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
0
-1
1
2
Applying , R¯2→R¯2-R¯1,R¯3→R¯3-R¯1 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