If `|{:(1+ax,1+bx,1+cx),(1+a_1x,1+b_1x,1+c_1x),(1+a_2x,1+b_2x,1+c_2x):}|=A_0+A_1x+A_2x^2+A ""_3x^3`, then `A_0` is

Options

A. (a) -1
B. (b) 0
C. (c) 1
D. (d) 2

Answer

To solve the given problem, we need to evaluate the determinant:

\[ D = \begin{vmatrix} 1 + ax & 1 + bx & 1 + cx \\ 1 + a_1x & 1 + b_1x & 1 + c_1x \\ 1 + a_2x & 1 + b_2x & 1 + c_2x \end{vmatrix} \]

We are tasked with finding the constant term \( A_0 \) in the polynomial expansion of this determinant, which is expressed as:

\[ D = A_0 + A_1x + A_2x^2 + A_3x^3 \]

### Step 1: Simplifying the Determinant

To simplify the determinant, we can perform column operations. We will subtract the first column from the second and third columns:

\[ D = \begin{vmatrix} 1 + ax & (1 + bx) - (1 + ax) & (1 + cx) - (1 + ax) \\ 1 + a_1x & (1 + b_1x) - (1 + a_1x) & (1 + c_1x) - (1 + a_1x) \\ 1 + a_2x & (1 + b_2x) - (1 + a_2x) & (1 + c_2x) - (1 + a_2x) \end{vmatrix} \]

This simplifies to:

\[ D = \begin{vmatrix} 1 + ax & (b - a)x & (c - a)x \\ 1 + a_1x & (b_1 - a_1)x & (c_1 - a_1)x \\ 1 + a_2x & (b_2 - a_2)x & (c_2 - a_2)x \end{vmatrix} \]

### Step 2: Factoring Out \( x \)

We can factor out \( x \) from the second and third columns:

\[ D = x^2 \begin{vmatrix} 1 + ax & b - a & c - a \\ 1 + a_1x & b_1 - a_1 & c_1 - a_1 \\ 1 + a_2x & b_2 - a_2 & c_2 - a_2 \end{vmatrix} \]

### Step 3: Evaluating the Determinant

Now, we need to evaluate the determinant:

\[ D = x^2 \begin{vmatrix} 1 + ax & b - a & c - a \\ 1 + a_1x & b_1 - a_1 & c_1 - a_1 \\ 1 + a_2x & b_2 - a_2 & c_2 - a_2 \end{vmatrix} \]

The first column contains terms that are linear in \( x \). Therefore, when we expand this determinant, all terms will be multiplied by \( x^2 \) or higher powers of \( x \).

### Step 4: Finding \( A_0 \)

Since \( A_0 \) is the constant term of the polynomial, and we see that every term in the expansion of the determinant contains \( x^2 \) or higher, it follows that there are no constant terms independent of \( x \).

Thus, we conclude that:

\[ A_0 = 0 \]

### Final Answer

The value of \( A_0 \) is:

\[ \boxed{0} \]