Determinant with Rows Transposed

Theorem
If two rows (or columns) of a matrix with determinant $D$ are transposed, its determinant becomes $-D$.

Proof
Let $\mathbf A = \left[{a}\right]_{n}$ be a square matrix of order $n$.

Let $\det \left({\mathbf A}\right)$ be the determinant of $\mathbf A$.

Let $1 \le r < s \le n$.

Let $\rho$ be the permutation on $\N^*_n$ which transposes $r$ and $s$.

From Parity of K-Cycle, $\operatorname{sgn} \left({\rho}\right) = -1$.

Let $\mathbf A' = \left[{a'}\right]_{n}$ be $\mathbf A$ with rows $r$ and $s$ transposed.

By the definition of a determinant:
 * $\displaystyle \det \left({\mathbf A'}\right) = \sum_{\lambda} \left({\operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n a'_{k \lambda \left({k}\right)}}\right)$

By Permutation of Determinant Indices:
 * $\displaystyle \det \left({\mathbf A'}\right) = \sum_{\lambda} \left({\operatorname{sgn} \left({\rho}\right) \operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n a_{\rho \left({k}\right) \lambda \left({k}\right)}}\right)$

We can take $\operatorname{sgn} \left({\rho}\right) = -1$ outside the summation because it is constant, and so we get:


 * $\displaystyle \det \left({\mathbf A'}\right) = \operatorname{sgn} \left({\rho}\right) \sum_{\lambda} \left({\operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n a_{\rho \left({k}\right) \lambda \left({k}\right)}}\right) = - \sum_{\lambda} \left({\operatorname{sgn} \left({\lambda}\right) \prod_{k \mathop = 1}^n a_{k \lambda \left({k}\right)}}\right)$

hence the result.

From Determinant of Transpose‎, it follows that the same applies to columns as well.