Property of Determinant

Question

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.

Answer 1

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.

Answer 2

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