The value of ∣∣ ∣∣y+zxxyz+xyzzx+y∣∣ ∣∣ is equal to
4xyz
Given that,
Δ=∣∣ ∣∣y+zxxyz+xyzzx+y∣∣ ∣∣
Here you can either go for straight away evaluation of determinant along the first row or apply some properties and then evaluate. Since applying properties is going to make the final evaluations easier we will go with that method.
Δ=∣∣ ∣∣y+zxxyz+xyzzx+y∣∣ ∣∣ [R1→R1+R2+R3]
=2∣∣ ∣∣y+zx+zx+yyz+xyzzx+y∣∣ ∣∣ [R2→R2−R1][R3→R3−R1]
=2∣∣ ∣∣y+zx+zx+y−z0−x−y−x0∣∣ ∣∣ [R1→R1+R2+R3]
=2∣∣
∣∣0zy−z0−x−y−x0∣∣
∣∣
=2[−z(−xy)+y(xz)]
=4xyz