How can I obtain the Smith normal form of large matrix?

Question

Sorry that I could not include the matrix to the title, it goes over the limit of character number. The given matrix is \begin{bmatrix}x-\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2}\\-\frac{1}{2}&x-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}\\\frac{1}{2}&\frac{1}{2}&x-\frac{1}{2}&-\frac{1}{2}\\-\frac{1}{2}&\frac{3}{2}&-\frac{1}{2}&x-\frac{1}{2}\end{bmatrix} I tried to find appropriate pivot, and I got \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&2x&x^2-x-2\\0&0&-4x^2+2x&x^2-1\end{bmatrix} but now stuck here. I think I have done something wrong. My guess is that by the properties of normal form, it should be \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&x^4-2x^3+2x^2-1\end{bmatrix} Can you show several steps to get the answer?

Answer 1

We can see that $$ 1=\frac{x-1}{4}\cdot 2x -\frac{1}{2}\cdot (x^2-x-2) $$ so that we should replace $C_3$ by $\frac{x-1}{4}\cdot C_3 -\frac{1}{2}\cdot C_4$. [This is possible by a sequence of steps dictated by the extended Euclidean Algorithm that calculates the HCF.]

We then use the $1$ in the $(3,3)$ to clear the off diagonal entries. As all this preserves the determinant we'll get the correct entry in the $(4,4)$ place.