Dear Jack Son ,

There are at most 6 even entries such that det(A)=1.

If A3x3 with entries

(a,b,c

d,e,f,

g, h, i)

then it's determinant is det(A) = iae-ibd-cge-afh+bfg+cdh

There are at most 6 even entries such that det(A)=1.

Assume that we have seven even entries, then det(A) should be even

(see the equation of det(A) ).

The best example is the 3x3 identity it includes 6 even integers(zeros)

and 3 ones.

Best regards

Dear Dr. Kaddoura and Dr. Son,

Using the definition of a determinant of any square n x n matrix, Dr. Kadoura’s proof can be extended/generalized by the following statement.

“Let n be any positive integer. If A is a square n x n matrix such that det(A) = 1, then A contains at most n2 – n even entries. In particular, the unit matrix In of order n contains n2 – n even entries (equals to zero) and det(In) = 1.”

Nice observation Romeo Meštrović .

We can generalize your (generalization) by

“Let n be any positive integer. If A is a square n x n matrix with integer entries such that det(A) = r is an odd number, then A contains at most (n2 – n) even entries. In particular, the diagonal matrix with the product of its entries is r ”

Best wishes

Dear Dr. Kaddoura,

Of course! Moreover, it can be of interest to consider “the odd analogue(s)” of proposed question\problem motivated by the following fact.

“Let n be any positive integer. If A = [a_ij]n x n is a square n x n marrix whose all entries a_ij are +1 or – 1, then using standard elementary transformations (rules) over determinants, it can be proved that det(A) is divisible by 2n -1. ”

Accordingly, I propose the following “curious” question.

Question. If A = [aij]n x n is a square n x n matrix whose entries aij are +1 or –1. What (integer) values can be taken det(A)/ 2n -1?

Best regards

Dear Romeo Meštrović ,

I think there is a missing condition on the entries of your proposed matrix.

For example

The (one)- 4x4 matrix - Identity = J has determinant -3.

I mean

J = ( 0 1 1 1

1 0 1 1

1 1 0 1

1 1 1 0 )

has detJ = - 3 not divisible by 24-1 = 8 .

Best regards

Dear Dr. Kaddoura,

In my previous question, I wrote “…n x n matrix whose entries aij are +1 or – 1…”

For example, let A = [aij] be the square n x n matrix with entries aij = 1 if i ≤ j and

aij = - 1, otherwise. Then using standard elementary transformations over determinants, it can be proved that det(A) = 2n-1 (a familiar problem from my student’s days!) Hence, this matrix is a n x n matrix with minimal determinant (by absolute value) into a class of all n x n marrices whose entries are +1 or – 1.

Moreover, if A = [aij]n x n is any square n x n matrix whose all entries aij are +1 or – 1, then clearly, by definition of a determinant, det(A) < n! (n factorial).

Best regards

Dear Romeo Meštrović ,

"Or" means you can consider all are +1 , or all are -1 or a combination of both.

There were no conditions. So , to make the problem works, we need to set some conditions. For example:

The proposed matrix is as the following:

Mnxn=(aij ) , aij = +1 or -1

under the condition M = D + K

where D is diagonal ( identity or other form) and K is skew symmetric.

Or you may have more cases?

Best regards

Dear Dr.Kaddoura

Precisely, for any fixed order n, we consider determinants of square n x n matrices A = [a_ij] whose entries a_ij satisy the condition a_ij ϵ {-1, +1} for all i, j ϵ {1, 2,…,n}. For any such a matrix A, (a_ij + a_pq) ϵ {-2, 0, 2} and (a_ij - a_pq) ϵ {-2, 0, 2} for all i, j, p, q ϵ {1, 2,…,n}. Accordingly, in order to prove that det(A) is divisible by 2^(n - 1), using additions and/or subtractions of first row with other n – 1 rows of A, we obviously obtain the equivalent matrix B= [b_ij] such that b_11 = a_11, b_i1 = 0 for all i ϵ { 2,…,n} and b_ij ϵ {-2, 0, 2} for all i, j ϵ {2,…,n}. Then taking the common factor 2 from every i -th row of matrix B with i ϵ {2,…,n}, we obtain

det(A) = det(B) = a_11*2^(n-1) det(B’),

where B’ is the integer square (n -1) x (n -1) matrix (whose entries belong to the set {-1, 0, 1}). Hence, det(A) is divisible by 2^(n-1). Moreover, by definition of a determinant, we have | det(A)| < n! (n factorial).

