Here is what my analysis and I was not satisfied with any analysis so I posted this let me know if it is correct.
By seeing the matrix given, the sum of eigenvalues=5 and product(the determinant of the matrix)=0.
Let our five eigen values be $\lambda_1,\lambda_2,\lambda_3,\lambda_4,\lambda_5$
$\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5=5$
and
$\lambda_1.\lambda_2.\lambda_3.\lambda_4.\lambda_5=0$
Since, determinant is 0, atleast one of the eigen values is 0. I assume say $\lambda_5=0$
Now, I can rewrite $\lambda_1+\lambda_2+\lambda_3+\lambda_4=5$
one of the possible eigen values I can think of is 1,1,1,2 but is 1 really an eigen value? If yes, then for the matrix $A-1.I$, you must be able to find a non-zero vector X such that $|A-I|.X=0$, means the determinant of $|A-I|$ (where A is our matrix given in the question) must be 0.
$|A-I|=$$\begin{bmatrix} 0 & 0 & 0 & 0& 1\\ 0& 0 &1 & 1& 0\\ 0 & 1 & 0&1 &0 \\ 0 &1 &1 & 0 &0\\ 1 & 0 &0 &0 & 0 \end{bmatrix}$
In this, all the rows and columns are independent and hence for this determinant is not zero.Hence, 1 is not our eigen value.
is 2 an eigen value?
$|A-2I|=$$\begin{vmatrix} -1& 0 & 0&0 &1 \\ 0& -1 & 1 &1 &0 \\ 0 & 1& -1& 1 &0 \\ 0 &1 & 1 & -1& 0\\ 1 & 0 &0 &0 & -1 \end{vmatrix}$
To this, vector X=$\begin{bmatrix} 1\\ 0\\ 0\\ 0\\ 1 \end{bmatrix}$ is such that $|A-2I|.X=O$
Hence, 2 is one of the eigen value.
Now, say $\lambda_1=2$ and we assumed $\lambda_5=0$
Now, $\lambda_2+\lambda_3+\lambda_4=3$
The only way this is possible is( 1 is not an eigenvalue), if one of the eigenvalues is 3 and others are zero.
so let us assume $\lambda_2=3$ and $\lambda_3=\lambda_4=0$ and 0 is an eigenvalue for the given matrix so this combination validates our equation
$\lambda_1+\lambda_2+\lambda_3+\lambda_4+\lambda_5=5$ and
$\lambda_1.\lambda_2.\lambda_3.\lambda_4.\lambda_5=0$
Any other eigen values possible?, no because atleast 1 of them has to be 0 and 1 is not an eigen-value.!!
Hence. answer : $3*2=6$