Matrix Determinant n×n

Type in nine entries and get det A instantly, together with a clear verdict on whether the matrix is invertible and the linear system has a unique solution.

How the calculation works

The calculator takes a 3×3 matrix with entries a11…a33, where the first digit is the row and the second the column. Each entry can range from −1000 to 1000 and may be a decimal. A 2×2 determinant works too: put the matrix in the top-left corner, set a33 = 1 and a13, a23, a31, a32 = 0, and the result equals a11·a22 − a12·a21.

The determinant is found by cofactor (Laplace) expansion along the first row: det A = a11·(a22·a33 − a23·a32) − a12·(a21·a33 − a23·a31) + a13·(a21·a32 − a22·a31). This gives the same value as the rule of Sarrus. For larger matrices, Gaussian elimination to triangular form is the practical method; this tool does not handle sizes above 3×3.

Alongside det A you get the product of the main diagonal, which equals the determinant for triangular or diagonal matrices and makes a handy cross-check, and a status line: if det = 0 the matrix is singular, has no inverse, and Ax = b has no unique solution.

Worked example

Default matrix rows: [2, −1, 0], [1, 3, 2], [0, 1, 4]. det A = 2·(3·4 − 2·1) − (−1)·(1·4 − 2·0) + 0·(1·1 − 3·0) = 2·10 + 4 = 24. The diagonal product 2·3·4 is also 24, but that is a coincidence because the matrix is not triangular. Status: non-singular.

Things to keep in mind

• Watch the signs: the middle term of the expansion is subtracted, and a sign slip on a12 is the most common hand-calculation error.
• Swapping two rows flips the sign of det, multiplying a row by k multiplies det by k, and adding a multiple of one row to another leaves det unchanged.
• With decimal entries a tiny result such as 1e−13 is almost certainly zero plus rounding noise; the calculator treats |det| < 1e−12 as singular.
• Geometrically, |det| of a 3×3 matrix is the volume of the parallelepiped spanned by its row vectors.