The value of k, for which the pair of linear equations `k x + y = k^(2)` and `x + k y = 1` has infinitely many solution, is :

Options

A. `pm 1`
B. `-2`
C. `-1`
D. 2

Answer

To find the value of \( k \) for which the pair of linear equations \( kx + y = k^2 \) and \( x + ky = 1 \) has infinitely many solutions, we can follow these steps:

### Step 1: Write the equations in standard form
We start with the given equations:
1. \( kx + y = k^2 \) can be rewritten as \( kx + y - k^2 = 0 \).
2. \( x + ky = 1 \) can be rewritten as \( x + ky - 1 = 0 \).

### Step 2: Identify coefficients
From the rewritten equations, we identify the coefficients:
- For the first equation \( a_1 = k, b_1 = 1, c_1 = -k^2 \).
- For the second equation \( a_2 = 1, b_2 = k, c_2 = -1 \).

### Step 3: Set up the condition for infinitely many solutions
For the system of equations to have infinitely many solutions, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
Substituting the coefficients we found:
\[ \frac{k}{1} = \frac{1}{k} = \frac{-k^2}{-1} \]

### Step 4: Solve the equations
From the first two ratios:
\[ \frac{k}{1} = \frac{1}{k} \]
Cross-multiplying gives:
\[ k^2 = 1 \]
Thus, we have:
\[ k = 1 \quad \text{or} \quad k = -1 \]

Now, from the second ratio:
\[ \frac{1}{k} = \frac{k^2}{1} \]
Cross-multiplying gives:
\[ 1 = k^3 \]
Thus:
\[ k = 1 \]

### Step 5: Combine results
The values of \( k \) from both conditions are \( k = 1 \) and \( k = -1 \). Therefore, the values of \( k \) for which the pair of linear equations has infinitely many solutions are:
\[ k = 1 \quad \text{and} \quad k = -1 \]

### Final Answer
The value of \( k \) for which the pair of linear equations has infinitely many solutions is \( k = 1 \) or \( k = -1 \).

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