The characteristic polynomials are

det\left(A-tI_{3} \right) = \left|\begin{matrix} -t & -1 & 0 \\ 3 & 3-t & 1 \\ 1 & 1 & 1-t \end{matrix} \right| = -t\left(3-t\right)\left(1-t\right)-1+3\left(1-t\right)+t

=-\left(t-1\right) ^{2}\left(t-2\right),

det\left(B-tI_{3} \right) =-\left(t-1\right) ^{2}\left(t-2\right).

So both are the same and A and B have common eigenvalues 1, 1 and 2.

By computation,

\left(A-I_{3} \right)\left(A-2I_{3} \right)= \left[\begin{matrix} -1 & -1 & 0 \\ 3 & 2 & 1 \\ 1 & 1 & 0 \end{matrix} \right]\left[\begin{matrix} -2 & -1 & 0 \\ 3 & 1 & 1 \\ 1 & 1 & -1 \end{matrix} \right]

=\left[\begin{matrix} -1 & \ldots & \ldots \\ \ldots & \ldots & \ldots \\ \ldots & \ldots & \ldots \end{matrix} \right] \neq O,

\left(B-I_{3} \right)\left(B-2I_{3} \right)= \left[\begin{matrix} -2 & 4 & 2 \\ -1 & 2 & 1 \\ -1 & 2 & 1 \end{matrix} \right]\left[\begin{matrix} -3 & 4 & 2 \\ -1 & 1 & 1 \\ -1 & 2 & 1 \end{matrix} \right]=O.

So A has the minimal polynomial \left(t-1\right) ^{2}\left(t-2\right), while B  has \left(t-1\right)\left(t-2\right).

Therefore A and B are not similar and B is diagonalizable while A is not.

Equivalently , r\left(A-I_{3} \right)=2 implies that dim E_{1} =3-r\left(A-I_{3} \right)=1 which is not equal to the algebraic multiplicity 2 of 1, meanwhile r\left(B-I_{3} \right)=1 implies that dim E_{1} =3-r\left(B-I_{3} \right)=2. Thus A is not diagonalizable while B is and they are not similar.