Let u = i + j + k , v = i + j + k , w = i + j + k , and b = i + j + k . If the vector equation x u + y v + z w = b has infinitely many solutions for the scalars x, y, z , then the sum of ( + 2 )^2 over all possible ordered pairs ( , ) is

Step-by-step solution

Equating the components of the vectors, we get the system of linear equations: x + y + z = 1 x + y + z = x + y + z = 1 For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero: = vmatrix 1 & 1 & 1 & & 1 & 1 & 1 vmatrix = 0 Expanding the determinant: 1( - 1) - 1(1 - ) + (1 - ^2) = 0 2( - 1) - ( - 1)( + 1) = 0 ( - 1)(2 - ^2 - ) = 0 -( - 1)^2( + 2) = 0 Thus, = 1 or = -2 . Case 1: = 1 The equations become identical: x + y + z = 1 x + y + z = x + y + z = 1 For infinitely m