By the "volume" of "parallelepiped", I mean the Lebesgue measure of n-Parallelotope.

If I have $$\vec{v\_i}=\begin{bmatrix}a\_{1i} \\ a\_{2i} \\ \vdots \\ a\_{ni}\end{bmatrix} \qquad \text{ for } i\in\{1,2,3\ldots,n\}$$
and $$\mathbf A=\begin{bmatrix}\vec{v\_1} & \vec{v\_2} & \cdots & \vec{v\_n}\end{bmatrix}=\begin{bmatrix}a\_{11} & a\_{12} & \cdots & a\_{1n}\\ a\_{21} & a\_{22} & \cdots & a\_{2n}\\ \vdots & \vdots & \ddots & \vdots \\ a\_{n1} & a\_{n2} & \cdots & a\_{nn}\end{bmatrix}$$
Now there are two ways to define the determinant.

**Definition 1:** If $\mathbf C\_{ij}$ is the cofactor, then $$\det(\mathbf A)=\sum\_{k=1}^{n}a\_{ik}C\_{ik}=\sum\_{k=1}^{n}a\_{kj}C\_{kj} \qquad \text{for any } i,j\in\{1,2,3,\ldots,n\}$$
**Definition 2:** $\det(\mathbf A)$ is the Lebesgue measure of the fundamental n-Parallelotope spanned by the the column vectors $\vec{v\_i}\in\mathbb R^n$.

How do I prove that the two definitions are equivalent?
I personally like definition 2 because, I can visualize it but in definition 1, first we need to show that the summations are giving same values for all i and j.

I can use the second definition definition for n=2, first thing I noted was that column operations does not change the area of the parallelogram because of simple geometry properties. Thus, $$\begin{vmatrix}a & c\\ b & d\end{vmatrix}=\begin{vmatrix}a & c-a\frac{c}{a}\\ b & d-b\frac{c}{a}\end{vmatrix}=\begin{vmatrix}a & 0\\ b & \frac{ad-bc}{a}\end{vmatrix}=\begin{vmatrix}a-0\frac{ab}{ad-bc} & 0\\ b-\frac{ad-bc}{a}\frac{ab}{ad-bc} & \frac{ad-bc}{a}\end{vmatrix}=\begin{vmatrix}a & 0\\ 0 & \frac{ad-bc}{a}\end{vmatrix}$$ This turns it into a rectangle whose area can be calculated easily.
$$\begin{vmatrix}a & c\\ b & d\end{vmatrix}=ad-bc$$
But this is the same as we get from definition 1. Thus both definitions are equivalent for n=2.

The argument in finding determinant for n=2 by the second definition generalizes easily, but the method of computation feels completely different as compared to definition 1. Like for n=3, I got $$\begin{vmatrix}a&d&g\\b&e&h\\c&f&i\end{vmatrix}=\begin{vmatrix}\frac{a(ei-hf)-d(bi-ch)-g(ec-bf)}{ei-hf}&0&0\\0&\frac{ei-hf}{i}&0\\0&0&i\end{vmatrix}=a(ei-hf)-d(bi-ch)+g(bf-ec)$$ I can see a little bit connection for $i=1$ in definition 1.

I got to know that definition 1 is called [Laplace Expansion](https://en.wikipedia.org/wiki/Laplace_expansion#Proof), but the proof written on Wikipedia went above my tiny brain. I am in 11th grade and I know very little about Linear algebra(I know only the stuff Grant Sanderson told in his essence of LA playlist). After reading the answer [Determinant of transpose](https://math.stackexchange.com/questions/598258/determinant-of-transpose) I can make sense why row operations do not change the determinant as well. I would be really happy if someone proves definition 1 using definition 2.