Let and ( > ) be the distinct roots of the equation vmatrix x & 1 & 1 1 & x & 1 1 & 1 & x vmatrix = 0 . The number of values of k for which the system of linear equations x + y + z = 1 x + y + z = k x + y + z = has infinitely many solutions, is

Step-by-step solution

First, we solve the given determinant equation: vmatrix x & 1 & 1 1 & x & 1 1 & 1 & x vmatrix = 0 Applying C₁ C₁ + C₂ + C₃ : vmatrix x+2 & 1 & 1 x+2 & x & 1 x+2 & 1 & x vmatrix = 0 (x+2) vmatrix 1 & 1 & 1 1 & x & 1 1 & 1 & x vmatrix = 0 Applying R₂ R₂ - R₁ and R₃ R₃ - R₁ : (x+2) vmatrix 1 & 1 & 1 0 & x-1 & 0 0 & 0 & x-1 vmatrix = 0 (x+2)(x-1)^2 = 0 The distinct roots are 1 and -2 . Since > , we have = 1 and = -2 . Substituting these values into the given system of equations, we get: x + y + z = 1 x + y + z = k x +