If `|(1 +ax,1 +bx,1 + bx),(1 +a_(1) x,1 +b_(1) x,1 + c_(1) x),(1 + a_(2) x,1 + b_(2) x,1 + c_(2) x)| = A_(0) + A_(1) x + A_(2) x^(2) + A_(3) x^(3)`, then `A_(1)` is equal to

Options

A. abc
B. 0
C. 1
D. none of these

Answer

To solve the problem, we need to evaluate the determinant and find the coefficient \( A_1 \) of \( x \) in the expression given. The determinant is:

\[ D = \begin{vmatrix} 1 + ax & 1 + bx & 1 + bx \\ 1 + a_1 x & 1 + b_1 x & 1 + c_1 x \\ 1 + a_2 x & 1 + b_2 x & 1 + c_2 x \end{vmatrix} \]

### Step 1: Rewrite the determinant
We can rewrite the determinant by factoring out the common terms. Notice that each entry in the determinant can be expressed as \( 1 + kx \) for some constant \( k \).

### Step 2: Factor out \( x \)
Since \( x \) is common in each row, we can factor \( x \) out of each row:

\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} + x \cdot \begin{vmatrix} a & b & b \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} \]

### Step 3: Evaluate the constant determinant
The first determinant is a matrix of all ones, which equals zero:

\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0 \]

### Step 4: Focus on the second determinant
Now we need to evaluate the second determinant:

\[ D_1 = \begin{vmatrix} a & b & b \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} \]

### Step 5: Expand the determinant
Using the determinant expansion method (cofactor expansion), we can compute \( D_1 \).

### Step 6: Identify the coefficient of \( x \)
The expression for the determinant \( D \) can now be simplified to:

\[ D = x \cdot D_1 \]

Since we are looking for the coefficient \( A_1 \), we see that \( A_1 \) corresponds to the coefficient of \( x \) in \( D \).

### Conclusion
From the determinant, since the first part contributes zero and the second part contributes \( x \cdot D_1 \), we can conclude that:

\[ A_1 = 0 \]

Thus, the value of \( A_1 \) is:

\[ \boxed{0} \]