I've been asked to compute the determinant of a $3 \times 3$ matrix with complex entries. I have done so using the normal expansion along a row or column method that I would use were the entries real. My questions are:

1) is this what I should be doing?

2) I obtained a complex result - how do I interpret what this means? I've been given to understand that the absolute of the determinant of a $3 \times 3$ matrix would represent it's volume, but can a volume be complex?

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    $\begingroup$ For 1: yes. For 2: you certainly can't interpret your determinant as a volume anymore, but the definition of the determinant still works with complex matrices, and is sometimes useful, so we admit them. $\endgroup$ – J. M. isn't a mathematician May 3 '12 at 8:53
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    $\begingroup$ The interpretation as a volume no longer makes sense, but there are other interpretations of the determinant which still do. For instance, you can think of the determinant as the product of all eigenvalues, counted with multiplicity. $\endgroup$ – Dan Petersen May 3 '12 at 9:01
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    $\begingroup$ Is it really the case that we lose the notion of volume here? If we view the complex matrix as an endomorphism of a complex vector space and then perform an identification of the complex space with a real space of double the dimension, then in fact the square of the absolute value of the "complex determinant" is the scaling factor for volumes in real space acted upon by the endomorphism ... $\endgroup$ – GaryMak Apr 16 '13 at 17:05
  • $\begingroup$ So I would argue that we can interpret the determinant as a "volume" in the same way that we regard any complex number as a "scalar": in order to bring things back to something meaningful in a physical sense we need to invoke complex conjugates and take the norm; but meanwhile we are perfectly able to speak about complex numbers as though they were some sort of sensible extension of the reals, provided that we have a method of bringing them "back to earth" later ... $\endgroup$ – GaryMak Apr 18 '13 at 7:39
  • $\begingroup$ Hence "i" is a perfectly acceptable multiplication factor for real quantities (interpreting it if you like - but not necessarily - as taking us into a $\pi/2$-rotation of the state into some larger, "invisible" space) and its effects are measured by the distance it moves things once we distil the action back into the real realm. I would venture to say that the relation of the complex determinant to volume is exactly the same. $\endgroup$ – GaryMak Apr 18 '13 at 7:41

While you lose interpretation of "volume", but other information survives, such as:

  • the determinant is nonzero iff the matrix is nonsingular
  • $|det(A)|=1$ iff the matrix is a unitary matrix (The vertical bars are denoting complex modulus, here)

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