In my previous answer, for any positive integer n, it was presented the matrix A = [a_ij] with entries a_ij = 1 if i ≤ j and a_ij = - 1, otherwise; then det(A) = 2^(n-1) (for example if n = 3, then det[{{1,1,1},{-1,1,1},{-1,-1,1}}] = 4).

Best regards

Dear Romeo Meštrović ,

All possible answers of your question

Question. If A = [aij]n x n is a square n x n matrix whose entries aij are +1 or –1. What (integer) values can be taken det(A)/ 2n-1?

are 0, 1, -1, 2 and -2.

Best regards

Dear Dr. Kaddoura,

It seems that your assertion is true for the cases n = 2, 3, 4. However, if n = 5, then for example,

det[{{-1,1,1,1,1},{1,-1,1,1,1},{1,1,-1,1,1},{1,1,1,-1,1},{1,1,1,1,-1}}] = 48 (= 24*3).

More generally, for any fixed positive integer n > 1, let A = [aij] be the square n x n matrix with entries aij = - 1 if i = j and aij = 1, otherwise. Then using elementary transformations over determinants, it can be proved that det(A) = (n-2) *(- 2)n-1, and hence, det(A)/(2)n-1 = (n-2)(- 1)n-1.

Best regards

Dear Romeo Meštrović ,

According to your last result,

det|[1]matrix - 2(Inxn)| = (n-2)(-1)n-1 2n-1

where [1] is the matrix where aij =1 ( for all i and j), and Inxn is the identity.

This result turns my guess 0, +1, -1, 2 and -2, in general, is not valid.

Also, different patterns may provide the same conclusion, as an example:

the matrix with rows:

[{1,1,1,1,1},{1,1,1,-1,-1},{1,1,-1,1,-1},{1,-1,1,1,-1},{1,1,1,1,-1}]

Indeed the problem deserves more investigation.

Best regards

Dear Romeo Meštrović ,

With more investigation ,

the value |det(A)/(2)n-1 |= (n-3) in some cases,

and then one need to extend the expected value of

det(A)/(2)n-1 into more general case.

Best regards

Dear Dr. Kaddoura,

I believe that your previous observation is true and it can be extended for some othet values of |det(A)|/2n-1 for all values of n, as it can be seen in my examples given below (obtained in Mathematica 9), where I changes only the first row of the matrices (of orders 5, 6 and 7):

For Dr. Kaddoura:

For Dr. Kaddoura:

Examples for n=5

Det[{{-1,1,1,1,1},{1,-1,1,1,1},{1,1,-1,1,1},{1,1,1,-1,1},{1,1,1,1,-1}}]/16

3

Det[{{-1,-1,1,1,1},{1,-1,1,1,1},{1,1,-1,1,1},{1,1,1,-1,1},{1,1,1,1,-1}}]/16

2

Det[{{-1,-1,-1,1,1},{1,-1,1,1,1},{1,1,-1,1,1},{1,1,1,-1,1},{1,1,1,1,-1}}]/16

1

Det[{{-1,-1,-1,-1,1},{1,-1,1,1,1},{1,1,-1,1,1},{1,1,1,-1,1},{1,1,1,1,-1}}]/16

0

Det[{{-1,-1,-1,-1,-1},{1,-1,1,1,1},{1,1,-1,1,1},{1,1,1,-1,1},{1,1,1,1,-1}}]/16

-1

Examples for n=6

Det[{{-1,1,1,1,1,1},{1,-1,1,1,1,1},{1,1,-1,1,1,1},{1,1,1,-1,1,1},{1,1,1,1,-1,1},{1,1,1,1,1,-1}}]/32

