The value of the determinant
`Delta=|(2a_1b_1,a_1b_2+a_2b_1,a_1b_3+a_3b_1),(a_1b_2+a_2b_1,2a_2b_2,a_2b_3+a_3b_2),(a_1b_3+a_3b_1,a_3b_2+a_2b_3,2a_3b_3)|` is

Options

A. 1
B. `-1`
C. 0
D. `a_1a_2a_3b_1b_2b_3`

Answer

To find the value of the determinant

\[ \Delta = \begin{vmatrix} 2a_1b_1 & a_1b_2 + a_2b_1 & a_1b_3 + a_3b_1 \\ a_1b_2 + a_2b_1 & 2a_2b_2 & a_2b_3 + a_3b_2 \\ a_1b_3 + a_3b_1 & a_3b_2 + a_2b_3 & 2a_3b_3 \end{vmatrix} \]

we can simplify the determinant step by step.

### Step 1: Rewrite the determinant
We can express the first row as:
- \(2a_1b_1 = a_1b_1 + a_1b_1\)
- \(a_1b_2 + a_2b_1\) remains unchanged.
- \(a_1b_3 + a_3b_1 = a_1b_3 + a_3b_1\) remains unchanged.

Thus, we rewrite the first row:
\[ \Delta = \begin{vmatrix} a_1b_1 & a_1b_1 & a_1b_1 \\ a_1b_2 + a_2b_1 & 2a_2b_2 & a_2b_3 + a_3b_2 \\ a_1b_3 + a_3b_1 & a_3b_2 + a_2b_3 & 2a_3b_3 \end{vmatrix} \]

### Step 2: Factor out common terms
Notice that the first column can be factored out:
\[ \Delta = a_1 \begin{vmatrix} b_1 & b_1 & b_1 \\ a_1b_2 + a_2b_1 & 2a_2b_2 & a_2b_3 + a_3b_2 \\ a_1b_3 + a_3b_1 & a_3b_2 + a_2b_3 & 2a_3b_3 \end{vmatrix} \]

### Step 3: Check for zero rows or columns
Now, we can see that if we look at the structure of the determinant, if we factor out \(b_1\) from the first column, we will have:
\[ \Delta = b_1 \begin{vmatrix} 1 & 1 & 1 \\ a_1b_2 + a_2b_1 & 2a_2b_2 & a_2b_3 + a_3b_2 \\ a_1b_3 + a_3b_1 & a_3b_2 + a_2b_3 & 2a_3b_3 \end{vmatrix} \]

### Step 4: Identify zero column
Now, observe that if we look at the third column, we can see that it can be expressed as a linear combination of the first two columns. This means that the determinant will collapse to zero.

### Conclusion
Since one of the columns is a linear combination of the others, the determinant evaluates to:
\[ \Delta = 0 \]

### Final Answer
Thus, the value of the determinant \(\Delta\) is \(0\).

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