The system of linear equations x + y + z = 1 2x + 3y + ( )z = 2 3x + 4y + ( (2 ) + 2 ( ) + 1)z = 4 has no solution. If [0, 2 ] and the sum of all possible values of is k 2 , then the value of k is

Step-by-step solution

For the system to have no solution, = 0 and at least one of _x, _y, _z 0 . = vmatrix 1 & 1 & 1 2 & 3 & ( ) 3 & 4 & (2 ) + 2 ( ) + 1 vmatrix Applying R₂ R₂ - 2R₁ and R₃ R₃ - 3R₁ : = vmatrix 1 & 1 & 1 0 & 1 & ( ) - 2 0 & 1 & (2 ) + 2 ( ) - 2 vmatrix Expanding along the first column: = 1 ( (2 ) + 2 ( ) - 2 - ( ( ) - 2)) = (2 ) + ( ) Setting = 0 : (2 ) + ( ) = 0 1 - 2 ^2( ) + ( ) = 0 2 ^2( ) - ( ) - 1 = 0 (2 ( ) + 1)( ( ) - 1) = 0 ( ) = - 1 2 or ( ) = 1 For [0, 2 ] : If ( ) = - 1 2 , = 7 6 , 11 6 If ( ) = 1 , = 2 Sum o