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. -1
B. 0
C. 1
D. 2

Answer

To find \( A_0 \) from the determinant given by

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

we will follow these steps:

### Step 1: Rewrite the Determinant
We can rewrite the determinant by manipulating the columns. Let's denote the columns as \( C_1, C_2, C_3 \). We can express \( C_2 \) and \( C_3 \) in terms of \( C_1 \):

\[ C_2 = C_2 - C_1 \quad \text{and} \quad C_3 = C_3 - C_1 \]

This gives us:

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

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: Factor Out \( x \)
Now we can factor \( x \) 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: Evaluate the Determinant
Now, we need to evaluate the determinant. The determinant can be expanded, and we can find the constant term \( A_0 \) when \( x = 0 \):

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

### Step 4: Calculate \( A_0 \)
When \( x = 0 \), the determinant simplifies to:

\[ D = 0^2 \cdot \text{(some constant)} = 0 \]

Thus, \( A_0 \) is the value of the determinant when \( x = 0 \):

\[ A_0 = \begin{vmatrix} 1 & b - a & c - a \\ 1 & b_1 - a_1 & c_1 - a_1 \\ 1 & b_2 - a_2 & c_2 - a_2 \end{vmatrix} \]

### Final Result
After evaluating the determinant, we find that:

\[ A_0 = (b - a)(b_1 - a_1)(c_2 - a_2) + (c - a)(b_1 - a_1)(b_2 - a_2) + (b - a)(c_1 - a_1)(b_2 - a_2) - (c - a)(b_1 - a_1)(b_2 - a_2) - (b - a)(c_1 - a_1)(b_2 - a_2) - (b - a)(b_1 - a_1)(c_2 - a_2) \]