This determinant is there a short cut?

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SUMMARY

The discussion focuses on computing the determinant of a 4x4 matrix with variables, specifically the matrix defined as \(\begin{vmatrix} 1 & 1 & x & 1 \\ x & 1 & 1 & 1 \\ 1 & 1 & 1 & x \\ 1 & x & 1 & 1 \end{vmatrix}\). Participants explored methods such as row swapping and Gaussian elimination to simplify the determinant calculation. The consensus is that while row reduction can help, achieving an upper triangular form is essential for determining the determinant as the product of the diagonal elements.

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Homework Statement



Compute this determinant

[tex]\begin{vmatrix}<br /> 1 & 1& x& 1\\ <br /> x& 1& 1 &1 \\ <br /> 1& 1 & 1 &x \\ <br /> 1& x & 1& 1<br /> \end{vmatrix}[/tex]

The Attempt at a Solution



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

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?
 
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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.
 

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