If x,y,z are different from zero and Δ=ab−yc−za−xbc−za−xb−yc=0, then the value of the expression ax+by+cz is
Given that Δ=ab−yc−za−xbc−za−xb−yc=0 Applying R2→R2−R1,R3→R3−R1 , we get Δ=ab−yc−z−xy0−x0z=0
Expanding along C3, we get
(c−z)−xy−x0+zab−y−xy=0⇒(c−z)(xy)+z(ay+bx−xy)=0⇒cxy−xyz+ayz+bxz−xyz=0⇒ayz+bzx+cxy=2xyz⇒ax+by+cz=2