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  • + 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.)

  • + 0 comments
    print(-1)
    
  • + 0 comments

    -1 will be the answer

  • + 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.

    • + 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))
      
  • + 0 comments

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