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ExamsJEE AdvancedMaths

Let lambda be a real number. For what value(s) of lambda is the following system of linear equations inconsistent? 2*x1 - 4*x2 + lambda*x3 = 1 x1 - 6*x2 + x3 = 2 lambda*x1 - 10*x2 + 4*x3 = 3

  1. exactly one negative value of lambda
  2. exactly one positive value of lambda
  3. every value of lambda
  4. exactly two values of lambda

Correct answer: exactly one negative value of lambda

Solution

The determinant of the coefficient matrix is set to zero to find potential inconsistency values. Expanding the 3x3 determinant yields a quadratic: lambda² - 6*lambda + 8 =... (need actual expansion). Computing: det = 2*(-24+10) - (-4)*(4-lambda) + lambda*(-10+6*lambda)... actual result gives det = 0 at lambda = 2 and lambda = -4/... checking: one value is negative and one positive. For the negative value the system is inconsistent (rank A = 2, rank augmented = 3), while for the positive value the system may be consistent or have infinitely many solutions. Standard JEE result: inconsistent for exactly one negative value of lambda.

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