Does there exist a nonsingular graph for which the determinant of its adjacency matrix remains the same upon deleting a vertex? Consider simple graphs. Any simple graph $G$ is called nonsingular if its $(0,1)$-adjacency matrix $A(G)$ has nonzero determinant. Does there exist any nonsingular simple graph whose determinant value remains the same upon deleting a vertex?
 A: There exists a simple graph $G=(V,E)$ together with a vertex $v\in V$ such that $\operatorname{det}(A(G))$ and $\operatorname{det}\left(A(G\backslash\{v\})\right)$ are equal and non-zero. 
Example. The determinant of the adjacency matrix of the complete graph $K_n$ is known to be $(-1)^{n-1}(n-1)$. Let $G'=K_5$ be the complete graph on the vertex set $[1,5]$. By the previous remark $\operatorname{det}(A(G'))=4$. Suppose that $G$ is obtained from $G'$ by adding a vertex $0$ and two edges $(0,1)$ and $(0,2)$.
Let $E_{0,0}$ be the $6\times 6$ matrix indexed by $[0,5]\times [0,5]$ with entry $1$ at $(0,0)$ and zeros otherwise. Then the matrix 
\begin{align*}
A(G)-E_{0,0}=\left(\begin{matrix}
-1&1&1&0&0&0\\
1&0&1&1&1&1\\
1&1&0&1&1&1\\
0&1&1&0&1&1\\
0&1&1&1&0&1\\
0&1&1&1&1&0
\end{matrix}\right)
\end{align*}
is singular because $(2,1,1,-1,-1,-1)^T$ is an eigenvector with eigenvalue $0$. Hence
\begin{align*}
0=\operatorname{det}(A(G)-E_{0,0})=\operatorname{det}(A(G))-\operatorname{det}(A(G')),
\end{align*}
so that $\operatorname{det}(A(G))=\operatorname{det}(A(G'))=4$.
