Convention: No sum over repeated indices unless otherwise stated.

Definition: DetA=e(ijk)a1ia2ja3k, sum, e(ijk) is Levi-Civita symbol.

A common “proof”* of detA=detA^{T}, is incorrect. It is:

DetA=e(ijk)a1ia2ja3k=e(ijk)ai1aj2ak3=DetA^{T}, sum

For Ex, the term

e(312)a13a21a32 becomes e(312)a31a12a23;

to show this is a term of D, you have to rearrange the a’s to get the first indices in numerical order which gives:

e(312)a12a23a31, which is a term of D if e(312) = e(231), which has to be proved in general (for each term).

Theorem:

Given e(ijk)ai1aj2ak3. Rearrange a’s so that first indices are in numerical order. Then:

e(ijk)ailaj2ak3=e(lmn)a1la2ma3n.

Proof (Mirsky):

Consider:

e(123)e(ijk)ai1aj2ak3

For every interchange of a pair of a’s, simultaneously change the corresponding indices of the first and second e’s, with the result that:

e(123)e(ijk)ai1aj2ak3 = e(lmn)e(123)a1la2ma3n, or

e(ijk)ai1aj2ak3 = e(lmn)a1la2ma3n

End Proof.

Therefore:

DetA^{T}= e(ijk)ai1aj2ak3 = e(lmn)a1la2ma3n = DetA, sum

Note

Nothing changes for the general case ijk,……

* Determinant of Transpose - ProofWiki