InterviewSolution
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InterviewSolution
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If `|(x, x^2, x^3 +1), (y, y^2, y^3+1), (z, z^2, z^3+1)|` = 0 and x ,y and z are not equal to any other, prove that, xyz = -1 |
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Answer» Let the given determinant be `Delta`. Then, `Delta = |{:(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)):}|` `=|{:(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)):}|` `=|{:(x, x^(2), 1),(y, y^(2), 1), (z, z^(2), 1):}| +(xyz)(-1)^(2)* |{:(x, x^(2),1 ),(y, y^(2), 1), (z, z^(2), 1):}| ["interchanging the columns of the 2nd det. twice"]` `=(1+xyz)|{:(x, x^(2), 1),(y, y^(2), 1), (z, z^(2), 1):}| ` `=(1+xyz)|{:(" "x, " "x^(2), 1),((y-x), (y^(2)-x^(2)), 0), ((z-x), (z^(2)-x^(2)), 0):}| [R_(2) to (R_(2)-R_(1)), R_(3) to (R_(3) -R_(1))]` `=(1+xyz)(y-x)(z-x)|{:(x, " "x^(2), 1),(1, y+z, 0), (1, z+x, 0):}|` `=(1+xyz)(y-x)(z-x)*1*|{:(1, y+x),(1, z+x):}| ["expanding by"C_(3)]` `=(1+xyz)(y-x)(z-x)(z-y).` `therefore Delta = 0 rArr (1+xyz)(y-x)(z-x)(z-y) =0` `rArr (1+xyz)=0 [because (y-x) ne 0, (z-x) ne 0, (z-y)ne 0]` `rArr xyz = -1` Hence, xyz =-1 |
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