For the 2nd problem,
use det (A) = det ( A transpose ) and that your matrix in particular has
A = (-A)^transpose. That should show that the det = 0 for n=odd.
Only two determinant calculation had left!
I can't prove even one of them! I am been trying all the week, but I failed!
1. The diagonal is "a" and all the rest is "b"
a , b, ... , b
b , a,
.
.
.
b , ... , a
2. Matrix in which the diagonal is "0". Above the diagonal is "-1" and "1" is below the diagonal. (Need to separate to 2 cases when n is odd and when n is even)
0,-1,...,-1
1,0
.
.
.
1,... 0
I know that when n is odd so det=0 and when n is even so det=1, but how to prove it ?!
Thank you very much!
I am puzzled by the first question; how many rows & columns? Hmm, but A has to be square in order to be able to find its determinant. Problem here is that you don't know the values of "a" and "b". If for example b=0, you knew A is diagonal and the determinant of a diagonal matrix is the product of the diagonal elements. Maybe you need to treat this as the general case:
In which case determinant of A is defined to be the sum of all signed elementary products from A, and can be summerized as . So that the signed elementary products are summed over all possible permutations of the column indices. In the sign is +ve if the permutation is even and -ve if it's odd.
[quote=Also sprach Zarathustra;438061]Only two determinant calculation had left!
I can't prove even one of them! I am been trying all the week, but I failed!
1. The diagonal is "a" and all the rest is "b"
a , b, ... , b
b , a,
.
.
.
b , ... , a
Lemma:
Proof: By induction on n.
This is NOT an easy induction proof, but I think it is very beautiful and educative.
How did I reach the general formula? I did the cases n =1,2,3,4 and then recognized the general pattern (precisely from the case n =4 ), and then proved it by induction. A good idea, imo, for other similar cases.
Tonio