The value of the determinant `Delta = |(1 + a_(1) b_(1),1 + a_(1) b_(2),1 + a_(1) b_(3)),(1 + a_(2) b_(1),1 + a_(2) b_(2),1 + a_(2) b_(3)),(1 + a_(3) b_(1) ,1 + a_(3) b_(2),1 + a_(3) b_(3))|`, is

Options

A. `a_(1) a_(2) a_(3) + b_(1) b_(2) b_(3)`
B. `(a_(1) a_(2) a_(3)) (b_(1) b_(2) b_(3))`
C. `a_(1) a_(2) b_(1) b_(2) + a_(2) a_(3) b_(2) b_(3) + a_(3) a_(1) b_(3) b_(1)`
D. none of these

Explanation

To find the value of the determinant

\[ \Delta = \begin{vmatrix} 1 + a_1 b_1 & 1 + a_1 b_2 & 1 + a_1 b_3 \\ 1 + a_2 b_1 & 1 + a_2 b_2 & 1 + a_2 b_3 \\ 1 + a_3 b_1 & 1 + a_3 b_2 & 1 + a_3 b_3 \end{vmatrix} \]

we can simplify it step by step.

### Step 1: Rewrite the determinant
We can express the determinant as the sum of two parts:
\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} + \begin{vmatrix} a_1 b_1 & a_1 b_2 & a_1 b_3 \\ a_2 b_1 & a_2 b_2 & a_2 b_3 \\ a_3 b_1 & a_3 b_2 & a_3 b_3 \end{vmatrix} \]

### Step 2: Evaluate the first determinant
The first determinant is:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \]
This determinant is zero because all rows are identical.

### Step 3: Evaluate the second determinant
Now, we need to evaluate the second determinant:
\[ \begin{vmatrix} a_1 b_1 & a_1 b_2 & a_1 b_3 \\ a_2 b_1 & a_2 b_2 & a_2 b_3 \\ a_3 b_1 & a_3 b_2 & a_3 b_3 \end{vmatrix} \]
We can factor out \(a_1\), \(a_2\), and \(a_3\) from the rows and \(b_1\), \(b_2\), and \(b_3\) from the columns:
\[ = a_1 a_2 a_3 \begin{vmatrix} b_1 & b_2 & b_3 \\ b_1 & b_2 & b_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]
Again, this determinant is zero because all rows are identical.

### Step 4: Combine results
Thus, we have:
\[ \Delta = 0 + 0 = 0 \]

### Final Result
The value of the determinant \(\Delta\) is:
\[ \Delta = 0 \]