Third-order determinants

A 3 × 3 determinant is found by expanding along a row or a column: each entry of that line is multiplied by (−1)ⁱ⁺ʲ and by the 2 × 2 determinant left when its own row and column are struck out, and the three terms are added.

Along R₁, for rows (a, b, c), (d, e, f), (g, h, i): |A| = a(ei − fh) − b(di − fg) + c(dh − eg). Written out, this is six products of three entries, each taking one entry from every row and every column.

There are six ways to expand (three rows, three columns) and all give the same value.

The factor (−1)ⁱ⁺ʲ is +1 when i + j is even and −1 when it is odd, giving the sign pattern + − + / − + − / + − + across the matrix.

Choose the row or column with the most zeros. For the matrix with rows (1, 2, 4), (−1, 3, 0), (4, 1, 0), column 3 has two zeros, so ∆ = 4·(−1·1 − 3·4) = 4·(−13) = −52.

The matrix with rows (0, sin α, −cos α), (−sin α, 0, sin β), (cos α, −sin β, 0) has determinant 0 for all α and β: the two surviving terms of the expansion cancel.

Scaling: if A = kB for square matrices of order n, then |A| = kⁿ|B|. For order 2, A = [[2, 2], [4, 0]] = 2[[1, 1], [2, 0]] gives |A| = −8 = 2²·(−2). For order 3, |3A| = 27|A|.