Find matrix using its characteristic matrix

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so I'm given the following matrix: $$B= \left(\begin{matrix}x-1&0&0\\0&(x-1)(x-3)&0\\0&0&0\end{matrix}\right) $$ and I'm asked to find a normal form matrix $A$ whose characteristic matrix ($xI_n-A$) is similar to B. I used the rational form and got the matrix : $$ A= \left(\begin{matrix}1&0&0\\0&1&0\\0&0&3\end{matrix}\right) $$ to check my result I calculated the Smith Normal Form of A to compare it with B and I got the matrix: $$ S(A)= \left(\begin{matrix}1&0&0\\0&x-1&0\\0&0&(x-1)(x-3)\end{matrix}\right) $$ which is obviously not equivalent to B (compare rank or determinant for example). Could anyone tell me where the mistake is?