Laplace Expansion Along Any Column

Why a Column Can Be Chosen Freely

A determinant can often be computed efficiently by expanding along a row with many zeros. The corresponding freedom to choose a column is not a separate computational trick: it follows from the same permutation structure that defines the determinant. The key is to group the terms in the permutation formula according to which entry in the chosen column they use. Each group will turn out to be one entry multiplied by its cofactor.

Work over \(\mathbb{R}\), and let \(\mathbf{A}=(a_{ij})\) be an \(n\times n\) matrix. Recall the Leibniz formula: sum over all permutations \(\sigma\) of \(\{1,\dots,n\}\), with each permutation contributing its sign times one product of entries, one from each row and column.

Here \(\operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}\), where \(\operatorname{inv}(\sigma)\) is the number of inversions: pairs of positions \(r<s\) for which \(\sigma(r)>\sigma(s)\). For a fixed entry \(a_{ij}\), delete row \(i\) and column \(j\) from \(\mathbf{A}\). The determinant of the resulting \((n-1)\times(n-1)\) matrix is its minor.

The factor \((-1)^{i+j}\) records how the sign of a permutation changes when the row and column containing the chosen entry are removed. We establish that sign relationship first, then use induction to prove the expansion formula. For the \(1\times1\) case, use the convention that the determinant of the empty matrix is \(1\).

The Sign After Removing a Row and Column

Proof. Assume \(\sigma(i)=j\), and let \(L\) be the number of values less than \(j\) that occur before position \(i\). Among the \(i-1\) values before position \(i\), exactly \(i-1-L\) exceed \(j\). Among the \(j-1\) values less than \(j\), exactly \(j-1-L\) occur after position \(i\). The inversions involving position \(i\) therefore number \((i-1-L)+(j-1-L)=i+j-2-2L\), which has the same parity as \(i+j\). Removing position \(i\) and value \(j\) leaves all other inversion comparisons unchanged: relabelling values in increasing order preserves their order. Thus \(\operatorname{inv}(\sigma)\) and \(\operatorname{inv}(\tau)\) differ in parity by \(i+j\). Since each sign is \((-1)\) raised to the corresponding inversion count, the stated sign identity follows. \(\square\)

This count includes both kinds of inversions involving the selected entry: those formed with earlier positions whose values exceed \(j\), and those formed with later positions whose values are less than \(j\). Counting only one kind would not give the cofactor sign.

Column Expansion: Induction on Matrix Size

Proof. We use induction on \(n\), proving the formula for every column \(j\) at each size. If \(n=1\), then \(\mathbf{A}=[a_{11}]\), its determinant is \(a_{11}\), and the expansion gives \(a_{11}(-1)^{1+1}\det(\text{empty matrix})=a_{11}\). Now assume \(n\ge2\) and that the formula holds for matrices of size \(n-1\). Fix any column \(j\). In the Leibniz formula for \(\det\mathbf{A}\), group permutations according to the row \(i\) for which \(\sigma(i)=j\). Every permutation uses column \(j\) exactly once, so these groups partition all terms. In the group for row \(i\), each product contains the factor \(a_{ij}\). After removing that factor, the remaining entries correspond to a permutation of the rows other than \(i\) with the columns other than \(j\). In their original increasing orders, those entries form exactly the minor matrix whose determinant is \(M_{ij}\). By the lemma, each original permutation sign is \((-1)^{i+j}\) times the sign of its induced permutation. Summing the remaining signed products over this group therefore gives \((-1)^{i+j}M_{ij}\), by the Leibniz formula for that minor. Hence the whole group contributes \(a_{ij}(-1)^{i+j}M_{ij}\). Adding the groups for \(i=1,\dots,n\) proves \(\det\mathbf{A}=\sum_{i=1}^{n}a_{ij}(-1)^{i+j}M_{ij}\). The choice of \(j\) was arbitrary, so the result holds along every column. \(\square\)

The grouping is the essential step: it does not assume that a column is sparse or that its entries have any special values. It establishes the formula for each chosen column directly from the determinant definition. In particular, when a column contains zeros, the corresponding groups contribute zero and need not be evaluated.

Worked Column Expansions

Choosing and Applying the Expansion

The theorem permits any column; efficiency depends on the entries in that column. A zero contributes nothing, while each nonzero entry requires evaluating a determinant one size smaller. If two columns have the same number of nonzero entries, the minors may still differ in difficulty, so inspect their structure before choosing. The cofactor signs alternate down a fixed column, beginning with \((-1)^{1+j}\) in row \(1\). They begin positive when \(j\) is odd and negative when \(j\) is even.

Common Sign Pitfall

A frequent error is to multiply an entry by its minor without applying the cofactor sign. In the third-column example, the row \(2\) entry is \(5\), and its minor is \(-2\). The incorrect calculation \(3(-4)+5(-2)+6(4)=2\) omits the sign \((-1)^{2+3}=-1\). The correct cofactor is \((-1)(-2)=2\), so the row \(2\) term is \(5(2)=10\), and the determinant is \(-12+10+24=22\). The checkerboard sign belongs to the original position \((i,j)\), not to the signs of entries inside the minor.

Column expansion matters both as a way to compute determinants and as a consequence of their permutation definition. The induction proof explains why the formula works for every column, including one with no zeros: each possible selected entry gathers precisely the permutation terms that use it. For practical calculations, the theorem then lets the structure of the matrix, rather than a fixed rule about which direction to expand, guide the work.