This determinant is there a short cut?

[flyingpig]

Homework Statement

Compute this determinant

\begin{vmatrix}<br /> 1 &amp; 1&amp; x&amp; 1\\ <br /> x&amp; 1&amp; 1 &amp;1 \\ <br /> 1&amp; 1 &amp; 1 &amp;x \\ <br /> 1&amp; x &amp; 1&amp; 1<br /> \end{vmatrix}

The Attempt at a Solution

I tried swapping rows and eventually I got
-\begin{vmatrix}<br /> x &amp; 1&amp; 1&amp; 1\\ <br /> 1&amp; x&amp; 1 &amp;1 \\ <br /> 1&amp; 1 &amp; x &amp;1 \\ <br /> 1&amp; 1 &amp; 1&amp; x<br /> \end{vmatrix}

Now I could row reduce, but I could get two 0s at most and I still have to do two 3 x 3 determinants. Is there an easier way?

 

[I like Serena]

You're allowed to multiply a row with a constant and subtract the result from another row.
In this manner you can use Gaussian elimination to resolve the determinant (note that not all rules of Gaussian elimination apply though).
When you have reached an upper triangular matrix, the determinant is the same as the product of the numbers on the main diagonal.