If the system of linear equations x + 4y + (k+4)z = 0 x + y + (k+1)z = 0 (k-2)x + (3-2k)y = 0 has a non-trivial solution, then the sum of all possible values of k is

Step-by-step solution

For a homogeneous system of linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. vmatrix 1 & 4 & k+4 1 & 1 & k+1 k-2 & 3-2k & 0 vmatrix = 0 Applying the row operation R₁ R₁ - R₂ : vmatrix 0 & 3 & 3 1 & 1 & k+1 k-2 & 3-2k & 0 vmatrix = 0 Applying the column operation C₂ C₂ - C₃ : vmatrix 0 & 0 & 3 1 & -k & k+1 k-2 & 3-2k & 0 vmatrix = 0 Expanding the determinant along the first row: 3 [1(3 - 2k) - (-k)(k - 2)] = 0 3 - 2k + k^2 - 2k = 0 k^2 - 4k + 3 = 0 The roots of