Every way to compute a determinant

Cofactor expansion, row reduction, the triangular shortcut, and what the number actually means geometrically.

For a 2x2 or 3x3, use the direct formula — it's fastest. For anything triangular (or triangularized), read the determinant straight off the diagonal. For a general n x n, row-reduce to triangular form and track the sign flips and scaling from each operation; that beats cofactor expansion for anything bigger than 4x4. Cofactor expansion is the definition everything else is proved from, and it's still the right tool when a row or column is mostly zeros.

Small cases — direct formulas

Section

The 2x2 formula, ad - bc

Section

The 3x3 case, worked directly

Symbol

$\det$

determinant

The general definition

Term

cofactor expansion

See section →

Computing a determinant by summing entries times signed minors along any row or column.

Expand along any row or column — the definition that works for every size.

Theorem

Cofactor Expansion Theorem

Cofactor expansion →

Let $A$ be an $n \times n$ matrix with entries $a_{ij}$ , and for each $i, j$ let $M_{ij}$ denote the minor — the determinant of the $(n-1)\times(n-1)$ submatrix obtained by deleting row $i$ and column $j$ of $A$ — and let $C_{ij} = (-1)^{i+j} M_{ij}$ denote the corresponding cofactor. Then for any fixed row $i$ , $\det(A) = \sum_{j=1}^n a_{ij} C_{ij} = \sum_{j=1}^n (-1)^{i+j} a_{ij} M_{ij},$ and for any fixed column $j$ , $\det(A) = \sum_{i=1}^n a_{ij} C_{ij} = \sum_{i=1}^n (-1)^{i+j} a_{ij} M_{ij}.$

The formal cofactor-expansion statement.

Show proof

Start from the Leibniz definition of the determinant as a sum over every permutation $\sigma$ of $\{1, \dots, n\}$ : $\det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^n a_{k,\sigma(k)}.$ Fix a row $i$ and sort this sum by which column each permutation sends row $i$ to. Every $\sigma \in S_n$ satisfies $\sigma(i) = j$ for exactly one $j \in \{1, \dots, n\}$ , so the sum splits cleanly into $n$ disjoint groups: $\det(A) = \sum_{j=1}^n \sum_{\substack{\sigma \in S_n \\ \sigma(i)=j}} \operatorname{sgn}(\sigma) \prod_{k=1}^n a_{k,\sigma(k)}.$ In the inner sum, every term already carries the factor $a_{i,\sigma(i)} = a_{ij}$ , which does not depend on $\sigma$ beyond fixing $\sigma(i) = j$ , so it factors out: $\det(A) = \sum_{j=1}^n a_{ij} \left( \sum_{\substack{\sigma \in S_n \\ \sigma(i)=j}} \operatorname{sgn}(\sigma) \prod_{k \neq i} a_{k,\sigma(k)} \right).$ The claim is that the bracketed sum is exactly the cofactor $C_{ij}$ . Each permutation $\sigma$ with $\sigma(i) = j$ restricts to a bijection between the remaining $n-1$ row indices $\{1,\dots,n\}\setminus\{i\}$ and the remaining $n-1$ column indices $\{1,\dots,n\}\setminus\{j\}$ — call this restricted map $\tau$ . Relabeling those two index sets in increasing order identifies $\tau$ with a permutation $\bar\tau$ of $\{1, \dots, n-1\}$ , and the product $\prod_{k \neq i} a_{k,\sigma(k)}$ becomes exactly the corresponding product of entries of the submatrix obtained by deleting row $i$ and column $j$ — the submatrix whose determinant is $M_{ij}$ . Summing $\operatorname{sgn}(\bar\tau) \prod \bar\tau\text{-entries}$ over all such $\bar\tau$ gives $M_{ij}$ itself, provided the signs match up: that $\operatorname{sgn}(\sigma) = (-1)^{i+j} \operatorname{sgn}(\bar\tau)$ .

To see the sign correction, compare $\sigma$ to the permutation $\bar\tau$ extended back to $\{1,\dots,n\}$ by sending $i \mapsto j$ in the "obvious" position. Moving row $i$ up to row $1$ by swapping it past the $i-1$ rows above it takes $i - 1$ transpositions; moving column $j$ left to column $1$ by the same process takes $j - 1$ transpositions. Each such adjacent swap flips the sign of a permutation by $-1$ , so relating $\sigma$ 's sign to $\bar\tau$ 's sign (computed on the un-shifted remaining indices) costs a factor of $(-1)^{(i-1)+(j-1)} = (-1)^{i+j}$ . Hence $\sum_{\substack{\sigma \in S_n \\ \sigma(i)=j}} \operatorname{sgn}(\sigma) \prod_{k \neq i} a_{k,\sigma(k)} = (-1)^{i+j} M_{ij} = C_{ij}.$ Substituting back gives the row- $i$ expansion $\det(A) = \sum_{j=1}^n a_{ij} C_{ij}.$ The column- $j$ expansion follows by the identical argument applied to $A^T$ , using $\det(A) = \det(A^T)$ and the fact that transposing swaps the roles of rows and columns (and swaps $M_{ij} \leftrightarrow M_{ji}$ , while $(-1)^{i+j} = (-1)^{j+i}$ is unaffected).

Sanity check: for $n = 2$ , expanding along row $1$ gives $\det(A) = a_{11}C_{11} + a_{12}C_{12} = a_{11}(a_{22}) - a_{12}(a_{21}) = a_{11}a_{22} - a_{12}a_{21}$ — the familiar $2\times 2$ determinant formula, since each $1\times 1$ minor is just the surviving entry itself.

Term