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  2. Mathematics
  3. Linear Algebra Foundations
  4. Linear Algebra Foundations #8 - Systems of Equations
  5. Discussions

Linear Algebra Foundations #8 - Systems of Equations

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  • daniel_v_oconnor
     + 0 comments

    I did Gaussian elimination carefully and discovered that the coefficient matrix for the row reduced system of equations is invertible unless a = 2 or a = -1.

    Then I checked that there is in fact no solution when a = -1. (We know b is not 0, because otherwise the given system of equations would always have a solution consisting of all zeros.)

  • hoangziet22
     + 0 comments
    print(-1)

  • 12223348_Vikash
     + 0 comments

    -1 will be the answer

  • chisom213
     + 1 comment

    The determinant of the left hand side of the system reduces to D = 2a² - 2a - 4. For the system to have no solution, D = 0, solving the quadratic equation yields:

    a² - a - 2 = 0;

    (a + 1)(a - 2) = 0;

    a = -1, a = 2

    The least possible value of a is -1.

    • chisom213
       + 0 comments

      Python3:

      if __name__ == "__main__":
  # Compute the 3 x 3 determinant of the system
  # Determinant = 0, when system of equations has *NO SOLUTIONS*
  det = lambda a: a*(2*a - 1) - 1*(a - 2) + 2*(1 - 4)
  
  # Hold all possible alpha values
  lst = list()
  
  # Loop through several values that make det = 0
  for i in range(-100, 101):
    if det(i) == 0:
      lst += [i]
      
  # Return the minimum value
  print(min(lst))

  • armagedescu
     + 0 comments

    D = 2a² - 4 - 2a Regardless of b there is no solution for D=0 => a=[-1, 2]

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