Because of Determinant of Transpose, it is necessary to prove only one of these identities. Let: :$D = \begin {vmatrix} a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\  \vdots & \ddots &  \vdots & \ddots &  \vdots \\ a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\  \vdots & \ddots &  \vdots & \ddots &  \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix}$ First, note that from Determinant with Row Multiplied by Constant, we have: :$\begin{vmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{r 1} &       0 & \cdots &       0 \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end {vmatrix} = a_{r 1} \begin {vmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\  \vdots &  \vdots & \ddots &  \vdots \\       1 &       0 & \cdots &       0 \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end {vmatrix}$ and similarly: :$\begin {vmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\  \vdots &  \vdots & \ddots &  \vdots \\       0 & a_{r 2} & \cdots &       0 \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end {vmatrix} = a_{r 2} \begin {vmatrix} a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\  \vdots &  \vdots & \ddots &  \vdots \\       0 &       1 & \cdots &       0 \\  \vdots &  \vdots & \ddots &  \vdots \\ a_{n 1} & a_{n 2} & \cdots & a_{n n} \end{vmatrix}$ and so on for the whole of row $r$. From Determinant as Sum of Determinants: :$\displaystyle \begin {vmatrix} a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\  \vdots & \ddots &  \vdots & \ddots &  \vdots \\ a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\  \vdots & \ddots &  \vdots & \ddots &  \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix} = \sum_{k \mathop = 1}^n \paren {a_{r k} \begin {vmatrix} a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\  \vdots & \ddots &  \vdots & \ddots &  \vdots \\       0 & \cdots &       1 & \cdots &       0 \\  \vdots & \ddots &  \vdots & \ddots &  \vdots \\ a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n} \end {vmatrix} }$ Consider the determinant: :$\begin{vmatrix}              a_{1 1} & \cdots &              a_{1 \paren {k - 1} } &              a_{1 k} &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\               \vdots & \ddots &                             \vdots &               \ddots &                            \vdots  & \ddots &               \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} k} & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\                    0 & \cdots &                                  0 &                    1 &                                  0 & \cdots &                    0 \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} k} & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\               \vdots & \ddots &                             \vdots &               \ddots &                            \vdots  & \ddots &               \vdots \\              a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n k} &              a_{n \paren {k + 1} } & \cdots &              a_{n n} \end {vmatrix}$ Exchange rows $r$ and $r - 1$, then (the new) row $r - 1$ with row $r - 2$, until finally row $r$ is at the top. Row 1 will be in row 2, row 2 in row 3, and so on. This is permuting the rows by a $k$-cycle of length $r$. Call that $k$-cycle $\rho$. Then from Parity of K-Cycle: :$\map \sgn \rho = \paren {-1}^{r - 1}$ Thus: :$\begin {vmatrix}                    0 & \cdots &                    1 & \cdots &                    0 \\              a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\               \vdots & \ddots &               \vdots & \ddots &               \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\               \vdots & \ddots &               \vdots & \ddots &               \vdots \\              a_{n 1} & \cdots &              a_{n k} & \cdots & a_{n n} \end {vmatrix} = \paren {-1}^{r - 1} \begin {vmatrix}              a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\               \vdots & \ddots &               \vdots & \ddots &               \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\                    0 & \cdots &                    1 & \cdots &                    0 \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\               \vdots & \ddots &               \vdots & \ddots &               \vdots \\              a_{n 1} & \cdots &              a_{n k} & \cdots & a_{n n} \end {vmatrix}$ The same argument can be applied to columns. Thus: :$\begin {vmatrix}                    1 &                    0 & \cdots &                                  0 &                                  0 & \cdots &                    0 \\              a_{1 k} &              a_{1 1} & \cdots &              a_{1 \paren {k - 1} } &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\               \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\ a_{\paren {r - 1} k} & a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\ a_{\paren {r + 1} k} & a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\               \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\              a_{n k} &              a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n \paren {k + 1} } &              \cdots & a_{n n} \end {vmatrix} = \paren {-1}^{k-1}\begin {vmatrix}                    0 & \cdots &                    1 & \cdots &                    0 \\              a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\               \vdots & \ddots &               \vdots & \ddots &               \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\               \vdots & \ddots &               \vdots & \ddots &               \vdots \\              a_{n 1} & \cdots &              a_{n k} & \cdots &              a_{n n} \end {vmatrix}$ and so: :$\begin{vmatrix}                    1 &                    0 & \cdots &                                  0 &                                  0 & \cdots &                    0 \\              a_{1 k} &              a_{1 1} & \cdots &               a_{1 \paren {k - 1}} &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\               \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\ a_{\paren {r - 1} k} & a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\ a_{\paren {r + 1} k} & a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\               \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\              a_{n k} &              a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n \paren {k + 1} } & \cdots &              a_{n n} \end {vmatrix} = \paren {-1}^{r + k} \begin {vmatrix}              a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\               \vdots & \ddots &               \vdots & \ddots &               \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\                    0 & \cdots &                    1 & \cdots &                    0 \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\               \vdots & \ddots &               \vdots & \ddots &               \vdots \\              a_{n 1} & \cdots &              a_{n k} & \cdots &              a_{n n} \end {vmatrix}$ Then: :$\paren {-1}^{r + k} \begin {vmatrix}              a_{1 1} & \cdots &              a_{1 \paren {k - 1} } &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\ a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\ a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\              a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n \paren {k + 1} } & \cdots &              a_{n n} \end{vmatrix}$ is $A_{r k}$, the cofactor of $a_{r k}$ in $D$. But from Determinant with Unit Element in Otherwise Zero Row, we have: :$\begin {vmatrix}       1 &       0 & \cdots &       0 \\ b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\  \vdots &  \vdots & \ddots &  \vdots \\ b_{n 1} & b_{n 2} & \cdots & b_{n n} \end {vmatrix} = \begin {vmatrix} b_{2 2} & \cdots & b_{2 n} \\  \vdots & \ddots &  \vdots \\ b_{n 2} & \cdots & b_{n n} \end {vmatrix}$ Assembling all the pieces derived above, the result follows. {{qed}}