Intuitive understanding of determinants? 
For a $n \times n$ matrix $A$:$$\det (A) = \sum^{n}_{i=1}a_{1i}C_{1i}$$
where $C$ is the cofactor of $a_{1i}$. If the determinant is $0$, the matrix is not invertible.

Could someone an intuitive explanation of why a zero determinant means non-invertibility? I'm not looking for a proof, the book gives me one. I'm looking for intuition.
For example, consider the following properties of zero:
Property 1: If a factor of a polynomial is $0$ at $x$, then the polynomial is zero at $x$.
Intuition: Anything times zero, so if the remaining polynomial is being multiplied by zero, it has to be zero.
Property: $x + 0 = x$
Intuition: If you have something and don't change anything about it, it remains the same.
So why does having a zero determinant imply non-invertibility? Could someone give me a similar intuition for determinants?
Thanks.
 A: There are several different ways to think about this intuitively, and which one you prefer may differ depending on your algebraic/geometric inclination.
The first interpretation is the one that tylerc0816 mentions. Each matrix induces a corresponding linear mapping. For such a map, the standard basis vectors (i.e. the unit cube) gets mapped to a parallelotope formed by the columns of your matrix. In order for this map to be invertible, the parallelotope must be non-degenerate. What is meant by degeneracy here is that the parallelotope must be "full"-dimensional, i.e. you cannot have a 3D parallelepiped collapse into a 2D parallelogram.
If such a collapse does happen, then intuitively you have multiple points stacked on top of each other. If this happens, then the map cannot be invertible. The condition for this collapse is equivalent to the parallelotope having $0$ volume, i.e. for the matrix to have zero determinant.
Secondly, you may view the determinant in terms of how it changes under elementary row operations:


*

*If you add a row to another row, then the determinant is unchanged. 

*Multiplying a row by a non-zero scalar multiplies the determinant by the same non-zero scalar. 

*Switching two rows changes the sign of the determinant.
Notice that all of these operations preserve the "zero-ness" of the determinant, i.e. the determinant will be zero after an elementary row operation if and only if it was zero before. With this view in mind, we see that the determinant of a matrix will be zero if and only if the determinant of its Reduced Row Echelon Form is zero. The RREF is either identity for invertible matrices, in which case the determinant is $1$, or it has an all zero row, which necessarily forces the determinant zero.
Alternatively, you may wish to look at the determinant in terms of the characteristic polynomial. The determinant is precisely the value of the characteristic polynomial evaluated at $0$ (perhaps up to some factors of $-1$ depending on how you define the characteristic polynomial). If the characteristic polynomial is $0$ when evaluated at $0$, then the matrix must have a zero eigenvalue which means a non-trivial nullspace. Again this forces the matrix to be non-invertible.
A: The absolute value of the determinant is the volume of the mapping of the unit cube under $A$.  If the determinant is $0$, then there had to be some 'collapsing' under $A$ (think dimension).  Thus, $A$ is not injective, and thus not invertible.
A: There is a simple property for determinants:
$Det(A)Det(b)=Det(AB)$, so if you take:
$Det(A A^{-1})= Det(A)Det(A^{-1})= Det(I) = 1$, from this follows than $det(A)$ must be different from $0$ to be invertible.
A: The determinant is the product of the eigenvalues of the matrix.  If the matrix is invertible, the eigenvalues of that inverse matrix are the reciprocals of the eigenvalues of the original matrix -- this fails if one of them is zero.  Hence,
($A$ is non-invertible) $\leftrightarrow$ ($A$ has a zero eigenvalue) $\leftrightarrow$ (det $A=0$).
