[proofplan] We construct an auxiliary matrix $B$ by replacing row $k$ of $A$ with row $i$ of $A$. The matrix $B$ has two identical rows ($i$ and $k$), so $\det B = 0$. On the other hand, expanding $\det B$ along row $k$ recovers exactly the sum $\sum_{j=1}^{n} A_{ij}\, C_{kj}$, because the cofactors of row $k$ depend only on the rows other than $k$ — which are the same in $A$ and $B$. The column version follows by applying the row result to $A^\top$. [/proofplan] [step:Construct an auxiliary matrix with a repeated row] Let $B \in M_{n \times n}(F)$ be the matrix obtained from $A$ by replacing row $k$ with row $i$. That is, define $B$ by \begin{align*} B_{\ell j} &:= \begin{cases} A_{\ell j} & \text{if } \ell \neq k, \\ A_{ij} & \text{if } \ell = k, \end{cases} \end{align*} for $\ell = 1, \ldots, n$ and $j = 1, \ldots, n$. Since $i \neq k$, the matrix $B$ has two identical rows: row $i$ and row $k$ both equal $(A_{i1}, \ldots, A_{in})$. [guided] The strategy is to build a matrix whose determinant we can evaluate in two different ways. We define $B \in M_{n \times n}(F)$ by copying $A$ but overwriting row $k$ with the entries of row $i$: \begin{align*} B_{\ell j} &:= \begin{cases} A_{\ell j} & \text{if } \ell \neq k, \\ A_{ij} & \text{if } \ell = k, \end{cases} \end{align*} for $\ell = 1, \ldots, n$ and $j = 1, \ldots, n$. Because $i \neq k$, the original row $i$ of $A$ is untouched (it sits at position $i$ in $B$), and we have placed a copy of it at position $k$. So rows $i$ and $k$ of $B$ are identical: $B_{ij} = A_{ij} = B_{kj}$ for every $j$. Why this particular construction? Because we want the cofactors of row $k$ in $B$ to match the cofactors of row $k$ in $A$, while the entries in row $k$ of $B$ are the entries of row $i$ of $A$. This will let us relate the cofactor expansion of $\det B$ to the sum $\sum_j A_{ij} C_{kj}$. [/guided] [/step] [step:Observe that $\det B = 0$ because $B$ has two identical rows] By the [Properties of the Determinant](/theorems/917), if a matrix has two identical rows, its determinant is zero. Rows $i$ and $k$ of $B$ are identical, so \begin{align*} \det B &= 0. \end{align*} [/step] [step:Expand $\det B$ along row $k$ to recover $\sum_{j} A_{ij}\, C_{kj}$] Let $\hat{B}_{kj}$ denote the $(n-1) \times (n-1)$ matrix obtained by deleting row $k$ and column $j$ from $B$, and let $C_{kj}(B) = (-1)^{k+j} \det \hat{B}_{kj}$ be the $(k,j)$-cofactor of $B$. By the [Cofactor Expansion](/theorems/398) along row $k$: \begin{align*} \det B &= \sum_{j=1}^{n} B_{kj}\, C_{kj}(B). \end{align*} We now identify the two ingredients. First, by the definition of $B$, the entries in row $k$ are $B_{kj} = A_{ij}$ for each $j$. Second, the submatrix $\hat{B}_{kj}$ is obtained by deleting row $k$ and column $j$ from $B$. Since $B$ and $A$ agree on every row except row $k$, and row $k$ is the deleted row, we have $\hat{B}_{kj} = \hat{A}_{kj}$. Therefore $C_{kj}(B) = (-1)^{k+j} \det \hat{A}_{kj} = C_{kj}$, the $(k,j)$-cofactor of $A$. Substituting: \begin{align*} \det B &= \sum_{j=1}^{n} A_{ij}\, C_{kj}. \end{align*} [guided] We apply the [Cofactor Expansion](/theorems/398) along row $k$ to the matrix $B$. The cofactor expansion theorem states that for any $n \times n$ matrix $M$ and any row index $\ell$, \begin{align*} \det M &= \sum_{j=1}^{n} M_{\ell j}\, (-1)^{\ell + j} \det \hat{M}_{\ell j}, \end{align*} where $\hat{M}_{\ell j}$ is the submatrix obtained by deleting row $\ell$ and column $j$. Applying this to $M = B$ with $\ell = k$: \begin{align*} \det B &= \sum_{j=1}^{n} B_{kj}\, (-1)^{k+j} \det \hat{B}_{kj} = \sum_{j=1}^{n} B_{kj}\, C_{kj}(B). \end{align*} Now we make two identifications: 1. **Entries of row $k$ in $B$:** By construction, $B_{kj} = A_{ij}$ for each $j = 1, \ldots, n$. 