Number of possible Jordan canonical form without constructing the matrix

Question

Problem: Let $\text A$ be a square matrix of order $9$ and $1$ and $-2$ are Eigenvalues with algebraic multiplicity $5$ and $4$ respectively. And minimal polynomial of the matrix $A$ is $m(x) = ( − 1)^3( + 2)^2$ then number of possible Jordan canonical form is
$(a)~~9~~~~~~~~~~~~~~~~~(b)~~5~~~~~~~~~~~~~~~~~(c)~~4~~~~~~~~~~~~~~~~~(d)~~2$

My solution:
Characteristic polynomial $~\text{Ch}(x)=( − 1)^5( + 2)^4~$ and
Minimal polynomial $~~~~~~~~~~~\text{m}(x)=( − 1)^3( + 2)^2~$
Now we can write $~5=3+2=3+1+1~,$ so there are two possible choice.
Also we can weite $~4=2+2=2+1+1~,$ so there are again two possible choice.
Hence the number of possible Jordan canonical form is $~2\times2=4~.$

Question: Here I am trying to solve the problem without constructing the matrix. Is my way of thinking the solution of the problem is correct ? If not, please help to find out the solution.

Answer 1

Your working and answer are correct, except that you should multiply two and two to get the final answer of four – this is a matter of combinatorics. The selection of Jordan blocks for each distinct eigenvalue are independent.