If `f(x) = |(x + lamda,x,x),(x,x + lamda,x),(x,x,x + lamda)|, " then " f(3x) - f(x) =`

Options

A. `3x lamda^(2)`
B. `6x lamda^(2)`
C. `x lamda^(2)`
D. none of these

Explanation

To solve the problem step by step, we will calculate \( f(x) \) and then find \( f(3x) - f(x) \).

### Step 1: Define the function
We start with the function defined as:
\[ f(x) = \begin{vmatrix} x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda \end{vmatrix} \]

### Step 2: Apply column operations
We can simplify the determinant by applying the column operation \( C_1 \to C_1 + C_2 + C_3 \):
\[ f(x) = \begin{vmatrix} (x + \lambda) + x + x & x & x \\ x + (x + \lambda) + x & x + \lambda & x \\ x + x + (x + \lambda) & x & x + \lambda \end{vmatrix} \]
This simplifies to:
\[ f(x) = \begin{vmatrix} 3x + \lambda & x & x \\ 3x + \lambda & x + \lambda & x \\ 3x + \lambda & x & x + \lambda \end{vmatrix} \]

### Step 3: Factor out common terms
We can factor out \( 3x + \lambda \) from the first column:
\[ f(x) = (3x + \lambda) \begin{vmatrix} 1 & x & x \\ 1 & x + \lambda & x \\ 1 & x & x + \lambda \end{vmatrix} \]

### Step 4: Row operations
Now, we perform row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\[ f(x) = (3x + \lambda) \begin{vmatrix} 1 & x & x \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{vmatrix} \]

### Step 5: Calculate the determinant
The determinant can now be calculated easily:
\[ f(x) = (3x + \lambda) \cdot \lambda^2 \cdot 1 = (3x + \lambda) \lambda^2 \]

### Step 6: Find \( f(3x) \)
Now we substitute \( x \) with \( 3x \):
\[ f(3x) = (3(3x) + \lambda) \lambda^2 = (9x + \lambda) \lambda^2 \]

### Step 7: Calculate \( f(3x) - f(x) \)
Now we find \( f(3x) - f(x) \):
\[ f(3x) - f(x) = (9x + \lambda) \lambda^2 - (3x + \lambda) \lambda^2 \]
Factoring out \( \lambda^2 \):
\[ = \lambda^2 \left( (9x + \lambda) - (3x + \lambda) \right) \]
This simplifies to:
\[ = \lambda^2 (9x - 3x) = \lambda^2 \cdot 6x \]

### Final Answer
Thus, we have:
\[ f(3x) - f(x) = 6\lambda^2 x \]