2. **Cofactors of row $k$ in $B$ equal the cofactors of row $k$ in $A$:** The submatrix $\hat{B}_{kj}$ is formed by deleting row $k$ and column $j$ from $B$. The remaining rows of $B$ are rows $1, \ldots, k-1, k+1, \ldots, n$, and on each of these rows $B$ and $A$ agree (only row $k$ was modified). Therefore $\hat{B}_{kj} = \hat{A}_{kj}$, and so $C_{kj}(B) = (-1)^{k+j} \det \hat{A}_{kj} = C_{kj}$, the $(k,j)$-cofactor of the original matrix $A$. Substituting both identifications into the cofactor expansion: \begin{align*} \det B &= \sum_{j=1}^{n} A_{ij}\, C_{kj}. \end{align*} This is the key observation: the cofactor expansion "mixes" the entries of row $i$ with the cofactors of row $k$, producing exactly the sum we want to evaluate. [/guided] [/step] [step:Conclude the row identity and deduce the column version by transposition] Combining the two results: $\det B = 0$ and $\det B = \sum_{j=1}^{n} A_{ij}\, C_{kj}$, we obtain \begin{align*} \sum_{j=1}^{n} A_{ij}\, C_{kj} &= 0 \quad \text{for } i \neq k. \end{align*} For the column version, apply the row result to $A^\top$. By the [Properties of the Determinant](/theorems/917), $\det A^\top = \det A$, and the $(i,j)$-cofactor of $A^\top$ satisfies $C_{ij}(A^\top) = C_{ji}(A)$ (since deleting row $i$ and column $j$ from $A^\top$ yields the transpose of the matrix obtained by deleting row $j$ and column $i$ from $A$, and the determinant is invariant under transposition). The row identity applied to $A^\top$ with row index $j$ and distinct row index $k$ gives \begin{align*} 0 &= \sum_{i=1}^{n} (A^\top)_{ji}\, C_{ki}(A^\top) = \sum_{i=1}^{n} A_{ij}\, C_{ik}(A) = \sum_{i=1}^{n} A_{ij}\, C_{ik}, \end{align*} which is the desired column identity for $j \neq k$. [guided] From the previous two steps we have $\det B = 0$ (because $B$ has a repeated row) and $\det B = \sum_{j=1}^{n} A_{ij}\, C_{kj}$ (by cofactor expansion). Setting these equal: \begin{align*} \sum_{j=1}^{n} A_{ij}\, C_{kj} &= 0 \quad \text{for } i \neq k. \end{align*} This completes the row version. For the column analogue, we use the standard transposition trick. By the [Properties of the Determinant](/theorems/917), $\det A^\top = \det A$. The entries of $A^\top$ satisfy $(A^\top)_{ji} = A_{ij}$. The cofactor $C_{ki}(A^\top)$ involves the submatrix obtained by deleting row $k$ and column $i$ from $A^\top$, which is the transpose of the submatrix obtained by deleting row $i$ and column $k$ from $A$. Since the determinant is invariant under transposition, $C_{ki}(A^\top) = (-1)^{k+i} \det \hat{A}_{ik} = C_{ik}(A)$. Applying the row identity to $A^\top$ with row indices $j \neq k$: \begin{align*} 0 &= \sum_{i=1}^{n} (A^\top)_{ji}\, C_{ki}(A^\top) = \sum_{i=1}^{n} A_{ij}\, C_{ik}. \end{align*} This is precisely the column version of the theorem: $\sum_{i=1}^{n} A_{ij}\, C_{ik} = 0$ for $j \neq k$. Together, the row and column identities say that "cross-expanding" the determinant — pairing entries from one row (or column) with cofactors from a different row (or column) — always yields zero. Combined with the [Cofactor Expansion](/theorems/398) theorem itself (which gives $\sum_j A_{kj} C_{kj} = \det A$ when the indices match), these identities can be written compactly as \begin{align*} \sum_{j=1}^{n} A_{ij}\, C_{kj} &= (\det A)\, \delta_{ik}, \end{align*} which is exactly the $(i,k)$-entry of the [Adjugate Identity](/theorems/397) $A \cdot \operatorname{adj}(A) = (\det A)\, I_n$. [/guided] [/step]