For what value of 'k' the system of equation `kx + 3y = 1 , 12x + ky = 2` has no solution.

Answer

To determine the value of 'k' for which the system of equations has no solution, we need to analyze the given equations:

1. **Equations:**
- \( kx + 3y = 1 \) (Equation 1)
- \( 12x + ky = 2 \) (Equation 2)

2. **Rearranging the equations:**
We can rewrite both equations in the standard form \( Ax + By + C = 0 \):
- For Equation 1:
\[ kx + 3y - 1 = 0 \]
- For Equation 2:
\[ 12x + ky - 2 = 0 \]

3. **Identifying coefficients:**
From the equations, we can identify:
- For Equation 1: \( A_1 = k, B_1 = 3, C_1 = -1 \)
- For Equation 2: \( A_2 = 12, B_2 = k, C_2 = -2 \)

4. **Condition for no solution:**
The system of equations will have no solution if:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \quad \text{and} \quad \frac{A_1}{A_2} \neq \frac{C_1}{C_2} \]
This means the lines represented by the equations are parallel.

5. **Setting up the ratios:**
We can set up the ratios:
\[ \frac{k}{12} = \frac{3}{k} \]

6. **Cross-multiplying:**
Cross-multiplying gives:
\[ k^2 = 36 \]

7. **Solving for k:**
Taking the square root of both sides, we find:
\[ k = 6 \quad \text{or} \quad k = -6 \]

8. **Checking the condition for no solution:**
We need to check which of these values leads to the condition \( \frac{C_1}{C_2} \):
- For \( k = 6 \):
\[ \frac{C_1}{C_2} = \frac{-1}{-2} = \frac{1}{2} \quad \text{(this is equal)} \]
- For \( k = -6 \):
\[ \frac{C_1}{C_2} = \frac{-1}{-2} = \frac{1}{2} \quad \text{(this is not equal)} \]

9. **Conclusion:**
Therefore, the value of \( k \) for which the system of equations has no solution is:
\[ k = -6 \]