[Keshav Sharma]

I was wrong, I assumed that the enteries can be any real number.

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

[Keshav Sharma]

I 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 C

now 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=1

now 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

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