For which value(s) of `lamda`, do the pair of linear equations `lamda x+ y= lamda^(2) and x+ lamda y= 1` have
infinitely many solutions?

Answer

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

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

has infinitely many solutions, we will follow these steps:

### Step 1: Rewrite the equations in standard form
We can rewrite both equations in the standard form \( Ax + By + C = 0 \).

- For the first equation:
\[ \lambda x + y - \lambda^2 = 0 \]
Here, \( A_1 = \lambda \), \( B_1 = 1 \), and \( C_1 = -\lambda^2 \).

- For the second equation:
\[ x + \lambda y - 1 = 0 \]
Here, \( A_2 = 1 \), \( B_2 = \lambda \), and \( C_2 = -1 \).

### Step 2: Set up the condition for infinitely many solutions
For the system of equations to have infinitely many solutions, the ratios of the coefficients must be equal:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]

### Step 3: Write the ratios
From our equations, we have:
- \( \frac{A_1}{A_2} = \frac{\lambda}{1} = \lambda \)
- \( \frac{B_1}{B_2} = \frac{1}{\lambda} \)
- \( \frac{C_1}{C_2} = \frac{-\lambda^2}{-1} = \lambda^2 \)

Thus, we need to solve:
\[ \lambda = \frac{1}{\lambda} = \lambda^2 \]

### Step 4: Solve the first pair of ratios
From \( \lambda = \frac{1}{\lambda} \):
Cross-multiplying gives:
\[ \lambda^2 = 1 \]
This implies:
\[ \lambda = 1 \quad \text{or} \quad \lambda = -1 \]

### Step 5: Solve the second pair of ratios
From \( \frac{1}{\lambda} = \lambda^2 \):
Cross-multiplying gives:
\[ 1 = \lambda^3 \]
This implies:
\[ \lambda = 1 \]

### Step 6: Find the common values
The values obtained from the two conditions are:
- From the first condition: \( \lambda = 1 \) or \( \lambda = -1 \)
- From the second condition: \( \lambda = 1 \)

The common value is:
\[ \lambda = 1 \]

### Conclusion
Thus, the value of \( \lambda \) for which the pair of linear equations has infinitely many solutions is:
\[ \boxed{1} \]