Before I knew about determinant, I only thought that the column-view of a matrix is useful and meaningful. For example: given a linear operator $T$ on $V, \dim(V)=n, A=[T]_\beta\in M_{n\times n}(F),$ and $\textrm{rank}(A)$ is defined by the number of linearly independent columns of $A$. When $A$ be seen as columns: $[\mathbf{a_1}, \mathbf{a_2},\dots,\mathbf{a_n}],$ it's clear that the range of $T$ is span by these columns. Then I tried to learn the orientation of a given basis $\{\mathbf u,\mathbf v\}\in \mathbb{R^2}$, ($O$ for orientation):
$$O(A)=\frac{\det{u\choose v}}{\lVert \det{u\choose v}\rVert},$$
where $\mathbf u,\mathbf v$ are now seen as(/in form of) row vectors. It's hard to connect this with $\det(A)$ also, since for $A$ representing a linear operator there is no connection(in my current understanding) between columns.
I know the fact that the choice of basis of $V$ is independent of $\det(A)$ since given $Q=[I]_{\beta'}^\beta$: $$\det(Q^{-1}AQ)=\det(Q^{-1})\det(Q)\det(A)=\det(A),$$
so similar matrices all have the same non-zero determinant value. What did I miss?