How this instrument works
Cofactor expansion — also called Laplace expansion — computes a determinant by choosing one row (or column), multiplying each entry by its own cofactor, and adding the results. A cofactor is a signed minor: delete that entry's row and column, take the determinant of what's left, then attach a sign of (−1)^(i+j) for its position. Expanding a 2×2 matrix along row 1, deleting row 1 and column 1 leaves the single number d, so C₁₁ = +d; deleting row 1 and column 2 leaves c, but position (1,2) carries a minus, so C₁₂ = −c. Term 1 is a×d, Term 2 is b×(−c), and their sum is the determinant.
The plus and minus signs form a checkerboard: position (i,j) gets (−1)^(i+j), so (1,1) is positive, (1,2) is negative, (1,3) is positive, and the pattern keeps alternating in every direction across a larger grid. That alternation isn't decoration — it falls straight out of the determinant's definition as a signed sum over permutations, where swapping which column pairs with which row flips a term's sign. Writing the calculation as term1 + term2 instead of collapsing straight to ad − bc matters because that same multiply-then-sign step is exactly what a 3×3 or 10×10 determinant needs, just with each cofactor built from a smaller determinant instead of a lone number.
An entry of zero along the chosen row deletes an entire term for free — expand along whichever row or column has the most zeros and the arithmetic shrinks accordingly, which is the standard hand trick for anything larger than 2×2. The most common slip is treating C₁₂ as plain c instead of −c: skip that sign and the expansion silently returns the wrong number even though both multiplications were correct. Comparing against the plain ad − bc shortcut, when one is available, is a fast way to confirm the sign was applied correctly.
- Enter the top row of the matrix into a (row 1, col 1) and b (row 1, col 2).
- Enter the bottom row into c (row 2, col 1) and d (row 2, col 2).
- Read Term 1: a × C₁₁ — that is a multiplied by the cofactor +d.
- Read Term 2: b × C₁₂ — that is b multiplied by the cofactor −c.
- Read Determinant (term1 + term2) for the two signed terms added together.
Worked example — expanding [[1, 2], [3, 4]] along row 1
Take a = 1, b = 2, c = 3, d = 4 — the matrix [[1, 2], [3, 4]]. Deleting row 1 and column 1 leaves just d = 4, and position (1,1) is positive, so C₁₁ = +4; Term 1: a × C₁₁ comes out to 1 × 4 = 4. Deleting row 1 and column 2 leaves c = 3, but position (1,2) is negative, so C₁₂ = −3; Term 2: b × C₁₂ comes out to 2 × (−3) = −6.
Determinant (term1 + term2) adds those two signed products: 4 + (−6) = −2. That matches the plain ad − bc shortcut exactly (1×4 − 2×3 = −2), which is the point — cofactor expansion and the shortcut are two routes to the identical number, and the row-by-row version is the one that keeps working once a matrix grows past 2×2, where no such single-line shortcut exists.
Questions
What is cofactor expansion, and why does it use a plus and minus sign?
Cofactor expansion computes a determinant by multiplying each entry in a chosen row (or column) by its cofactor — a signed minor — and adding the products. The sign follows a checkerboard pattern, (−1)^(i+j): position (1,1) is positive, (1,2) is negative, and so on. For a 2×2 matrix expanded along row 1, that gives C₁₁ = +d and C₁₂ = −c, so term1 = a×d and term2 = b×(−c).
How is cofactor expansion different from the plain ad − bc shortcut?
They return the same number for a 2×2 matrix — this sheet's Determinant (term1 + term2) always equals ad − bc — but expansion keeps the two products visible as separate terms instead of collapsing them. That row-by-row structure is what generalizes: a 3×3 or larger matrix has no single shortcut formula, only cofactor expansion applied recursively, with each cofactor built from a smaller determinant.
Does the row or column you expand along change the answer?
No. Any row or column produces the identical determinant, because cofactor expansion is the same permutation sum grouped differently by which entries you pick together. Choosing a row or column with zero entries is still worth it by hand, since each zero contributes 0 × (its cofactor) and can be skipped — with a = 5, b = 0, c = 0, d = 5, Term 2 vanishes and only Term 1 = 25 remains.
What's the most common mistake when doing cofactor expansion by hand?
Forgetting to apply the sign on an odd-position cofactor. C₁₂ sits at position (1,2), where 1+2 = 3 is odd, so the sign is negative and C₁₂ = −c, not c. Using +c instead of −c is the single most common slip — Term 2 gets the wrong sign, and the final determinant ends up off by twice that term's size.
Does cofactor expansion work for matrices larger than 2×2?
Yes — this 2×2 case is the smallest instance of a fully general method, also called Laplace expansion. For a 3×3 matrix, each cofactor is itself a 2×2 determinant with a sign attached; for an n×n matrix, expansion recurses down to (n−1)×(n−1) determinants until only single entries remain. The 2×2 version here is the base case that recursion eventually bottoms out at.
Why is C₁₁ positive but C₁₂ negative?
Because the cofactor sign follows (−1)^(i+j) for row i, column j: (1,1) gives (−1)^2 = +1, while (1,2) gives (−1)^3 = −1. That alternating pattern comes directly from the determinant's definition as a signed sum over permutations — pairing a different column with a given row flips the sign of the term, and the checkerboard is just that flip written out position by position.