Determinant with Rows Transposed

Theorem
If two rows (or columns) of a determinant with value $$D$$ are transposed, its value 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, $$\sgn \left({\rho}\right) = -1$$.

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

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

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

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


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

hence the result.

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