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Question: Calculate the following determinant $$A=\begin{vmatrix} a&b&c&d\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{vmatrix}$$

Progress: So I apply $R1'=R1+R2+R3+R4$ and get $$A=(a+b+c+d)\begin{vmatrix} 1&1&1&1\\ b&a&d&c\\ c&d&a&b\\d&c&b&a\end{vmatrix}$$ Then, I apply $C2'=C2-C1,\, C3'=C3-C1$, etc to get $$A=(a+b+c+d)\begin{vmatrix}1&0&0&0\\b&a-b&d-b&c-b\\c&d-c&a-c&b-c\\d&c-d&b-d&a-d \end{vmatrix}$$ Thus, $$A=(a+b+c+d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\c-d&b-d&a-d \end{vmatrix}$$ But now I'm stuck here. I don't really want to expand the whole thing, because it is really messy. Is there any approach that don't require much calculation?

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    $\begingroup$ Have you tried the Leibniz formula for determinants ? $\endgroup$ – Lærne Feb 12 '18 at 9:46
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    $\begingroup$ This is a block matrix, you can calculate a determinant of 2x2 matrix made of 2x2 matrices, and then expand that $\endgroup$ – Yuriy S Feb 12 '18 at 9:47
  • $\begingroup$ @Lærne: Then I would have to expand the determinant, no? The point is I'm looking for a way to manipulate this determinant, so that I don't have to calculate much. $\endgroup$ – MS1902 Feb 12 '18 at 9:50
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    $\begingroup$ @Yuriy S: Then the matrix $\begin{bmatrix} c&d\\d&c\end{bmatrix}$ and $\begin{bmatrix} a&b\\b&a\end{bmatrix}$ must be commute, right? $\endgroup$ – MS1902 Feb 12 '18 at 9:52
  • $\begingroup$ @MS1902, and no, as you can see these two matrices do not have to commute. We are squaring the matrices, not multiplying them together $\endgroup$ – Yuriy S Feb 12 '18 at 10:42
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\begin{align}A&=(a+b+c+d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\c-d&b-d&a-d \end{vmatrix} \\&=(a+b+c+d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\0 &a+b-c-d&a+b-c-d \end{vmatrix} \\ &= (a+b+c+d)(a+b-c-d)\begin{vmatrix}a-b&d-b&c-b\\d-c&a-c&b-c\\0 &1&1\end{vmatrix} \\ &= (a+b+c+d)(a+b-c-d)\begin{vmatrix}a-b&d-b&c-b\\a+d-b-c&a+d-b-c& 0\\0 &1&1\end{vmatrix} \\ &=(a+b+c+d)(a+b-c-d)(a+d-b-c)\begin{vmatrix}a-b&d-b&c-b\\1&1& 0\\0 &1&1\end{vmatrix} \\ &=(a+b+c+d)(a+b-c-d)(a+d-b-c)\begin{vmatrix}a-b&d-c&0\\1&1& 0\\0 &1&1\end{vmatrix} \\ &=(a+b+c+d)(a+b-c-d)(a+d-b-c)(a-b+c-d) \end{align}

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This is a block matrix, you can calculate a determinant of a $2 \times 2$ matrix made of $2 \times 2$ matrices, and then expand that: $$A=\begin{vmatrix} M&N\\ N&M\end{vmatrix}$$ where $$M=\begin{bmatrix} a&b\\ b&a\end{bmatrix}$$ and $$N=\begin{bmatrix} c&d\\ d&c\end{bmatrix}$$

All matrices are symmetric, this is really simple (you don't have to think about commutativity of the matrix products).

$$A=\begin{vmatrix} M&N\\ N&M\end{vmatrix}=|M^2-N^2|$$

$$M^2=\begin{bmatrix} a^2+b^2&2ab\\ 2ab&a^2+b^2\end{bmatrix}$$

$$N^2=\begin{bmatrix} c^2+d^2&2cd\\ 2cd&c^2+d^2\end{bmatrix}$$

$$A=(a^2+b^2-c^2-d^2)^2-4(ab-cd)^2$$

The formula is correct, as can be checked by substituting some numbers.


Edit

A more general case of the determinant of a block matrix. (See here for example).

If $C$ and $D$ commute.

$$|M|= \begin{vmatrix} A & B \\C & D \end{vmatrix}=|AD-BC|$$


Edit 2

Just in case there are any more people claiming this is wrong, here's the check:

Simplify[Det[{{a, b, c, d}, {b, a, d, c}, {c, d, a, b}, {d, c, b, a}}] - (a^2 + b^2 - c^2 - d^2)^2 + 4 (a b - c d)^2]

enter image description here

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