Why LU decomposition is not used to give elementary proof of $\det (AB) = \det (A) \det (B) $ for square matrices?

It seems for any square matrix $ A $, $ A = PLU$

P is permutation matrix, L lower triangular matrix, U upper triangular matrix.

Using elementary row operations :

$ \det (PB) = \det (B) \det (P) $,

$\det (LB) = \det (L) \det (B) $,

$\det (UB) = \det (U) \det (B) $

Regarding the last 2 identities: You can obtain $ LB $ by performing row additions to the matrix $ G=TB $, where $ T $ is a diagonal matrix such that $ T_{ii} = L_{ii} $.

Starting from the last row, take each row of $ G $ and add multiples of rows above it to obtain the rows of $ LB $, this implies $ det (LB) = det (TB) =det (T) det (B)= det (L) det (B) $

Similar trick works for $ UB $ by adding multiples of rows below a given row starting from the first row.

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    $\begingroup$ It is not entirely clear to me why should $\det(LB)=\det L\det B$ or $\det (UB)=\det U\det B$ $\endgroup$ – Gae. S. Jul 12 at 15:56
  • $\begingroup$ And how is it usually proved? The proof we were given in the class was basically a repeat of LU proof. (I guess it is always useful repeat such important stuff for students). But i personally love to think just of $\mathrm{det}$ as of constant by which volumes gets multiplied, then (at least for $\mathbb{R}^{n}$) this formula becomes obvious, and if we wish to extend it for other commutative rings, there is an algebraic trick. $\endgroup$ – Rybin Dmitry Jul 12 at 15:58
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    $\begingroup$ Well, I wish I knew a proper answer which I could share with you, but I suspect that a proof involving LU decomposition might be considered non-elementary. There is no proof more elementary than one which uses little more than the definition of a determinant. See chapter 8 of math.ucdavis.edu/~anne/linear_algebra/mat67_course_notes.pdf and chapter 6 of cip.ifi.lmu.de/~grinberg/primes2015/…. $\endgroup$ – Thomas Winckelman Jul 12 at 16:02
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    $\begingroup$ The usual intro proof uses elementary matrices/operations, which you would need anyway to prove your three identities. Seems like you're inserting a middle-man. $\endgroup$ – Randall Jul 12 at 16:03
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    $\begingroup$ I agree with @Randall $\endgroup$ – Thomas Winckelman Jul 12 at 16:05

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