No. If the matrix is singular, i.e. if $\det A=0$, the system has $0$ solution or an infinity of solutions.
The general criterion is based on the augmented matrix $[A|B]$:
Let $A$ be an $m\times n$ matrix, $r$ its rank, $B$ an $m\times 1$ matrix. The system of equations $AX=B$ has a solution if and only if $\operatorname{rank}A=\operatorname{rank}[A|B]$.
Furthermore the set of solutions is an affine subspace of $\mathbf R^n$ of dimension $n-r$.
Simplest way:
These equations are also Eq. of planes in 3D.
If any two panes are parallel then no solution.
If two of them are coincident.Then planes meet in a line and there are many solutions.
If all three planes are coincident, then many solutions.
Else,
let $z=k$ and solve first two equations get $x,y$ in terms of $k$, put them in third equation. Three mutully exclusive things can happen.
1-$k$ gets determined, so unique solution: Planes meeting in a point.
2-$k$ disappears leaving a true statement like $3=3$, so many solutions
possible for any real value of $k$: Planes meeting in a line.
3-$k$ disappears leaving a flase statement like $3=4$, so no solutions possible for any real value of $k$:Planes forming an open prism.
These situation can also be told in terms of Cramer determinants, adjoint or rank. of a matrix. When no solutions, system is called inconsistent. If unique or many solutions, the system is called consistent.
Best Answer
No. The system$$\left\{\begin{array}{l}x+y+z=0\\x+y+z=1\end{array}\right.$$has no solutions.
However, if you are talking about homogeneous linear equations (that is, equations of the type $a_1x_1+a_2x_2+\cdots+a_nx_n=0$), then you are right: a system with less than $n$ of those equations in $n$ variables always has infinitely many solutions. Actually, the set of solutions is a vector space whose dimension is greater than $0$.