Easier way of calculating the determinant for this matrix I have to calculate the determinant of this matrix:
$$
\begin{pmatrix}
a&b&c&d\\b&c&d&a\\c&d&a&b\\d&a&b&c
\end{pmatrix}
$$
Is there an easier way of calculating this rather than the long regular way?
 A: A pedestrian's solution (experimentX's suggestion below the question). 
Add the first three columns to the fourth:
\begin{align*}
\begin{vmatrix}
a&b&c&d\\b&c&d&a\\c&d&a&b\\d&a&b&c
\end{vmatrix} &
=(a+b+c+d)\begin{vmatrix}
a&b&c&1\\b&c&d&1\\c&d&a&1\\d&a&b&1
\end{vmatrix} 
\end{align*}
Subtract the second row from the first row, the third row from the second row and the fourth row from the third row; develop after the fourth column:
\begin{align*} 
&=(a+b+c+d)
\begin{vmatrix}
a-b&b-c&c-d \\
b-c&c-d&d-a \\
c-d&d-a&a-b 
\end{vmatrix} \\
\end{align*}
Add the first column to the third column:
\begin{align*}
&=(a+b+c+d)(a-b+c-d)
\begin{vmatrix}
a-b&b-c&1 \\
b-c&c-d&-1 \\
c-d&d-a&1 
\end{vmatrix} 
\end{align*}
Add the second row to the first row and the third row to the second row:
\begin{align*}
&=(a+b+c+d)(a-b+c-d)
\begin{vmatrix}
a-c&b-d&0 \\
b-d&c-a&0 \\
c-d&d-a&1 
\end{vmatrix} \\
&=-(a+b+c+d)(a-b+c-d)[(a-c)^2+(b-d)^2].
\end{align*}
A: You can easily transform your matrix to a circulant matrix
$$
M=\left(\begin{array}{cccc}
a&b&c&d\\
d&a&b&c\\
c&d&a&b\\
b&c&d&a
\end{array}\right)
$$
by carrying out obvious row swaps. The eigenvalue theory of circulant matrices is completely known. In the $4\times 4$ case the eigenvalues of $M$ are
$$
\lambda_1=a+b+c+d,\ \lambda_2=a+bi-c-di,\ \lambda_3=a-b+c-d\ \text{and}\ \lambda_4=a-bi-c+di.
$$
The determinant of a matrix is the product of its eigenvalues so
$$
\det M=\lambda_1\lambda_2\lambda_3\lambda_4.
$$
