First of all, they said that a matrix A is invertible (there exists an $n \times n$ square matrix B such that $AB = BA= I_{n}$) if and only if its determinant
is nonzero. So, there some kind of scalar value that determines if a matrix if invertible.
Assuming that you have knowledge in row reducing a matrix, we known that there exist a criteria for determining if a matrix is invertible. That criteria is that when row reducing a matrix its entry $a_{nn}$ must not equal zero. That means $a_{nn} \neq 0.$
So let's see the case for a $3 \times 3$ matrix:
$$ \begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix} $$
So for row reducing first we have to get zeros in the first column except in the first column, so we can use the row operation of multiplying a row by a scalar. In this case we multiply row 2 and 3 with the scalar $a_{11}$ and then we use the row operation of summing a row multiplied by a scalar, in this case for the second row it will $-a_{21} \cdot R1$ and for third row $-a_{31} \cdot R1$.
$$\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{11}a_{21} & a_{11}a_{22} & a_{11}a_{23} \\
a_{11}a_{31} & a_{11}a_{32} & a_{11}a_{33}
\end{bmatrix} \sim
\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{11}a_{21} - a_{11}a_{21} & a_{11}a_{22} - a_{12}a_{21} & a_{11}a_{23} - a_{13}a_{21} \\
a_{11}a_{31} - a_{11}a_{31} & a_{11}a_{32} - a_{12}a_{31} & a_{11}a_{33} - a_{13}a_{31}
\end{bmatrix}
$$
$$ \sim \begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
0 & a_{11}a_{22} - a_{11}a_{21} & a_{11}a_{23} - a_{11}a_{21} \\
0 & a_{11}a_{32} - a_{11}a_{31} & a_{11}a_{33} - a_{11}a_{31}
\end{bmatrix} $$
So we can summarize this steps in a simple algorithm:
This algorithm for getting zeros in the first column can be generalize for an $n \times n$ matrix such that the first step will be:
Multiplying all the rows except the first one by the first entry of
the first column.
Adding to all rows except the first one by the $-a_{i1} \cdot R1.$
So we can apply again this algorithm to the next submatrix $A_{22}$:
After applying the algorithm to the submatrix $A_{22}$ we get th echelon form of the matrix A:
$$\sim \begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
0 & a_{11}a_{22} - a_{12}a_{21} & a_{11}a_{23} - a_{13}a_{21} \\
0 & 0 & (a_{11}a_{33} - a_{11}a_{31})(a_{11}a_{22} - a_{12}a_{21}) - (a_{11}a_{32} - a_{12}a_{31})(a_{11}a_{23} - a_{13}a_{21})
\end{bmatrix} $$
So we known $a_{nn} \neq 0.$, then:
$$(a_{11}a_{33} - a_{11}a_{31})(a_{11}a_{22} - a_{12}a_{21}) - (a_{11}a_{32} - a_{12}a_{31})(a_{11}a_{23} - a_{13}a_{21})$$
$$=a_{11}^{2}a_{22}a_{33}-a_{11}a_{12}a_{21}a_{33}-a_{11}a_{13}a_{22}a_{31}+a_{12}a_{13}a_{21}a_{31}-[a_{11}^{2}a_{23}a_{32}-a_{11}a_{13}a_{21}a_{32}-a_{11}a_{12}a_{23}a_{31}+a_{12}a_{13}a_{21}a_{31}]$$
$$=a_{11}[a_{11}a_{22}a_{33}+a_{13}a_{21}a_{32}+a_{12}a_{23}a_{31}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}]+a_{12}a_{13}a_{21}a_{31}-a_{12}a_{13}a_{21}a_{31}$$
$$=a_{11}[a_{11}a_{22}a_{33}+a_{13}a_{21}a_{32}+a_{12}a_{23}a_{31}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}]+0 \neq 0$$
This is what we call determinant of a $3 \times 3$ matrix. Because it determines if a matrix is invertible or not.
$$\Delta = a_{11}a_{22}a_{33}+a_{13}a_{21}a_{32}+a_{12}a_{23}a_{31}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}$$
This proccess can be replicated for $n \times n$ matrices.
This algo have a geometrical meaning for $2 \times 2$ matrices and $3 \times 3$ matrices.
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