Suppose `a_(1)+a_(2)+a_(3)+1.3,b_(1)+b_(2)+b_(3)=2.1,c_(1)+c_(2)+c_(3)=1.6` and
`Delta(x)=|(1+a_(1)x,1+b_(1)x,1+c_(1)x),(1+a_(2)x,1+b_(2)x,1+c_(3)x),(1+a_(3)x,1+b_(3)x,1+c4x)|`
`=A_(0)+A_(1)x+A_(2)x^(2)+A_(2)x^(3)`,
then `A_(1)=`_________

Answer

To solve the problem, we need to find the coefficient \( A_1 \) in the expression for \( \Delta(x) \). Let's break down the steps systematically.

### Step 1: Understand the given equations
We have the following equations:
1. \( a_1 + a_2 + a_3 = 1.3 \)
2. \( b_1 + b_2 + b_3 = 2.1 \)
3. \( c_1 + c_2 + c_3 = 1.6 \)

### Step 2: Write the determinant
The determinant \( \Delta(x) \) is given by:
\[ \Delta(x) = \begin{vmatrix} 1 + a_1 x & 1 + b_1 x & 1 + c_1 x \\ 1 + a_2 x & 1 + b_2 x & 1 + c_2 x \\ 1 + a_3 x & 1 + b_3 x & 1 + c_3 x \end{vmatrix} \]

### Step 3: Expand the determinant
To find \( A_1 \), we need to compute the determinant and identify the coefficient of \( x \). We can expand the determinant using the properties of determinants.

### Step 4: Calculate the determinant
Using the determinant properties:
\[ \Delta(x) = \begin{vmatrix} 1 + a_1 x & 1 + b_1 x & 1 + c_1 x \\ 1 + a_2 x & 1 + b_2 x & 1 + c_2 x \\ 1 + a_3 x & 1 + b_3 x & 1 + c_3 x \end{vmatrix} \]

This can be simplified by factoring out the \( x \) terms:
\[ = \begin{vmatrix} 1 & 1 & 1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + x \cdot \begin{vmatrix} 0 & 1 & 1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \]

### Step 5: Identify the coefficient of \( x \)
The coefficient of \( x \) in the expansion will be:
\[ A_1 = b_2 + c_3 - b_3 - c_2 + a_2 + c_3 - a_3 - b_2 + a_2 + b_3 - a_3 - b_2 \]

### Step 6: Substitute the values
Now we substitute the values we have:
1. \( a_1 + a_2 + a_3 = 1.3 \)
2. \( b_1 + b_2 + b_3 = 2.1 \)
3. \( c_1 + c_2 + c_3 = 1.6 \)

### Step 7: Calculate \( A_1 \)
After substituting and simplifying, we find:
\[ A_1 = (b_2 + c_3 - b_3 - c_2) + (a_2 + c_3 - a_3 - b_2) + (a_2 + b_3 - a_3 - b_2) \]

### Final Result
After simplification, we find that \( A_1 \) is equal to:
\[ A_1 = 0 \]