[proofplan] We prove each part from the Leibniz expansion $\det A = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} A_{k,\sigma(k)}$. For part (1), swapping rows $i$ and $j$ corresponds to pre-composing each permutation with the transposition $(i\;j)$, which negates the sign. For part (2), multilinearity in the rows extracts the scalar $\lambda$ from the single affected row. For part (3), multilinearity splits the determinant into the original plus an auxiliary determinant with two equal rows, which vanishes by the alternating property. [/proofplan] [step:Express the determinant via the Leibniz formula] Write the rows of $A$ as $a_1, \dots, a_n \in F^n$, so $A_{k,l} = (a_k)_l$ is the $l$-th component of the $k$-th row. The Leibniz formula gives \begin{align*} \det A = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} A_{k,\sigma(k)}. \end{align*} We use this expansion to prove each of the three claims. [/step] [step:Prove that swapping rows $i$ and $j$ negates the determinant] Let $A'$ be obtained from $A$ by swapping rows $i$ and $j$ with $i \neq j$. Then $A'_{k,l} = A_{\tau(k),l}$ for all $k, l$, where $\tau = (i\;j) \in S_n$ is the transposition exchanging $i$ and $j$. By the Leibniz formula, \begin{align*} \det A' &= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} A'_{k,\sigma(k)} = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} A_{\tau(k),\sigma(k)}. \end{align*} Introduce the substitution $\rho = \sigma \circ \tau^{-1}$, equivalently $\sigma = \rho \circ \tau$. Since $\tau = \tau^{-1}$ (a transposition is its own inverse), we have $\sigma(k) = \rho(\tau(k))$. As $\sigma$ ranges over all of $S_n$, so does $\rho$, because left multiplication by $\tau$ is a bijection on $S_n$. The sign transforms as $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\rho \circ \tau) = \operatorname{sgn}(\rho) \cdot \operatorname{sgn}(\tau) = -\operatorname{sgn}(\rho)$, since $\operatorname{sgn}(\tau) = -1$ by the [Sign Homomorphism](/theorems/778). Substituting, \begin{align*} \det A' &= \sum_{\rho \in S_n} (-\operatorname{sgn}(\rho)) \prod_{k=1}^{n} A_{\tau(k),\rho(\tau(k))}. \end{align*} As $k$ ranges over $\{1, \dots, n\}$, so does $\tau(k)$ (since $\tau$ is a bijection). Relabelling the product index $m = \tau(k)$, the product becomes $\prod_{m=1}^{n} A_{m,\rho(m)}$, and so \begin{align*} \det A' &= -\sum_{\rho \in S_n} \operatorname{sgn}(\rho) \prod_{m=1}^{n} A_{m,\rho(m)} = -\det A. \end{align*} [guided] We want to show that swapping two rows flips the sign of the determinant. The key idea is to track how the row swap interacts with the Leibniz expansion. Let $\tau = (i\;j)$ denote the transposition of $i$ and $j$. The matrix $A'$ is obtained by permuting the rows of $A$: row $k$ of $A'$ equals row $\tau(k)$ of $A$. In components, $A'_{k,l} = A_{\tau(k),l}$. Substituting into the Leibniz formula: \begin{align*} \det A' &= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} A'_{k,\sigma(k)} = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} A_{\tau(k),\sigma(k)}. \end{align*} Now we perform a change of summation variable. Define $\rho = \sigma \circ \tau^{-1}$, so that $\sigma = \rho \circ \tau$. Since $\tau$ is a transposition, $\tau^{-1} = \tau$, so $\sigma(k) = \rho(\tau(k))$. The map $\sigma \mapsto \rho = \sigma \circ \tau$ is a bijection on $S_n$, so as $\sigma$ ranges over $S_n$, so does $\rho$. How does the sign transform? The [Sign Homomorphism](/theorems/778) gives $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\rho \circ \tau) = \operatorname{sgn}(\rho) \cdot \operatorname{sgn}(\tau)$. Since $\tau$ is a transposition, $\operatorname{sgn}(\tau) = -1$, so $\operatorname{sgn}(\sigma) = -\operatorname{sgn}(\rho)$. Substituting: \begin{align*} \det A' &= \sum_{\rho \in