Tunction/ J(x,y) = (x+2y.x-y). Does there exist an affine function g: R² R², such that fog: R² R² satisfies (fog)(0,1)=
Posted: Thu Jul 07, 2022 8:29 pm
Tunction/ J(x,y) = (x+2y.x-y). Does there exist an affine function g: R² R², such that fog: R² R² satisfies (fog)(0,1)= (6,3), (fog)(-1,1)=(1,1) and (fog)(-1,0)=(1,-5)?