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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?

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    $\begingroup$ I think $R_4 \to R_1 + R_2 + R_3 + R_4, C_1 \to C_4 - C_1 , C_2 \to C_4 - C_2, C_3 \to C_4 - C_3$ makes a bit easier by introducing 3 zeros at first, second, third column of fourth row. $\endgroup$
    – S L
    Mar 21, 2013 at 13:00
  • $\begingroup$ @experimentX This way is much nicer than the solution I posted (and deleted) and actually correct, my block diagonal thought turns out to be false. $\endgroup$
    – muzzlator
    Mar 21, 2013 at 13:12
  • $\begingroup$ @muzzlator sorry, i couldn't see ... when I clicked load answer, it was already deleted. $\endgroup$
    – S L
    Mar 21, 2013 at 13:12

2 Answers 2

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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 $$ \det M=\lambda_1\lambda_2\lambda_3\lambda_4. $$

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  • $\begingroup$ Observe that (in case $a,b,c,d$ are all real) the eigenvalues $\lambda_2$ and $\lambda_4$ are compex conjugates, so their product is always real. $\endgroup$ Mar 21, 2013 at 13:15
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    $\begingroup$ The eigenvalue theory comes from the observation that the eigenvalues of the matrix $$T=\pmatrix{0&1&0&0\cr0&0&1&0\cr0&0&0&1\cr1&0&0&0\cr}$$ are $1,i,-1$ and $-i$. Hardly a surprise as $T^4=I_4$. Here $$M=aI+bT+cT^2+dT^3,$$ so the claim follows. $\endgroup$ Mar 21, 2013 at 13:19
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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*}

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