Define $A=\begin{pmatrix} 0 & 0 & 3\\ 1 & 0 & 1\\ 0 & 1 & -3 \end{pmatrix}$ over $\mathbb F_p$.I am asking to find whether $A$ is diagonalizable when $p=2$.
I have find out characteristics polynomial $c_A(x)=-(x+3)(x-1)(x+1)=-(x+1)^3$ when $p=2$. So the algebraic multiplicity $a(-1)=3$. But I don't know the corresponding matrix for $c_A=-(x+1)^3$. How can I compute geometric multiplicity $g(-1)$?
There's no need to compute multiplicities.
You have shown that $A$ has $1$ as an eigenvalue with multiplicity $3$. If $A$ is diagonalizable, then $A$ must be similar to $I$ (the identity matrix), which would imply that $A = I$. However, it is clear that $A \neq I$. Thus, $A$ cannot be diagonalizable.
If you still want to compute a geometric multiplicity (as the solution apparently recommends), then the usual approach is to compute the dimension of the null space of $A + I$. We have $$ A - I = \pmatrix{-1 & 0 & 3\\ 1 & -1 & 1\\ 0 & 1 & -4} = \pmatrix{1&0&1\\1&1&1\\0&1&0}. $$ This matrix has rank at least two because its first two columns are linearly independent. Thus, its null space has dimension at most $3 - 2 = 1$ by the rank-nullity theorem. So, the geometric multiplicity of $-1$ is at most $1$. However, because $-1$ is an eigenvalue of $A$, its geometric multiplicity is also at least $1$.
So, the multiplicity of $-1$ is $1$.