If `Delta=|(a_1x+b_1y,a_2x+b_2y,a_3x+b_3y),(b_1x+a_1y,b_2x+a_2y,b_3x+a_3y),(b_1x+a_1,b_2x+a_2,b_3x+a_3)|` then `Delta` is equal to

Options

A. `a_1a_2a_3 x^(2) +b_1b_2b_3y^2`
B. `x^2+y^2`
C. 0
D. None

Answer

To solve the determinant given by

\[ \Delta = \begin{vmatrix} a_1x + b_1y & a_2x + b_2y & a_3x + b_3y \\ b_1x + a_1y & b_2x + a_2y & b_3x + a_3y \\ b_1x + a_1 & b_2x + a_2 & b_3x + a_3 \end{vmatrix} \]

we will follow these steps:

### Step 1: Rewrite the determinant
We can express the determinant in a clearer format:

\[ \Delta = \begin{vmatrix} a_1x + b_1y & a_2x + b_2y & a_3x + b_3y \\ b_1x + a_1y & b_2x + a_2y & b_3x + a_3y \\ b_1x + a_1 & b_2x + a_2 & b_3x + a_3 \end{vmatrix} \]

### Step 2: Factor out common terms
Notice that we can factor out \(x\) from the first row and \(y\) from the second row:

\[ \Delta = x \cdot y^2 \cdot \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]

### Step 3: Analyze the determinant
Now, we observe that the second and third rows of the determinant are identical:

\[ \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ b_1 & b_2 & b_3 \end{vmatrix} \]

Since two rows of a determinant are identical, the value of this determinant is zero:

\[ \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = 0 \]

### Step 4: Conclude the value of Delta
Thus, we have:

\[ \Delta = x \cdot y^2 \cdot 0 = 0 \]

Therefore, the value of \(\Delta\) is

\[ \Delta = 0 \]

### Summary of Steps:
1. Rewrite the determinant in a clearer format.
2. Factor out common terms from the rows.
3. Analyze the determinant for identical rows.
4. Conclude that the value of the determinant is zero.