For `x ne y ne z , |{:(1+x^(3),,x^(2),,1),(1+y^(3),,y^(2),,1),(1+z^(3),,z^(2),,1):}|=0 ` if xyz is

Options

A. 0
B. positive
C. negative
D. none of these

Answer

A

Explanation

We have, `|(x,x^(2),1 + x^(3)),(y,y^(2),1 + y^(3)),(z,z^(2),1 + z^(3))|`
`= |(x,x^(2),1),(y,y^(2),1),(z,z^(2),1)| + |(x,x^(2),x^(3)),(y,y^(2),y^(3)),(z,z^(2),z^(3))|" " [("Since each element of third"),("column is sum of two elements")]`
`= |(x,x^(2),1),(y,y^(2),1),(z,z^(2),1)| + xyz |(1,x,x^(2)),(1,y,y^(2)),(1,z,z^(2))| " " [("Taking x,y and z common"),(C_(1)"," C_(2) and C_(3) " in second det")]`
`= - |(x,1,x^(2)),(y,1,y^(2)),(z,1,z^(2))| + xyz |(1,x,x^(2)),(1,y,y^(2)),(1,z,z^(2))| " " [("Interchanging " C_(2)),(and C_(3) " in first det")]`
`= |(1,x,x^(2)),(1,y,y^(2)),(1,z,z^(2))| + xyz |(1,x,x^(2)),(1,y,y^(2)),(1,z,z^(2))| " " [("Interchanging" C_(1)),(and C_(2) " in first det")]`
`= |(1,x,x^(2)),(1,y,y^(2)),(1,z,z^(2))| (1 + xyz) = 0 " " [ :' xyz = -1]`