Answer
See below
Work Step by Step
Since $A$ has eigenvalues $\lambda=3,3,3,3,9,9$, it will have seven Jordan canonical forms, such as:
$$\begin{bmatrix}
3 & 0 & 0 & 0 & 0 & 0\\
0 & 3 & 0 & 0 & 0 & 0\\
0 & 0 & 3 & 0 & 0 & 0\\
0 & 0 & 0 & 3 & 0 & 0 \\
0 & 0 & 0 & 0 & 9& 0\\
0 & 0 & 0 & 0 & 0 & 9
\end{bmatrix}$$
$$\begin{bmatrix}
3 & 1 & 0 & 0 & 0 & 0\\
0 & 3 & 0 & 0 & 0 & 0\\
0 & 0 & 3 & 0 & 0 & 0\\
0 & 0 & 0 & 3 & 0 & 0 \\
0 & 0 & 0 & 0 & 9& 0\\
0 & 0 & 0 & 0 & 0 & 9
\end{bmatrix}$$
$$\begin{bmatrix}
3 & 1 & 0 & 0 & 0 & 0\\
0 & 3 & 0 & 0 & 0 & 0\\
0 & 0 & 3 & 1 & 0 & 0\\
0 & 0 & 0 & 3 & 0 & 0 \\
0 & 0 & 0 & 0 & 9& 0\\
0 & 0 & 0 & 0 & 0 & 9
\end{bmatrix}$$
$$\begin{bmatrix}
3 & 1 & 0 & 0 & 0 & 0\\
0 & 3 & 1 & 0 & 0 & 0\\
0 & 0 & 3 & 0 & 0 & 0\\
0 & 0 & 0 & 3 & 0 & 0 \\
0 & 0 & 0 & 0 & 9& 0\\
0 & 0 & 0 & 0 & 0 & 9
\end{bmatrix}$$
$$\begin{bmatrix}
3 & 1 & 0 & 0 & 0 & 0\\
0 & 3 & 1 & 0 & 0 & 0\\
0 & 0 & 3 & 1 & 0 & 0\\
0 & 0 & 0 & 3 & 0 & 0 \\
0 & 0 & 0 & 0 & 9& 0\\
0 & 0 & 0 & 0 & 0 & 9
\end{bmatrix}$$