Finding the Jordan canonical form of A and Choose the correct option Let $$ A = \begin{pmatrix}
  0&0&0&-4 \\ 1&0&0&0 \\ 0&1&0&5 \\ 0&0&1&0
 \end{pmatrix}$$
Then a Jordan canonical form of  A is 
Choose the correct option
$a) \begin{pmatrix}
  -1&0&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
 \end{pmatrix}$
$b) \begin{pmatrix}
  -1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
 \end{pmatrix}$
$c) \begin{pmatrix}
  1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
 \end{pmatrix}$
$d) \begin{pmatrix}
  -1&1&0&0 \\ 0&-1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
 \end{pmatrix}$
My attempt : I know that  Determinant   of  A = product  of eigenvalues of A,  as  option c and d  is  not correct because  Here Determinant of A = 4
that is $ \det A =  -(-4) \begin{pmatrix}1 & 0 &0\\0& 1 & 0\\ 0&0&1\end{pmatrix}$
I'm  in  confusion  about  option  a) and  b).......how  can I find the  Jordan canonical form of A ?
PLiz  help  me.
Any hints/solution will be appreciated.
Thanks in advance

Another answer that requires only a very minimal amount of computations.
The trace of $A$ is preserved under similarity transformations. $\operatorname{Tr}(A) = 0+0+0+0 = \lambda_1+\lambda_2+\lambda_3+\lambda_4$ is enough to exclude (c) and (d). As others have noted, (b) isn't even a Jordan canonical form (and by the way the matrix (b) is similar to (a) anyway). 
So (a) is the only remaining option and must be correct.
