I have an implementation which is working on a matrix $A$. Actually I am checking it's Gramian matrix. If my understanding is correct, a linear time invariant system with adjacency matrix $A$, $\dot x(t) = Ax(t) + Bu$ is controllable if the controllability matrix
$$C = \begin{bmatrix} B & AB & A^2B & \dots & A^{n-1}B\end{bmatrix}$$
has full row rank. Then we have the controllability Gramian
$$\textit{W} = \int^{t_1}_{t_0} e^{At}BB^Te^{A^Tt} \, \mathrm d t$$
If the systme is controllable, the controllability Gramian matrix is positive definite. If controllability matrix $C$ does not satisfy the full row rank requirement the system is not controllable and the controllability Gramian $W$ is singular and not invertible.
But, I have a matirx $A$ which it's controllability matrix $C$ has full rank and shows that system is controllable but the Gramian matrix $W$ is singular and not invertible. What does it mean ? and what condition makes this state?
For example. suppose the adjacency matrix $A$ is: $$A= \begin{bmatrix} 0& 0& 1 \\ 1& 0& 1 \\ 1& 1& 0 \\ \end{bmatrix}$$ and the input matrix $B$ is: $$B= \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}$$ So the controllability matrix $C$: $$C= \begin{bmatrix} 1& 0& 1 \\ 0& 1& 1 \\ 0& 1& 0 \\ \end{bmatrix}$$ has full rank, but the Gramian matrix is singular. $$W= \begin{bmatrix} 1.56& 1.09& 1.09 \\ 1.09& 0.93& 0.93 \\ 1.09& 0.93& 0.93 \\ \end{bmatrix}$$
The problem is that you have miscalculated the controllability matrix. We find that $$ C = \pmatrix{1&0&1\\0&1&1\\0&1&1}. $$ The second and third rows are identical, so we see that $C$ fails to have full row rank. So, the system is not controllable, and it makes sense that the controllability Gramian that you attained was singular.
An explanation of why $C$ the system is not controllable in this case: note that the column-vector $B$ is a linear combination of the eignvectors of $A$ associated with eigenvalues $\lambda_{\pm} = \frac{1 \pm \sqrt{5}}{2}$, namely $$ v_{\pm} = \left(\frac{-1 \pm \sqrt{5}}{2}, 1,1 \right). $$ We see that $(1,0,0) = \frac 1{\sqrt{5}}(v_+ - v_-)$.