Find the value of k for which the simultaneous equation x + y + z = 3; x + 2y + 3z = 4 and x + 4y + kz = 6 will not have a unique solution.
A.0
B.5
C.6
D.7

Hint: We need to have a basic idea of solving the system of equations in three variables to solve this problem. Use the determinant of a matrix to solve this problem.

The given equations are
x + y + z = 3
x + 2y + 3z = 4
x + 4y + kz = 6
we can represent the given system of equations in matrix form using the coefficients of the variables.
$ \Rightarrow \left| {\matrix
   1 & 1 & 1 \\
   1 & 2 & 3 \\
   1 & 4 & k \\

 \endmatrix } \right|$
The given system of equations will be consistent with unique solution, when
$\left| {\matrix
   1 & 1 & 1 \\
   1 & 2 & 3 \\
   1 & 4 & k \\

 \endmatrix } \right| \ne 0$
Finding the determinant of the above matrix, we get
$ \Rightarrow 1(2k - 12) + 1(3 - k) + 1(4 - 2) \ne 0$
On simplification,
$ \Rightarrow k - 12 + 3 + 2 \ne 0$
$ \Rightarrow k - 7 \ne 0$
$ \Rightarrow k \ne 7$
For k = 7, the given simultaneous equations will not have a unique solution. Hence option D is the correct answer.

Note:
To solve a system of equations we have different methods available: substitution method, graph method, elimination method. The system of equations in three variables are either dependent, independent or inconsistent. Dependent systems of equations have an infinite number of solutions. Independent system of equations has only one solution. Inconsistent systems of equations have no solution. If the determinant of a matrix is zero it represents a linearly dependent system.