed image) or infinitely many solutions (b is inside, but every preimage along the null space works). Either way, you cannot pin down a unique x, which is what an inverse would let you do.

For a 2x2 matrix [[a, b], [c, d]], the determinant is ad - bc. This is the Leibniz formula written out directly: multiply along the main diagonal, subtract the product along the anti-diagonal. As an example, for [[1, 2], [3, 4]] the determinant is 1*4 - 2*3 = 4 - 6 = -2. The same formula shows up in the closed-form 2x2 inverse: the inverse is (1 / det) * [[d, -b], [-c, a]], which only makes sense when det is non-zero, matching the general invertibility rule.

Sarrus' rule is a rote diagonal trick for 3x3 matrices only. Write the matrix and then copy the first two columns to the right of it. Sum the three full down-right diagonals, subtract the three full down-left diagonals. For [[a, b, c], [d, e, f], [g, h, i]] that gives det = aei + bfg + cdh - ceg - bdi - afh, which rearranges to the cofactor form a(ei - fh) - b(di - fg) + c(dh - eg). Sarrus is fast to memorize and quick to apply by hand, but it does not generalize: there is no 4x4 or higher-order Sarrus rule, so for larger matrices you have to fall back on cofactor expansion or row reduction.

Cofactor expansion is the general n x n formula: pick any row (or column), and write det(A) as the alternating sum over that row of entry * cofactor, where each cofactor is plus or minus the determinant of the (n-1) x (n-1) minor obtained by deleting that entry's row and column. Recursing all the way down, you end up evaluating roughly n! terms, so the cost grows as factorial in n. By contrast, LU decomposition (Gaussian elimination with partial pivoting) computes the determinant as the product of pivots in O(n^3) time. That is why this tool only goes up to 5x5 with cofactor expansion (5! = 120 terms is fine, 10! = 3,628,800 is not), and why every serious linear-algebra library uses LU at higher dimensions.