Answer to Question #223142 in Abstract Algebra for Peter

Question

Consider A, B, and C three points of the plan. Show that det(AB, AC) = det(BC, BA) = det (CA, CB)

a) Using a geometrical approach

b) Using the property of antisymmetry of the determinant

Answer 1

a)

The magnitude of the cross product can be interpreted as the positive area of the parallelogram having a and b as sides.

in our case:

$det(AB, AC) = det(BC, BA) = det (CA, CB)=2(area\ of\ \Delta ABC)$

b)

antisymmetry property of determinants states that interchanging two rows of a determinant changes the sign of the determinant

then:

$det(AB, AC)=det(AB, AB+BC)=det(AB, AB)+det(AB,BC)=$

$=det(AB,BC)$

$det(BC, BA)=-det(BC, AB)=det(AB,BC)$

$det (CA, CB)=-det (CA, BC)=-det (-(AB+BC), BC)=$

$=det (AB+BC, BC)=det(AB,BC)+det(BC,BC)=det(AB,BC)$

so, $det(AB, AC) = det(BC, BA) = det (CA, CB)$