Find the value of k in which the system of equations `kx + 2y = 5 and 3x + y = 1` has no solution ?

Options

A. 0
B. 3
C. 6
D. 15

Explanation

To find the value of \( k \) for which the system of equations

\[ kx + 2y = 5 \]
\[ 3x + y = 1 \]

has no solution, we will use the concept of determinants from linear algebra.

### Step 1: Write the system of equations in matrix form

The given equations can be represented in the form of a matrix equation \( AX = B \), where:

\[ A = \begin{pmatrix} k & 2 \\ 3 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 5 \\ 1 \end{pmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix \( A \)

The determinant of matrix \( A \) is calculated as follows:

\[ \text{det}(A) = k \cdot 1 - 2 \cdot 3 = k - 6 \]

### Step 3: Set the determinant equal to zero for no solution

For the system of equations to have no solution, the determinant must be equal to zero:

\[ k - 6 = 0 \]

### Step 4: Solve for \( k \)

Now, we solve the equation:

\[ k - 6 = 0 \implies k = 6 \]

### Conclusion

Thus, the value of \( k \) for which the system of equations has no solution is

\[ \boxed{6} \]