[Math] Why does a singular matrix imply that it does not have a solution

linear algebramatrices

(Trying to learn linear algebra over here)

The augmented matrix in question:

$$\begin{bmatrix}0 & 1 &5 & -4\\1 & 4 & 3 & 2\\2 & 7 & 1 & -2\end{bmatrix}$$

So I tried to solve the matrix above but I couldn't. I decided to see what happened when I pushed it through Numpy (Python):

numpy.linalg.linalg.LinAlgError: Singular matrix

So I went back to the definition for a singular matrix:

A square matrix that is not invertible is called singular or
degenerate.

The book simply says it is inconsistent. So, now I'm wondering: How does a singular matrix relate to it not having a solution?

Best Answer

Reducing an augmented matrix $[\mathbf{A}|\mathbf{b}]$ is equivalent to solving the system $\mathbf{Ax} = \mathbf{b}$. However, there is a unique solution to this system if and only if $\mathbf{A}$ is invertible. The solution is $\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}$.

If the matrix $\mathbf{A}$ is not invertible, there are either zero or an infinite number of solutions.