Q:

Interchanging two columns of a matrix will lead to change in the sign of determinant

Interchanging two columns of a square matrix changes the sign of the determinant.
I know that this is true, and I do understand how it works. But is there any proof for this statement?

A:

Any proof of this result depends on a definition of determinant. Let us define it using permutations: $\det(A) = \sum_{\tau \in S_n}\operatorname{sgn}(\tau)\,a_{1,\tau(1)}a_{2,\tau(2)} \ldots a_{n,\tau(n)},\;$ where the sum is over all $n!$ permutations of the columns by elements in the symmetric group $S_n.\;$ See the question about a determinant definition.
Let $A^\sigma$ be the result of rearranging the columns of $A$ using a permutation $\sigma.\;$ This replaces all the $\tau$ in the summation by $\sigma\tau$, the product of two permutations. Now $\;\operatorname{sgn}(\sigma\tau)=\operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)\;$ and, by distributivity, the common $\operatorname{sgn}(\sigma)$ comes out of the summation. Thus, $\;\det(A^\sigma)=\operatorname{sgn}(\sigma)\det(A).$
In our case, interchanging any two columns is a transposition and these all have signature $-1$, and so multiplying the determinant by $-1$ changes its sign. QED.

A:

Determinant of the matrix 
$$J = \begin{bmatrix}0&1\\1&0\end{bmatrix}$$ is $-1$. 
Determinant of block diagonal matrix $\det{\textrm{diag}(A,B)}$ is $(\det A)\times(\det B)$
determinant of a product of two matrices $AB$ is again $(\det A)\times(\det B)$

Now a row swap is equivalent to 
$$
\begin{bmatrix}
I_k & & & \\ 
  &0&1& \\
  &1&0& \\
  & & &I_l
\end{bmatrix}A = PA
$$
The determinant of $P$ is first by using the second property $\det(I_k)\det(J)\det(I_l)$. Now by using the first identity we have $\det(P) = 1\times-1\times1 = -1$. 
Now using the third property $\det(PA) = -\det(A)$.
If you work from the right side of $A$ this corresponds to a column swap.

