# [Math] Nonsingular matrix must have a solution proof

linear algebra

Im reading the book linear algebra from http://joshua.smcvt.edu/linearalgebra/.
In the book it says that a system has no solution if there is no particular solution and that a system has 1 solution if there is a particular solution and the homogenous solution is unique and that a system has many solution if there is a particular solution and there are many homogenous solution.

Then later on they say a nonsingular will always have a solution because it has a unique homogenous solution… but then what about the particular solution… what if the system has a unique homogenous solution but no particular solution?

If you want to solve a linear system $A(x)=b$, where $A$ is an $n\times n$ matrix. The fact that $A$ is non singular means that $A$ is invertible, thus $A^{-1}(b)$ is a solution of the equation.
When $A$ is not invertible, you need first to determine the space of homogeneous solutions which are the solution of $A(x)=0$. Denote by $Ker(A)$ this space of homogeneous solutions, and let $x_0$ be a particular solution, $A(x_0)=b$, the solutions are $x_0+a,a\in ker(A)$.