InterviewSolution
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InterviewSolution
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The system of equations `alphax+y+z=alpha-1, x+alphay+z=alpha-1, x+y+alphaz=alpha-1` has no solution if alpha is (A) 1 (B) not -2 (C) either -2 or 1 (D) -2A. 1B. not `-2`C. either `-2 or 1`D. `-2` |
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Answer» Correct Answer - D For `alpha = 1`, the system reduces to a homogeneous system which is always consistent. So, `alpha != 1`. For `alpha != 1`, we have `D = |(alpha,1,1),(1,alpha,1),(1,1,alpha)| = |(alpha + 2 ,alpha + 2,alpha + 2),(1,alpha,1),(1,1,alpha)| ["Applying" R_(1) rarr R_(1) + R_(2) + R_(3)]` `rArr D = (alpha + 2) |(1,1,1),(1,alpha,1),(1,1,alpha)| = (alpha + 2) |(1,0,0),(1,alpha -1,0),(1,0,alpha -1)| ["Applying " C_(2) rarr C_(2) - C_(1), C_(3) rarr C_(3) - C_(1)]` `rArr D = (alpha +2) (alpha -1)^(2)` and `D_(1) = |(alpha - 1,1,1),(alpha -1,alpha,1),(alpha -1,1,alpha)| = (alpha -1) |(1,1,1),(1,alpha,1),(1,1,alpha)|` `rArr D_(1) = (alpha -1) |(1,0,0),(1,alpha -1,0),(1,0,alpha -1)| [("Applying " C_(2) rarr C_(2) - C_(1)),(C_(3) rarr C_(3) - C_(1))]` `rArr D_(1) = (alpha -1)^(3)` Clearly, `D = 0 " for " alpha = -2 " but " D_(1) != 0` So, the system is inconsistent for `alpha = -2` |
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