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$\begingroup$ We may finish the proof after your second matrix, since the determinant of a block-triangular matrix is a product of determinants of diagonal blocks. $\endgroup$

Fedor Petrov
– Fedor Petrov

2016-05-21 18:34:37 +00:00
Commented May 21, 2016 at 18:34
1

$\begingroup$ @FedorPetrov I only remembered that after typing my answer and was wondering whether to edit it or not. I will edit my answer now, in order to shorten it and save readers' time. $\endgroup$

Rodrigo de Azevedo
– Rodrigo de Azevedo

2016-05-21 18:42:47 +00:00
Commented May 21, 2016 at 18:42

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