For which value(s) of `lamda`, do the pair of linear equations `lamda x+ y= lamda^(2) and x+ lamda y= 1` have
no solution?

Answer

To determine the value(s) of \( \lambda \) for which the pair of linear equations

1. \( \lambda x + y = \lambda^2 \)
2. \( x + \lambda y = 1 \)

have no solution, we need to analyze the conditions under which a pair of linear equations does not have a solution. This occurs when the lines represented by the equations are parallel, which can be determined using the coefficients of the equations.

### Step-by-Step Solution:

1. **Identify the coefficients**:
- From the first equation \( \lambda x + y - \lambda^2 = 0 \), we have:
- \( a_1 = \lambda \)
- \( b_1 = 1 \)
- \( c_1 = -\lambda^2 \)
- From the second equation \( x + \lambda y - 1 = 0 \), we have:
- \( a_2 = 1 \)
- \( b_2 = \lambda \)
- \( c_2 = -1 \)

2. **Set up the condition for no solution**:
- For the equations to have no solution, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \]
- This translates to:
\[ \frac{\lambda}{1} = \frac{1}{\lambda} \quad \text{and} \quad \frac{\lambda}{1} \neq \frac{-\lambda^2}{-1} \]

3. **Solve the first equation**:
- From \( \frac{\lambda}{1} = \frac{1}{\lambda} \):
\[ \lambda^2 = 1 \]
- This gives us:
\[ \lambda = 1 \quad \text{or} \quad \lambda = -1 \]

4. **Solve the second condition**:
- From \( \frac{\lambda}{1} \neq \frac{\lambda^2}{1} \):
\[ \lambda \neq \lambda^2 \]
- Rearranging gives:
\[ \lambda^2 - \lambda \neq 0 \quad \Rightarrow \quad \lambda(\lambda - 1) \neq 0 \]
- This means \( \lambda \neq 0 \) and \( \lambda \neq 1 \).

5. **Combine the results**:
- From the first condition, we have \( \lambda = 1 \) or \( \lambda = -1 \).
- From the second condition, \( \lambda \) cannot be 1.
- Therefore, the only value that satisfies both conditions is:
\[ \lambda = -1 \]

### Conclusion:
The value of \( \lambda \) for which the pair of linear equations has no solution is \( \lambda = -1 \).