Trace = 5. (Sum of diagonal elements)
There will be 5 (not necessarily distinct) eigenvalues, because this is a $5\times 5$ matrix.
Let the eigen values be $a, b,c,d,e$
$a+b+c+d+e=5$
The rank of this matrix is 2.
How?
$R_5 \leftarrow R_5-R_1$. Last row becomes 0. So, rank isn't 5.
For any $4\times 4$ matrix, we'll still not get non-zero determinant.
Same for $3\times 3$ matrix.
But for this $2\times 2$ matrix
$\begin{bmatrix} 0 & 1\\ 1& 0 \end{bmatrix}$
Determinant is non-zero.
So, rank =2.
Since rank = 2, we only have two non-zero eigenvalues. So, the equation reduces to
$a+b=5$
Now, eigenvalues can be $(1,4)$ or $(2,3)$
=> The answer could be 4 or 6.