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I am looking for a non-permutational definition of determinant. The definition should have these properties:

1: Calculational power (easily applicable, it cold be used for practical calculations).

2: It should hold the usual properties (otherwise it wouldn't be a determinant, right!)

3: No permutation, No permutation, please no permutations.

I would also appreciate if you apply the definition and calculate the determinant of a $4×4$ matrix.

For all I care the definition could be from an obsolete parchment, but it needs to have those three properties.

Appreciate all the help!

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    $\begingroup$ Alternating scalar valued tensor with value 1 at the point $(e_1,...,e_n)$. It is equivalent to the permutation definition, so there is no escaping it. $\endgroup$ – copper.hat Jan 22 at 21:38
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    $\begingroup$ You cannot escape the permutations. $\endgroup$ – copper.hat Jan 22 at 21:38
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    $\begingroup$ Then take the Leibniz formula as in my answer. The permutations are not directly visible, but still in the background, like copper.hat says. $\endgroup$ – Dietrich Burde Jan 22 at 21:39
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    $\begingroup$ The determinant of a triangular matrix is the product of the diagonal coefficients. For the general matrix, perform Gauss-Jordan reduction ( en.m.wikipedia.org/wiki/Gaussian_elimination ) till you get a triangular matrix, the determinant of each step being easy to compute. You may also want to look at math.stackexchange.com/questions/668/… . $\endgroup$ – Mindlack Jan 22 at 21:45
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    $\begingroup$ @BertrandWittgenstein'sGhost: At some level, permutations are the essence of the determinant... Being multlinear means there are potentially $2^n$ computations, by being alternating it reduces to $n!$. A smaller number means you are dealing with a special case (Hessenberg, triangular, etc.) $\endgroup$ – copper.hat Jan 22 at 22:04
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For the $3\times 3$-determinant we can use the Rule of Sarrus. Then the $4\times 4$ determinant reduces to the $3\times 3$ determinant, because of $${\begin{vmatrix}a&b&c&d\\e&f&g&h\\i&j&k&l\\m&n&o&p\end{vmatrix}}=a\,{\begin{vmatrix}f&g&h\\j&k&l\\n&o&p\end{vmatrix}}-b\,{\begin{vmatrix}e&g&h\\i&k&l\\m&o&p\end{vmatrix}}+c\,{\begin{vmatrix}e&f&h\\i&j&l\\m&n&p\end{vmatrix}}-d\,{\begin{vmatrix}e&f&g\\i&j&k\\m&n&o\end{vmatrix}}.$$

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  • $\begingroup$ Thanks for that, but it is not general. I need a general definition. The $4×4$ calculation was just so I can see how the def. Works when applied. $\endgroup$ – Bertrand Wittgenstein's Ghost Jan 22 at 21:40
  • $\begingroup$ The general definition is, as already said, by the Weiserstrass axioms. From there $1\times 1$, $2\times 2$, $3\times 3$ are obvious. Then we also have $4\times 4$ from the above formula. $\endgroup$ – Dietrich Burde Jan 23 at 9:05

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