Let E bet the set of matrices A in M(2n,R) such that
\( a_{ii}=0 \) for any \( 1\leq i\leq 2n \)
\( a_{ij} \) is -1 or 1 for any \( 1\leq i\neq j\leq 2n \)
1/ For A in E prove det(A) is an odd integer.
2/ Find the maximum of \( |\det(A)| \) for any matrix \( A \in E \) and find the minimum of \( |\det(A)| \) for any \( A \in E \).
Invertible matrix of even size
Moderators: Bogdan Posa, Laurian Filip, Beniamin Bogosel, Radu Titiu, Marius Dragoi
-
Beniamin Bogosel
- Co-admin
- Posts: 710
- Joined: Fri Mar 07, 2008 12:01 am
- Location: Timisoara sau Sofronea (Arad)
- Contact:
1) Pentru paritatea determinantului, observam ca nu conteaza daca schimbam un semn. (daca ne uitam, de exemplu in formula determinantului). Atunci putem considera matricea cu toate elementele 1 in afara de diagonala primcipala, unde sunt 0.
Acum adunam toate coloanele la prima coloana, si obtinem \( 2n-1 \) pe prima coloana peste tot. Dam factor pe \( 2n-1 \), si obtinem ca determinantul matricii noastre are aceeasi paritate cu matricea initiala, la care punem un 1 in coltul de sus, in stanga. Scadem prima coloana din celelalte si obtinem (in afara de prima coloana care ramane 1) -1 pe diagonala principala si 0 in rest, deci determinantul este -1, care este impar. Deci si matricea initiala are determinantul impar.
(we can consider all elements 1 and on the diagonal 0; this doesn't affect the parity of the determinant. Then we add all colunms to the first and factor \( 2n-1 \). Now we have a matrix with the first column 1. We substract this column from the others and we obtain -1 on the diagonal and 0 in rest, so the determinant is -1, odd, so the initial determinant is odd)
Imi pare rau pentru solutia povestita, dar nu am vreme sa scriu matrici asa mari
acuma
Acum adunam toate coloanele la prima coloana, si obtinem \( 2n-1 \) pe prima coloana peste tot. Dam factor pe \( 2n-1 \), si obtinem ca determinantul matricii noastre are aceeasi paritate cu matricea initiala, la care punem un 1 in coltul de sus, in stanga. Scadem prima coloana din celelalte si obtinem (in afara de prima coloana care ramane 1) -1 pe diagonala principala si 0 in rest, deci determinantul este -1, care este impar. Deci si matricea initiala are determinantul impar.
(we can consider all elements 1 and on the diagonal 0; this doesn't affect the parity of the determinant. Then we add all colunms to the first and factor \( 2n-1 \). Now we have a matrix with the first column 1. We substract this column from the others and we obtain -1 on the diagonal and 0 in rest, so the determinant is -1, odd, so the initial determinant is odd)
Imi pare rau pentru solutia povestita, dar nu am vreme sa scriu matrici asa mari
acuma