<s>[INST] 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
 [/INST] All the other answers to this question so far overlook the fact that the matrix in option (b) is not in Jordan canonical form in the first place, so you can eliminate that option without doing any work at all. After eliminating (c) and (d) as you’ve done, that leaves only (a).
</s>