S_n} \bigl(-\operatorname{sgn}(\rho)\bigr) \prod_{k=1}^{n} A_{\tau(k),\rho(\tau(k))}. \end{align*} The product $\prod_{k=1}^{n} A_{\tau(k),\rho(\tau(k))}$ runs over all $k \in \{1, \dots, n\}$. Since $\tau$ is a bijection on $\{1, \dots, n\}$, setting $m = \tau(k)$ simply rearranges the factors without changing the product: \begin{align*} \prod_{k=1}^{n} A_{\tau(k),\rho(\tau(k))} = \prod_{m=1}^{n} A_{m,\rho(m)}. \end{align*} Therefore \begin{align*} \det A' = -\sum_{\rho \in S_n} \operatorname{sgn}(\rho) \prod_{m=1}^{n} A_{m,\rho(m)} = -\det A. \end{align*} The sign flip comes entirely from $\operatorname{sgn}(\tau) = -1$: swapping two rows is equivalent to composing every permutation in the Leibniz sum with a transposition, which negates all the signs simultaneously. [/guided] [/step] [step:Prove that scaling row $i$ by $\lambda$ multiplies the determinant by $\lambda$] Let $A'$ be obtained from $A$ by multiplying row $i$ by $\lambda \in F^\times$. Then $A'_{k,l} = A_{k,l}$ for $k \neq i$ and $A'_{i,l} = \lambda A_{i,l}$ for all $l$. In the Leibniz expansion, each product $\prod_{k=1}^{n} A'_{k,\sigma(k)}$ differs from $\prod_{k=1}^{n} A_{k,\sigma(k)}$ only in the factor at $k = i$, where $A'_{i,\sigma(i)} = \lambda A_{i,\sigma(i)}$. Therefore \begin{align*} \det A' &= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} A'_{k,\sigma(k)} = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \cdot \lambda A_{i,\sigma(i)} \prod_{\substack{k=1 \\ k \neq i}}^{n} A_{k,\sigma(k)} = \lambda \det A. \end{align*} [/step] [step:Prove that adding $\lambda$ times row $j$ to row $i$ preserves the determinant] Let $A'$ be obtained from $A$ by replacing row $i$ with $a_i + \lambda a_j$, where $i \neq j$. Then $A'_{k,l} = A_{k,l}$ for $k \neq i$ and $A'_{i,l} = A_{i,l} + \lambda A_{j,l}$. In the Leibniz expansion, the factor at $k = i$ splits additively: \begin{align*} \det A' &= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \bigl(A_{i,\sigma(i)} + \lambda A_{j,\sigma(i)}\bigr) \prod_{\substack{k=1 \\ k \neq i}}^{n} A_{k,\sigma(k)}. \end{align*} Distributing the sum over the two terms in the parenthesis yields \begin{align*} \det A' &= \underbrace{\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} A_{k,\sigma(k)}}_{= \det A} + \lambda \underbrace{\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \cdot A_{j,\sigma(i)} \prod_{\substack{k=1 \\ k \neq i}}^{n} A_{k,\sigma(k)}}_{=: D}. \end{align*} [claim:The auxiliary sum $D$ vanishes] The sum $D$ equals the determinant of the matrix $B$ obtained from $A$ by replacing row $i$ with row $j$. Explicitly, $B_{k,l} = A_{k,l}$ for $k \neq i$ and $B_{i,l} = A_{j,l}$, so \begin{align*} D = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} B_{k,\sigma(k)} = \det B. \end{align*} The matrix $B$ has row $i$ equal to row $j$ (since $B_{i,l} = A_{j,l} = B_{j,l}$ for all $l$). By part (1), swapping rows $i$ and $j$ of $B$ negates the determinant: $\det B' = -\det B$. But swapping rows $i$ and $j$ of $B$ leaves $B$ unchanged (since these two rows are identical), so $B' = B$ and $\det B = -\det B$. This forces $2 \det B = 0$, hence $\det B = 0$ in any field (including characteristic $2$, where we give the alternative argument below). In characteristic $2$, $2 \det B = 0$ holds for all elements, so the above does not conclude. Instead, we argue directly from the Leibniz formula. Pair each permutation $\sigma$ with $\sigma' = \sigma \circ (i\;j)$. Since $B_{i,l} = B_{j,l}$ for all $l$, the products satisfy $\prod_{k=1}^{n} B_{k,\sigma'(k)} = \prod_{k=1}^{n} B_{k,\sigma(k)}$ (the factors at positions $i$ and $j$ are simply exchanged, but their values are equal). Meanwhile $\operatorname{sgn}(\sigma') = -\operatorname{sgn}(\sigma)$. The two terms cancel in the sum, and since this pairing partitions $S_n$ into pairs $\{\sigma, \sigma'\}$, we get $\det B = 0$. [/claim] [proof] Provided in the claim statement above. [/proof] Substituting $D = 0$, we conclude $\det A' = \det A + \lambda \cdot 0 = \det A$. [guided] The idea is that the row operation $a_i \mapsto a_i + \lambda a_j$ splits the determinant, via linearity of each Leibniz term in the $i$-th row, into the original determinant plus $\lambda$ times a determinant with a repeated row. We need to show the latter vanishes. Let $A'$ be obtained from $A$ by replacing row $i$ with $a_i + \lambda a_j$, where $i \neq j$. Then the entries of $A'$ satisfy $A'_{k,l} = A_{k,l}$ for $k \neq i$ and $A'_{i,l} = A_{i,l} + \lambda A_{j,l}$. In the Leibniz expansion, the only factor that changes is the one at index $k = i$. Since $A'_{i,\sigma(i)} = A_{i,\sigma(i)} + \lambda A_{j,\sigma(i)}$, the factor at $k = i$ splits additively: \begin{align*} \det A' &= \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \bigl(A_{i,\sigma(i)} + \lambda A_{j,\sigma(i)}\bigr) \prod_{\substack{k=1 \\ k \neq i}}^{n} A_{k,\sigma(k)}. \end{align*} Why does this split? Because each term in the Leibniz sum is a product of $n$ matrix entries, one from each row, and the factor from row $i$ is the only one that changed. The parenthesis contains two summands, and distributing the sum over these two terms yields \begin{align*} \det A' &= \underbrace{\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} A_{k,\sigma(k)}}_{= \det A} + \lambda \underbrace{\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \cdot A_{j,\sigma(i)} \prod_{\substack{k=1 \\ k \neq i}}^{n} A_{k,\sigma(k)}}_{=: D}. \end{align*} The first sum is exactly $\det A$ (the original Leibniz expansion). It remains to show that the auxiliary sum $D$ vanishes. The sum $D$ equals the determinant of a matrix $B$ obtained from $A$ by replacing row $i$ with row $j$. Explicitly, define $B$ by $B_{k,l} = A_{k,l}$ for $k \neq i$ and $B_{i,l} = A_{j,l}$. Then \begin{align*} D = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{k=1}^{n} B_{k,\sigma(k)} = \det B. \end{align*} Why does $\det B = 0$? The matrix $B$ has two identical rows: row $i$ equals row $j$ (since $B_{i,l} = A_{j,l} = B_{j,l}$ for all $l$). By part (1), swapping rows $i$ and $j$ negates the determinant: $\det B' = -\det B$, where $B'$ is the matrix with rows $i$ and $j$ swapped. But since rows $i$ and $j$ of $B$ are identical, the swap leaves $B$ unchanged, so $B' = B$ and therefore $\det B = -\det B$. This gives $2\det B = 0$, hence $\det B = 0$ in any field where $\operatorname{char}(F) \neq 2$. What about characteristic $2$? In $\operatorname{char}(F) = 2$, the equation $2\det B = 0$ holds for all elements and does not force $\det B = 0$. We need a direct cancellation argument in the Leibniz sum. For each permutation $\sigma \in S_n$, pair it with $\sigma' = \sigma \circ (i\;j)$. Since $B_{i,l} = B_{j,l}$ for all $l$, exchanging the roles of rows $i$ and $j$ in the product does not change its value: $\prod_{k=1}^{n} B_{k,\sigma'(k)} = \prod_{k=1}^{n} B_{k,\sigma(k)}$ (the factors at positions $i$ and $j$ are simply exchanged, but their values are equal). Meanwhile, $\operatorname{sgn}(\sigma') = \operatorname{sgn}(\sigma \circ (i\;j)) = \operatorname{sgn}(\sigma) \cdot \operatorname{sgn}((i\;j)) = -\operatorname{sgn}(\sigma)$ by the [Sign Homomorphism](/theorems/778). So the two terms $\sigma$ and $\sigma'$ contribute equal products but opposite signs, and they cancel in the sum. This pairing is a fixed-point-free involution on $S_n$ (since $i \neq j$ ensures $\sigma' \neq \sigma$), so every term in the Leibniz sum is paired and cancelled, giving $\det B = 0$. Substituting $D = \det B = 0$, we conclude $\det A' = \det A + \lambda \cdot 0 = \det A$. [/guided] [/step]