Question: > 
> $$
>  \left( \begin{array}
> {ccc}1 & 1 & x \\
> 1 & x & 1 \\
> x & 1 & 1 \\
> \end{array} \right) $$
> 
> 
> Find the values of $x$ such that the matrix above has rank $1$, $2$ and $3$.
> 
> 
> 

Can I get the matrix to reduced row echelon form, or is there another approach?

Answer: 
There are many methods, one is to compute its characteristic polynomial.

$$\chi(t) = -t^3 + (x+2) t^2 + (x-1)^2 t - (x-1)^2 (2 + x)$$
Then we see we have rank $3$ when $x \neq 1, -2$, rank $2$ when $x = -2$ and rank $1$ when $x = 1$.
Hint. Compute the determinant by using [Sarrus' rule](https://en.wikipedia.org/wiki/Rule_of_Sarrus). You will find a third degree polynomial which has $1$ as a root (for $x=1$ the matrix has rank $1$). If the determinant is not zero then the rank is $3$. Then consider the case when the determinant is zero.

P.S. The determinant is $3x-x^3-2=-(x-1)^2(x+2)$. Therefore for $x\not in \{1,-2\}$ the rank is $3$. If $x=1$ then the rank is $1$. For $x=-2$ the matrix is
$$\left(\begin{array}{ccc}1 & 1 & 2 \\
1 & 2 & 1 \\
2 & 1 & 1 \\
\end{array} \right)$$
which has rank $2$ because the minor $\left(\begin{array}{cc}1 & 1\\
1 & 2
\end{array} \right)$ has anon-zero determinat.