1. ## Calculating determinants.

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!

2. ## partial help

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.

3. 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:

$\displaystyle A = \begin{bmatrix}a_{11} & ... & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & ... & a_{nn}\end{bmatrix}$

In which case determinant of A is defined to be the sum of all signed elementary products from A, and can be summerized as $\displaystyle det(A)= \sum \pm a_1j_1 a_2j_2 ...a_nj_n$. So that the signed elementary products are summed over all possible permutations $\displaystyle \{ j_1,...,j_n \}$ of the column indices. In $\displaystyle \pm a_1j_1 a_2j_2...a_nj_n$ the sign is +ve if the permutation is even and -ve if it's odd.

4. [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: $\displaystyle det\,\begin{pmatrix} a&b&b&\ldots &b\\b&a&b&\ldots&b\\\ldots&\ldots&\ldots&\ldots&\l dots\\b&b&\ldots&b&a\end{pmatrix}\,=(a-b)^{n-1}(a+(n-1)b)$

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