Simpler proof that $\det A\neq 0$ if $A$ is invertible? One excercise asked me to "Prove that the determinant of an inversible matrix can't be 0". I couldn't remember the proof the teacher gave and I didn't want to "cheat" because I'm practising for an exam, so after some thinking I came up with this.
I'd like to know if someone has a simpler and/or shorter proof than mine, and if someone can find any mistakes in my proof. Oh, and if someone can suggest a way to shorten it or maybe use more symbols and less natural language.
I lost the sheet of the proof the teacher provided, and Googling/searching books I found nothing (only proof that $\det{(A ^{-1})} = 1/\det A$). Maybe I didn't word it properly because English is not my native language.
Proof
First, proof that the determinant of an elementary matrix $[E]$ can't be 0:
An elementary matrix $[E]$ is defined as a matrix obtained by applying a single row operation to the identity matrix. $|I|=1$, so:
$|E|=1$ if it comes from adding a row to another one
$|E|=-1$ if it comes from swapping two rows
$|E|=k/k\in\Bbb R \land k≠0$ if it comes from multiplying a row by $k$
$|E|≠0$
Now, let's say we have a square matrix $[A]_{n\times n}$ that is invertible. To obtain the inverse of $[A]$ we have to apply row operations to the augmented matrix $[A|I]$ until we get $[I|A^{-1}]$. Let's say $E_1*E_2*...E_n$ are the elementary matrixes we premultiply our initial augmented matrix by to get the identity and the inverse. Then we get:
$$E_1*E_2*...E_n * A = I$$ and $$E_1*E_2*...E_n * I = A^{-1}$$
Let's take the first one, and apply determinant to both sides of the equality
$$|E_1*E_2*...E_n * A| = |I|$$
$$|E_1*E_2*...E_n|*|A| = 1$$
Now let's suppose $|A|$ could be $0$. We would get:
$$|E_1*E_2*...E_n|*0 = 1$$
$$0 = 1$$
Which is absurd, and thus the determinant of an invertible matrix can't be 0.
 A: Our OP El Menduko's proof looks fine to me.
As per his (I assume the masculine is apropos here, based on the appearance of "El" in the OP's user name.) request, here is a shorter proof, as indicated in the comments:
If $A$ is invertible, there is a matrix $B$ with
$AB = I; \tag 1$
then
$\vert AB \vert = \vert I \vert = 1; \tag 2$
but
$\vert AB \vert = \vert A \vert \vert B \vert; \tag 3$
thus
$\vert A \vert \vert B \vert = I, \tag 4$
which of course prohibits $\vert A \vert = 0$; we are left with the sole option
$\vert A \vert \ne 0. \tag 5$
Note:  I had just composed and begun to type this answer when the comment of user 296602 appeared.  I guess that person deserves some credit as well, so I tip my hat.  I prefer to answer some of these questions to help cut down on the number of questions without answers, even if they do have the equivalent in the comments.  End of Note.
A: Recall
$$ det(AB) = det(A)det(B) $$
$$ AA^{-1}=I \iff \text{ $A$ is invertible.} $$
Suppose $A$ is an invertible matrix. Note that if $A$ is invertible, then it follows that $A \neq 0$. So
$$ det(A)det(A^{-1}) = 1. $$
Suppose $ det(A) = 0 $. Then
$$ 0 = 1 $$
Which is a contradiction. Then $det(A) \neq 0$.
