If any two rows or columns of a determinant are interchanged, the sign of the value of the determinant is changed.
What is the intuition of this? I can see from algebra perspective thats right, but I want a geometrical visualization. I tried to think in terms of linear transformation but I was not able to visualize that transformation.
Geometrically, the determinant gives you the signed volume of the image of the unit cube under the linear transformation described by the matrix.
Let $R_{ij}$ denote an identity matrix with its $i$th and $j$th rows are switched. Note that for a compatible column vector, $R_{ij}x$ switches the $i$th and $j$th entry of $R$. Correspondingly, going from $A$ to $R_{ij}A$ switches the $i$th and $j$th entry of each column of $A$, which is to say that it switches the $i$th and $j$th rows of $A$. The determinant we are interested in is the determinant of the product $R_{ij}A$.
Recall that $R_{ij}A$ is a matrix corresponding to the application of two successive transformations; the first transformation is described by $A$, the second is described by $R_{ij}$. Geometrically, $R_{ij}$ is a refection across the hyperplane $x_i = x_j$. For example, in two dimensions, switching the $x$ and $y$ coordinates reflects across the line $y = x$.
A reflection will not change the absolute volume of any shape, but it will change the sign of any volume: the image of a positive volume under $R_{ij}$ is negative, and the image of a negative volume under $R_{ij}$ is positive. Now, the determinant of $A$ is the signed area of the unit cube under the first transformation. Applying a reflection to this volume will maintain its absolute value but change its sign, which means that the determinant of $R_{ij}A$ indeed has the same absolute value as the determinant of $A$ but has the opposite sign.
The determinant of a matrix is the $n$-dimensional signed volume of a parallelepiped whose edge vectors are the columns of the matrix. If we move one of the edge vector in the plane of the other $n-1$ vectors, the altitude on the base of the $n-1$ other vectors has not changed, so the altitude times the area of the base is unchanged. That is,
$$
\begin{align}
\det\begin{bmatrix}
u_1&v_1&\cdots&w_1\\
u_2&v_2&\cdots&w_2\\
\vdots&\vdots&\ddots&\vdots\\
u_n&v_n&\cdots&w_n
\end{bmatrix}
&=\det\begin{bmatrix}
u_1+v_1&v_1&\cdots&w_1\\
u_2+v_2&v_2&\cdots&w_2\\
\vdots&\vdots&\ddots&\vdots\\
u_n+v_n&v_n&\cdots&w_n
\end{bmatrix}\tag1\\[6pt]
&=\det\begin{bmatrix}
u_1+v_1&-u_1&\cdots&w_1\\
u_2+v_2&-u_2&\cdots&w_2\\
\vdots&\vdots&\ddots&\vdots\\
u_n+v_n&-u_n&\cdots&w_n
\end{bmatrix}\tag2\\[6pt]
&=\det\begin{bmatrix}
v_1&-u_1&\cdots&w_1\\
v_2&-u_2&\cdots&w_2\\
\vdots&\vdots&\ddots&\vdots\\
v_n&-u_n&\cdots&w_n
\end{bmatrix}\tag3\\[6pt]
&=-\det\begin{bmatrix}
v_1&u_1&\cdots&w_1\\
v_2&u_2&\cdots&w_2\\
\vdots&\vdots&\ddots&\vdots\\
v_n&u_n&\cdots&w_n
\end{bmatrix}\tag4
\end{align}
$$
Explanation:
$(1)$: slide the first vector in the plane of the remaining $n-1$ vectors
$\phantom{\text{(1):}}$ (add the second column to the first)
$(2)$: slide the second vector in the plane of the remaining $n-1$ vectors
$\phantom{\text{(2):}}$ (subtract the first column from the second)
$(3)$: slide the first vector in the plane of the remaining $n-1$ vectors
$\phantom{\text{(3):}}$ (add the second column to the first)
$(4)$: the determinant is linear in each column
