Strange determinant inequation

Question

A is an all-one, 3 by 3 matrix.

Prove det(C+ xA) det(C-xA) <=(less or equal) (det C)² , for any C 3 by 3 matrix and x any real number.

Answer 1

Without loss of generality, assume that C is a 3x3 matrix characterized by the elements C(i,j) for row i and column j.  The determinant of a 3x3 matrix can be manually calculated as:

 

det(C) = C(1,1)C(2,2)C(3,3) - C(1,1)C(2,3)C(3,2) - C(2,1)C(1,2)C(3,3) + C(2,1)C(1,3)C(3,2) + C(3,1)C(1,2)C(2,3) - C(3,1)C(1,3)C(2,2)

 

Using the same process and simplifying the results, the determinants of C+xA and C-xA can be calculated as follows:

 

det(C+xA) = det(C) + Dx

det(C-xA) = det(C) - Dx

 

where D = C(1,1)C(3,3)+C(1,1)C(2,2)-C(1,1)C(2,3)-C(1,1)C(3,2)+C(2,1)C(1,3)+C(2,1)C(3,2)-C(2,1)C(1,2)-C(2,1)C(3,3)+C(3,1)C(1,2)+C(3,1)C(2,3)-C(3,1)C(1,3)-C(3,1)C(2,2)+C(1,2)C(2,3)-C(1,2)C(3,3)-C(1,3)C(2,2)+C(1,3)C(3,2)+C(2,2)C(3,3)-C(2,3)C(3,2)

 

Then det(C+xA)det(C-xA) = (det(c)+Dx)*(det(c)-Dx) = det(C)² - (Dx)².  Since (Dx)² >= 0 for all real numbers x, then det(C+xA)det(C-xA) <= (det(C))² for any 3x3 matrix C and any real number x.