Prove that:
∣∣
∣
∣∣y2z2yzy+zz2x2zxz+xx2y2xyx+y∣∣
∣
∣∣=0
We have to prove,
∣∣
∣
∣∣y2z2yzy+zz2x2zxz+xx2y2xyx+y∣∣
∣
∣∣=0
∴ LHS =∣∣
∣
∣∣y2z2yzy+zz2x2zxz+xx2y2xyx+y∣∣
∣
∣∣=1xyz∣∣
∣
∣∣xy2z2xyzxy+xzx2yz2xyzyz+xyx2y2zxyzxz+yz∣∣
∣
∣∣
[∵R1→xR1,R2→yR2,R3→zR3]
=1xyz(xyz)2∣∣
∣∣yz1xy+xzxz1yz+xyxy1xz+yz∣∣
∣∣
[taking (xyz) common from C1 and C2]
=xyz∣∣
∣∣yz1xy+yz+zxxz1xy+yz+zxxy1xy+yz+zx∣∣
∣∣[C3→C3+C1]
=xyz(xy+yz+zx)∣∣
∣∣yz11xz11xy11∣∣
∣∣
[taking (xy+yz+zx) common from C3]
=0 [since, C2 and C3 are equal]
=RHS
Hence proved.