TITLE: What's an intuitive way to think about the determinant?
QUESTION [763 upvotes]: In my linear algebra class, we just talked about determinants. So far I’ve been understanding the material okay, but now I’m very confused. I get that when the determinant is zero, the matrix doesn’t have an inverse. I can find the determinant of a $2\times 2$ matrix by the formula. Our teacher showed us how to compute the determinant of an $n \times n$ matrix by breaking it up into the determinants of smaller matrices. Apparently there is a way by summing over a bunch of permutations. But the notation is really hard for me and I don’t really know what’s going on with them anymore. Can someone help me figure out what a determinant is, intuitively, and how all those definitions of it are related?

REPLY [2 votes]: I will try to explain this intuitively. But first you must understand certain concepts. I recommend 3b1b videos for intuition in "linear combinations". Anyways, it's not a difficult concept to understand, I will slightly introduce it.
First of all, let's start with an example and then try to generalize. So imagine we have the matrix $A=\left[  
\begin{matrix}
3 & 1
 \\
1.5 & 2
\end{matrix}
\right]$.
Now let's take the column vectors of this matrix,  $\left[  
\begin{matrix}
3 \\ 1.5
\end{matrix}
\right]$
and
$\left[  
\begin{matrix}
1
 \\ 2
\end{matrix}
\right]$. The linear combination of this vectors is what we call the Column Space - Col(A), all the possible combinations of this vectors. Graphically it looks something like this:

Also we have the Row Space - Row(A), identically, defined as the linear combinations of the row vectors $r_{1}=\left[  
\begin{matrix}
3 & 1
\end{matrix}
\right]$
and
$r_{2}=\left[  
\begin{matrix}
1.5 & 2
\end{matrix}
\right]$. They can be represented graphically the same way as with Col(A).

So basically, the determinant is the area created by the parallelogram defined by the row vectors (column vector generates the same area, but for convention let's use the row vectors). In the image it's represented by the blue parallelogram. So, Area of Parallelogram $= Determinant(A) = Det(A)$.
So, how can we calculate this area? For understanding this part you should have basic knowledge in "row operations" and "area of a parallelogram".
Let's call "$r_{1}$" the first row and "$r_{2}$" the second row. One of the basic row operations consist on add to one row other row scaled. So imagine row operating on $r_{1}$ as $r_{1}'=r_{1}+kr_{2}$, $k$ any real number. Don't desperate if you don't understand why we are row operating, thing are going to be clear right away.
So, let's call B the new matrix generated after replacing $r_{1}$ by $r_{1}+kr_{2}$. So, $r_{1}'$ with a quotation mark is going to be the transformed version of $r_{1}$. What would happen to Row(A) and to Det(A)? See what happens to Row(B) and Det(B) when we changed $r_{1}$ to $r_{1}'=r_{1}+kr_{2}$ with different values for $k$:

So, we can see that $r_{1}'$ moves parallel to $r_{2}$ which is obvious because we are adding a scaled version of $r_{2}$ to $r_{1}$.
Assuming you have knowledge in "parallelogram areas", you can verified that the base and the height doesn't change. That means that the area stays constant when adding one row scaled to another row because we doesn't change the height never cause we move parallel. Thus Det(A)=Det(B).
So here's come the MAGICAL PART, we should find a $k$ such that we eliminate the y-component of $r_{1}$ row vector (y-component$=A_{12}=0$). So applying row operation with $k=\frac{1}{2}$ such that $A_{12}=0$, the transformed matrix would be:
$$\left[  
\begin{matrix}
3 & 1
 \\
1.5 & 2
\end{matrix}
\right]
\xrightarrow{r_1-\frac{1}{2}r_2}
\left[  
\begin{matrix}
2.25 & 0
 \\
1.5 & 2
\end{matrix}
\right]$$
So our matrix B have a triangular form, Row(B) looks like:

Now we have a parallelogram with base length $=2.25$ and height length $=2$. Thus by definition of parallelogram area, we have that $Det(A)=Det(B)=2.25*2=4.5$. So the determinant is just the product of the diagonal elements of the triangular matrix form, we call it the echelon form. MAGIC
We could search for a rectangle that have the same area as Det(A) by repeating this process but this time applying the row operation to $r_{2}$ such that we eliminate it's x-component (x-component$=A_{21}=0$) such that we get a rectangle with area Det(B) that have the equivalent area as Det(A), but this is completely unnecessary since it doesn't change the base and height of the parallelogram. Anyways for intuition of this proccess $r_{2}'=r_{2}+kr_{1}'$ would look like:

So $k=-\frac{2}{3}\approx-0.66$.
$$\left[  
\begin{matrix}
2.25 & 0
 \\
1.5 & 2
\end{matrix}
\right]
\xrightarrow{r_{2}-\frac{2}{3}r_{1}'}
\left[  
\begin{matrix}
2.25 & 0
 \\
0 & 2
\end{matrix}
\right]$$
We have that the base is $2.25$ and the height is $2$, so the area of the rectangle is $Det(B)=4.5=Det(A)$. The determinant is just the product of the diagonal elements of the diagonal matrix.
So we've seen that the product of the diagonal elements of a converted matrix in triangular form gives us the determinant of the matrix. Why triangular form? Imagine $x_{i}$ being the $i$ dimension, so every row vector in echelon form of the matrix adds a new component to the $i$ dimension, so in geometric terms it adds a height to the dimension.
The great think of this technique is that it can be applied to any n-dimensions and maintains intuition of what you are doing. I would like to present the 3-d graphical proof but it would be a lot of work that I think you could do with a little of imagination.
I try to make an intuitive and geometrically process using this Bibliography:

Linear Algebra Series - 3Blue1Brown (Youtube Channel) - Intuition on determinants.
A Derivation of Determinants - Mark Demers. http://faculty.fairfield.edu/mdemers/linearalgebra/documents/2019.03.25.detalt.pdf - For mathematical rigurosity and intuition in how to calculate determinants.