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Tagged: Combinatorics, contest math, Determinant, discrete math, matrix

This topic contains 3 replies, has 2 voices, and was last updated by Keshav Sharma 1 week, 2 days ago.

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- November 6, 2019 at 10:32 am #41407
Suddenly this question came in my mind thinking one of my own grad problems

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You must be logged in to view attached files.November 6, 2019 at 9:10 pm #41439I don’t know whether I am correct or not. but this is what i have done

A(k) be the 3*3 matrix with property that sum of each row is k and determinant is k

now consider the elements of first row to be a11, a12, a13 that of second a21, a22,a23 and that of third a31, a32, a33, now we know that here a11+a12+a13=a21+a22+a23=a31+a32+a33. now consider the determinant of A(k)

take out k(assume it to be non-zero) from the first row now we have new matrix call it B with a11′ = a11/k and a12’=a12/k and a13’=a13/k

also now a11’+a12’+a13’=(a11+a12+a13)/k=1

and det(A(k))=k*det(B)

now consider the row reduction of B, first R2->R2-(k-1)

*R1 and second R3->R3-(k-1)*R1 call the corresponding matrix Cnow again rename the elements of this matrix C using the an additional dash(‘) with corresponding elemetns of A

now a21′ = a21 – (k-1)*a11′

similarly other

thus now we get, a21′ + a22′ + a23’ = a21+a22+a23 – (k-1)

*(a11’+a12’+a13′)=k-(k-1)*1=1now we want det(A(k))=k so we want our final matrix (corresponding to the determiant) to be 1 hence it is one of the possible A(1)

so it is enough to find all the matrix of type A(1)

also we have a Null matrix and other trivial solutions for when sum is zero and determiant is zero

November 7, 2019 at 12:38 pm #41488I don’t know whether I understand your method correctly or not! But please keep in mind that A is a matrix having natural no entries. So when you take out k the first row would be a rational number.

Now at last ”

now we want det(A(k))=k so we want our final matrix (corresponding to the determinant) to be 1 hence it is one of the possible A(1)” it is true.

“so it is enough to find all the matrix of type A(1)” why is it true? If you are free sometime we can have a discussion over skype. Let me know.

November 7, 2019 at 7:20 pm #41527I was wrong, I assumed that the enteries can be any real number.

Yes we can discuss over Skype. Whenever you have time sir.

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