Intuition behind Matrix being invertible iff determinant is non-zero

Here's an explanation for three dimensional space ($3 \times 3$ matrices). That's the space I live in, so it's the one in which my intuition works best :-).

Suppose we have a $3 \times 3$ matrix $\mathbf{M}$. Let's think about the mapping $\mathbf{y} = f(\mathbf{x}) = \mathbf{M}\mathbf{x}$. The matrix $\mathbf{M}$ is invertible iff this mapping is invertible. In that case, given $\mathbf{y}$, we can compute the corresponding $\mathbf{x}$ as $\mathbf{x} = \mathbf{M}^{-1}\mathbf{y}$.

Let $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ be 3D vectors that form the columns of $\mathbf{M}$. We know that $\det{\mathbf{M}} = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$, which is the volume of the parallelipiped having $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ as its edges.

Now let's consider the effect of the mapping $f$ on the "basic cube" whose edges are the three axis vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$. You can check that $f(\mathbf{i}) = \mathbf{u}$, $f(\mathbf{j}) = \mathbf{v}$, and $f(\mathbf{k}) = \mathbf{w}$. So the mapping $f$ deforms (shears, scales) the basic cube, turning it into the parallelipiped with sides $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$.

Since the determinant of $\mathbf{M}$ gives the volume of this parallelipiped, it measures the "volume scaling" effect of the mapping $f$. In particular, if $\det{\mathbf{M}} = 0$, this means that the mapping $f$ squashes the basic cube into something flat, with zero volume, like a planar shape, or maybe even a line. A "squash-to-flat" deformation like this can't possibly be invertible because it's not one-to-one --- several points of the cube will get "squashed" onto the same point of the deformed shape. So, the mapping $f$ (or the matrix $\mathbf{M}$) is invertible if and only if it has no squash-to-flat effect, which is the case if and only if the determinant is non-zero.

I know this is pretty old, but for the people who might find this in a google search (I know I did), I thought I'd add this.

Remember that the space of all $n \times n$ matrices is isomorphic to the space of all operators on an $n$-dimensional vector space. In other words, the matrix is just some linear operator. Recall that the determinant is the product of the eigenvalues. Both $ \mathbb{C} $ and $ \mathbb{R} $ are integral domains so if the determinant is $0$ then that means you have a zero eigenvalue. That means, if your matrix is $A$, there exists a vector $0 \neq \overline{v} $ such that $ (A-0I)\overline{v}=\overline{0} $ which means $ A\overline{v}=\overline{0} $. Clearly $ \overline{0} $ gets sent to $ \overline{0} $ but so does some other non-zero vector. So the transformation isn't injective and, thus, non-invertible. This is just the intuition though.

The absolute value of the determinant of a matrix is the volume of the parallelepiped spanned by the column vectors of that matrix.

Michael

Another classical way is more understandable: note that a determinant is not changed if we add one row to other and one column to another. Thus we obtain a diagonal matrix $B$. This matrix differs from $A$ by matrix-multipliers which correspond to elementary transformations and are invertible. So $A$ is invertible iff $B$ is invertible iff $\det(B) \neq 0$ iff $\det(A) \neq 0$.