Repeated Rows Force A Zero Determinant

From Permutation Signs to Determinants

The determinant’s permutation formula expresses it as a signed sum: each term selects one entry from every row and every column, and the sign depends on the permutation that specifies those choices. This makes it possible to prove a basic fact about rows directly from the definition. Exchanging two rows pairs the terms in the sum with terms of opposite sign, so the determinant changes sign. If the two rows were already identical, the exchange leaves the matrix unchanged; the determinant must then equal its own negative and therefore be zero.

Throughout, let \(\mathbf{A}=(a_{ij})\) be an \(n\times n\) matrix over \(\mathbb{R}\), with \(n\geq 1\). A permutation \(\sigma\) of \(\{1,\ldots,n\}\) is a bijection from that set to itself. Its inversions are pairs \((r,s)\) with \(r<s\) and \(\sigma(r)>\sigma(s)\). The sign of \(\sigma\), written \(\operatorname{sgn}(\sigma)\), is \(1\) if the number of inversions is even and \(-1\) if it is odd. The Leibniz formula is

where the sum is over all permutations of \(\{1,\ldots,n\}\). In each product there is one entry from each row, and the permutation ensures that the chosen column indices are all distinct. We will track what happens to these terms when two row positions are exchanged.

Why Exchanging Two Positions Reverses a Permutation’s Sign

Fix distinct positions \(i\) and \(j\), and let \(\tau\) be the permutation that exchanges \(i\) and \(j\) and fixes every other position. Composing a permutation with \(\tau\) on the right exchanges the values at positions \(i\) and \(j\) in its ordered list. We first justify that this reverses its sign.

Proof. Assume, after naming the two positions in increasing order, that \(i<j\). In the list of values of \(\rho\), write \(x=\rho(i)\) and \(y=\rho(j)\). Exchanging the positions changes the status of the pair \((i,j)\): it is an inversion before the exchange exactly when it is not an inversion afterward. Thus this pair changes the inversion count by an odd number. For any position \(k\) outside the interval from \(i\) to \(j\), the two pairs involving \(k\) and the positions \(i,j\) have the same total inversion count before and after: the same two values \(x,y\) are being compared with \(\rho(k)\). For a position \(k\) strictly between \(i\) and \(j\), put \(z=\rho(k)\). The total contribution of \((i,k)\) and \((k,j)\) before the exchange is \([x>z]+[z>y]\), and after it is \([y>z]+[z>x]\), where each bracket is \(1\) when its inequality holds and \(0\) otherwise. The sum of these two totals is \(2\), since the values \(x,y,z\) are distinct. Consequently, the before and after totals have the same parity. All inversion contributions other than the pair \((i,j)\) therefore preserve parity, while that pair reverses it. The total inversion count changes parity, so the permutation sign changes from \(1\) to \(-1\) or from \(-1\) to \(1\). Hence \(\operatorname{sgn}(\rho\circ\tau)=-\operatorname{sgn}(\rho)\). \(\square\)

Exchanging Two Rows Reverses the Determinant’s Sign

Let \(\mathbf{B}\) be the matrix obtained from \(\mathbf{A}\) by exchanging rows \(i\) and \(j\), and let \(\tau\) exchange those row indices. Its entries satisfy \(b_{k\ell}=a_{\tau(k),\ell}\). The next result follows by reindexing the permutation sum; it does not require a determinant expansion along any particular row.

Proof. By the Leibniz formula and the definition of the entries of \(\mathbf{B}\),

In each product, set \(r=\tau(k)\). Because \(\tau\) exchanges two indices and fixes the others, \(\tau^{-1}=\tau\), so \(k=\tau(r)\). Define \(\rho=\sigma\circ\tau\); then \(\sigma=\rho\circ\tau\). As \(\sigma\) ranges over all permutations, so does \(\rho\), because composition with \(\tau\) is a bijection on the set of permutations. The product becomes \(\prod_{r=1}^{n}a_{r,\rho(r)}\), and the lemma gives \(\operatorname{sgn}(\sigma)=\operatorname{sgn}(\rho\circ\tau)=-\operatorname{sgn}(\rho)\). Therefore

This proves the claimed sign change. \(\square\)

Identical Rows Force a Zero Determinant

Suppose two distinct rows of a square matrix are identical. Exchanging those rows produces exactly the original matrix, entry by entry. At the same time, the row-swap theorem says the determinant of the exchanged matrix is the negative of the original determinant. These two facts give an equation in which a real number equals its negative. The fact that the scalars are real matters: if \(d=-d\), then \(2d=0\), and \(2\ne0\) in \(\mathbb{R}\), so \(d=0\).

Proof. Let rows \(i\) and \(j\) be identical, with \(i\ne j\), and let \(\mathbf{B}\) result from exchanging them. Since the two rows have the same entries, \(\mathbf{B}=\mathbf{A}\), so \(\det\mathbf{B}=\det\mathbf{A}\). The row-swap theorem also gives \(\det\mathbf{B}=-\det\mathbf{A}\). Thus \(\det\mathbf{A}=-\det\mathbf{A}\), which implies \(2\det\mathbf{A}=0\). Since \(2\ne0\) in \(\mathbb{R}\), division by \(2\) gives \(\det\mathbf{A}=0\). When \(n=1\), there are no two distinct rows, so the hypothesis cannot occur. \(\square\)

Common Mistake: Treating a Row Exchange as Sign-Preserving

A common error is to assume that the determinant stays the same because a row exchange preserves the collection of rows. For example, \(\mathbf{D}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\) has determinant \(1(4)-2(3)=-2\). Exchanging its rows gives \(\mathbf{E}=\begin{bmatrix}3&4\\1&2\end{bmatrix}\), whose determinant is \(3(2)-4(1)=2\). The incorrect calculation would assign \(-2\) to \(\det\mathbf{E}\), treating the exchange as sign-preserving. The correction is to track the sign: \(\det\mathbf{E}=-\det\mathbf{D}=2\). The row-swap theorem applies whether or not the rows are equal; only in the equal-row case does the exchanged matrix coincide with the original.

The permutation-sum proof explains why the sign change is systematic rather than a special feature of a small example. Reindexing pairs each determinant term with a term that uses the exchanged row positions, and the associated permutation sign is reversed. Identical rows then give the zero result because the matrix is unchanged by that exchange. This argument uses neither a determinant test for invertibility nor any row-reduction rule for computing determinants.