Laplace Expansion Along Any Row

Why Expand Along a Row Other Than the First?

The recursive definition of the determinant uses a cofactor expansion along the first row. In a calculation, however, another row may be more convenient: it might contain zeros, or its minors might be easier to evaluate. To use that choice with confidence, we need a reason that the result is still the same determinant. The Laplace expansion provides that reason. Its proof connects the recursive definition to the Leibniz formula by grouping permutations according to the entry selected in a chosen row.

All matrices in this tutorial are square matrices over \(\mathbb{R}\). For a matrix \(\mathbf{A}=(a_{ij})\in\mathbb{R}^{n\times n}\), the minor \(M_{ij}\) is the determinant of the \((n-1)\times(n-1)\) matrix formed by deleting row \(i\) and column \(j\). The cofactor in position \((i,j)\) is \(C_{ij}=(-1)^{i+j}M_{ij}\). The signs alternate in a checkerboard pattern, beginning with a positive sign in position \((1,1)\).

The formula is a statement about the determinant already defined by first-row cofactor expansion; it does not define a different determinant. We will prove that the two agree for every row. The case \(n=1\) can be read using the convention that the determinant of the empty matrix, which occurs as the sole minor, is \(1\). Then its only cofactor expansion gives \(a_{11}\).

The Sign When a Permutation Is Restricted

The Leibniz formula expresses a determinant as a sum over permutations. For \(\mathbf{A}\in\mathbb{R}^{n\times n}\), it is

Here \(S_n\) is the set of permutations of \(\{1,\ldots,n\}\), and \(\operatorname{sgn}(\sigma)\) is \(1\) or \(-1\) according as the permutation is even or odd. We need to compare the sign of a permutation with the sign of the smaller permutation left after one row and one column are removed.

Proof. Write \(\sigma\) as its list of values \(\sigma(1),\ldots,\sigma(n)\). Deleting the value at position \(r\) and then relabelling the remaining values in increasing order leaves the inversions among the remaining entries unchanged; these are exactly the inversions of \(\tau\). It remains to count the inversions involving the deleted value \(j=\sigma(r)\). Let \(x\) be the number of values less than \(j\) that occur before position \(r\), and let \(y\) be the number of values greater than \(j\) that occur after position \(r\). The inversions involving \(j\) number \(A+B\), where \(A\) is the number of values greater than \(j\) before \(r\), and \(B\) is the number of values less than \(j\) after \(r\). Since there are \(r-1\) entries before \(r\) and \(j-1\) values less than \(j\), \(A=r-1-x\) and \(B=j-1-x\). Therefore \(A+B=r+j-2-2x\), which has the same parity as \(r+j\). The inversion counts of \(\sigma\) and \(\tau\) thus differ in parity by \(r+j\), giving the stated sign relation. \(\square\)

This sign relation accounts for the checkerboard factor in a cofactor. After deleting row \(r\) and column \(j\), the terms that remain form a determinant of size \(n-1\); the sign needed to restore the original permutation is exactly \((-1)^{r+j}\).

Induction and the Expansion Along Any Row

Proof. We prove the result by induction on \(n\), first connecting the recursive determinant to the Leibniz formula. For \(n=1\), the determinant is \(a_{11}\), which is the sole term in the Leibniz formula. Now suppose that every determinant of size \(n-1\) equals its Leibniz sum, and let \(\mathbf{A}\) have size \(n\). By its defining first-row expansion,

where \(\mathbf{A}_{1j}\) is obtained by deleting row \(1\) and column \(j\). By the induction hypothesis, each \(\det\mathbf{A}_{1j}\) is the Leibniz sum over permutations of the remaining ordered row and column indices. For each fixed \(j\), those smaller permutations correspond exactly to the permutations \(\sigma\in S_n\) with \(\sigma(1)=j\). By the lemma with \(r=1\), their signs satisfy \(\operatorname{sgn}(\sigma)=(-1)^{1+j}\operatorname{sgn}(\tau)\). Consequently, the terms in \((-1)^{1+j}a_{1j}\det\mathbf{A}_{1j}\) are precisely the terms of the \(n\times n\) Leibniz formula whose permutations send row \(1\) to column \(j\). Summing over all \(j\) includes every permutation exactly once. Thus the recursively defined determinant equals the Leibniz formula for size \(n\).

It remains to expand along an arbitrary row \(r\). In the size-\(n\) Leibniz formula, group the terms by the value \(j=\sigma(r)\). For a fixed \(j\), the remaining assignments are permutations of the rows and columns left after deleting \(r\) and \(j\). The lemma says that each original permutation sign is \((-1)^{r+j}\) times the sign of that smaller permutation. The sum of the smaller signed products is the Leibniz formula for \(M_{rj}\), which equals that minor’s determinant by the induction result just proved. The group of terms with \(\sigma(r)=j\) is therefore \((-1)^{r+j}a_{rj}M_{rj}\). Summing these groups over \(j=1,\ldots,n\) proves the claimed expansion. For \(n=1\), the statement holds directly from the convention \(M_{11}=1\). \(\square\)

The proof has two distinct jobs. Induction establishes that the recursive determinant agrees with the permutation formula at every size. Grouping the permutation terms by the chosen row then gives the expansion along that row. The minor must carry the determinant of the matrix with both the selected row and selected column removed; the cofactor sign records the change in permutation sign.

Worked Expansions and Checks

Common Sign Pitfall

A frequent error is to use the column’s checkerboard sign without including the row index. For row 2 of \(\mathbf{B}\), this mistake treats the sign at \((2,3)\) as positive because column 3 is odd, and would calculate \(1(2)+2(-6)=-10\). That uses \(M_{23}\) directly instead of its cofactor. The correct sign is \((-1)^{2+3}=-1\), so \(C_{23}=-M_{23}=6\), and the expansion is \(1(2)+2(6)=14\). The sign depends on both indices, not on the column alone.

Another useful check is to compare two different row expansions when practical. They should agree because each is a grouping of the same Leibniz sum. A disagreement usually signals an arithmetic error in a minor or an incorrect checkerboard sign; it does not mean the determinant depends on which row was chosen.

Choosing a Row and Interpreting the Formula

Every row gives a valid expansion, but not every row is equally efficient. If a row has zeros, each zero removes a minor calculation because its term is zero. If two rows have the same number of zeros, the remaining minors may still differ in difficulty, so inspect their entries before expanding. The theorem justifies this choice: it is a computational convenience, not a change to the determinant.

Keep the roles of the three factors in each term distinct: \(a_{rj}\) is the matrix entry in the chosen row, \(M_{rj}\) is the determinant after deleting that row and column, and \((-1)^{r+j}\) supplies the cofactor sign. Writing these separately before simplifying makes sign errors easier to catch. For a row with a zero entry, the corresponding term vanishes regardless of its minor or sign.