`Delta= |{:(1,,1+ac,,1+bc),(1,,1+ad,,1+bd),(1,,1+ae,,1+be):}|` is independent of

Options

A. a
B. b
C. c,d,e
D. none of these

Explanation

To solve the given determinant problem step by step, we will analyze the determinant and simplify it. The determinant is given as:

\[ \Delta = \begin{vmatrix} 1 & 1 + ac & 1 + bc \\ 1 & 1 + ad & 1 + bd \\ 1 & 1 + ae & 1 + be \end{vmatrix} \]

### Step 1: Apply Column Operations
We will perform column operations to simplify the determinant. Specifically, we will subtract the first column from the second and third columns.

\[ \Delta = \begin{vmatrix} 1 & (1 + ac) - 1 & (1 + bc) - 1 \\ 1 & (1 + ad) - 1 & (1 + bd) - 1 \\ 1 & (1 + ae) - 1 & (1 + be) - 1 \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} 1 & ac & bc \\ 1 & ad & bd \\ 1 & ae & be \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Next, we can factor out common terms from the second and third columns. We will take 'a' out from the second column and 'b' out from the third column.

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & d & b \\ 1 & e & e \end{vmatrix} \cdot ab \]

### Step 3: Evaluate the Determinant
Now, we can evaluate the determinant:

\[ \Delta = ab \cdot \begin{vmatrix} 1 & 1 & 1 \\ 1 & d & b \\ 1 & e & e \end{vmatrix} \]

Notice that the first row is identical, which means the determinant will be zero:

\[ \Delta = 0 \]

### Conclusion
Thus, we conclude that the value of the determinant \(\Delta\) is independent of \(a\), \(b\), \(c\), \(d\), and \(e\). The determinant is zero.

### Final Answer
The determinant \(\Delta\) is independent of \(a\), \(b\), \(c\), \(d\), and \(e\).

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