-4

Det[{{-1,-1,1,1,1,1},{1,-1,1,1,1,1},{1,1,-1,1,1,1},{1,1,1,-1,1,1},{1,1,1,1,-1,1},{1,1,1,1,1,-1}}]/32

-3

Det[{{-1,-1,-1,1,1,1},{1,-1,1,1,1,1},{1,1,-1,1,1,1},{1,1,1,-1,1,1},{1,1,1,1,-1,1},{1,1,1,1,1,-1}}]/32

-2

Det[{{-1,-1,-1,-1,1,1},{1,-1,1,1,1,1},{1,1,-1,1,1,1},{1,1,1,-1,1,1},{1,1,1,1,-1,1},{1,1,1,1,1,-1}}]/32

-1

Det[{{-1,-1,-1,-1,-1,1},{1,-1,1,1,1,1},{1,1,-1,1,1,1},{1,1,1,-1,1,1},{1,1,1,1,-1,1},{1,1,1,1,1,-1}}]/32

0

Det[{{-1,-1,-1,-1,-1,-1},{1,-1,1,1,1,1},{1,1,-1,1,1,1},{1,1,1,-1,1,1},{1,1,1,1,-1,1},{1,1,1,1,1,-1}}]/32

1

Examples for n=7

Det[{{-1,1,1,1,1,1,1},{1,-1,1,1,1,1,1},{1,1,-1,1,1,1,1},{1,1,1,-1,1,1,1},{1,1,1,1,-1,1,1},{1,1,1,1,1,-1,1},{1,1,1,1,1,1,-1}}]/64

5

Det[{{-1,-1,1,1,1,1,1},{1,-1,1,1,1,1,1},{1,1,-1,1,1,1,1},{1,1,1,-1,1,1,1},{1,1,1,1,-1,1,1},{1,1,1,1,1,-1,1},{1,1,1,1,1,1,-1}}]/64

4

Det[{{-1,-1,-1,1,1,1,1},{1,-1,1,1,1,1,1},{1,1,-1,1,1,1,1},{1,1,1,-1,1,1,1},{1,1,1,1,-1,1,1},{1,1,1,1,1,-1,1},{1,1,1,1,1,1,-1}}]/64

3

Det[{{-1,-1,-1,-1,1,1,1},{1,-1,1,1,1,1,1},{1,1,-1,1,1,1,1},{1,1,1,-1,1,1,1},{1,1,1,1,-1,1,1},{1,1,1,1,1,-1,1},{1,1,1,1,1,1,-1}}]/64

2

Det[{{-1,-1,-1,-1,-1,1,1},{1,-1,1,1,1,1,1},{1,1,-1,1,1,1,1},{1,1,1,-1,1,1,1},{1,1,1,1,-1,1,1},{1,1,1,1,1,-1,1},{1,1,1,1,1,1,-1}}]/64

1

Det[{{-1,-1,-1,-1,-1,-1,1},{1,-1,1,1,1,1,1},{1,1,-1,1,1,1,1},{1,1,1,-1,1,1,1},{1,1,1,1,-1,1,1},{1,1,1,1,1,-1,1},{1,1,1,1,1,1,-1}}]/64

0

Det[{{-1,-1,-1,-1,-1,-1,-1},{1,-1,1,1,1,1,1},{1,1,-1,1,1,1,1},{1,1,1,-1,1,1,1},{1,1,1,1,-1,1,1},{1,1,1,1,1,-1,1},{1,1,1,1,1,1,-1}}]/64

-1

The above examples suggest the fact that |det(A)|/2n-1attains all values n –k with k = 2, 3, …,n.

However, it seems that the determination of all possible values of |det(A)|/2n-1 is very difficult combinatorial problem.

Best regards