If A = `[{:((b+c)^(2),a^(2), a^(2)),(b^(2) , (c+ a)^(2), b^(2)),(c^(2),c^(2) ,(a + b)^(2)):}|` , then det A is

Options

A. 2`(a + b + c)^(2)`
B. 2abc `(a + b + c)^(2)`
C. 2 abc `(a + b + c)^(3)`
D. abc `(a + b + c)^(2) `

Answer

To find the determinant of the matrix \( A \) given by:

\[ A = \begin{pmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{pmatrix} \]

we will follow these steps:

### Step 1: Apply Column Operations
We will perform column operations to simplify the determinant. Specifically, we will replace the second column \( C_2 \) with \( C_2 - C_1 \) and the third column \( C_3 \) with \( C_3 - C_1 \).

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

This gives us the new matrix:

\[ A' = \begin{pmatrix} (b+c)^2 & a^2 - (b+c)^2 & a^2 - (b+c)^2 \\ b^2 & (c+a)^2 - b^2 & b^2 - b^2 \\ c^2 & c^2 - c^2 & (a+b)^2 - c^2 \end{pmatrix} \]

### Step 2: Simplify the Entries
Now we simplify the entries in the new matrix:

1. For \( a^2 - (b+c)^2 = a^2 - (b^2 + 2bc + c^2) = a^2 - b^2 - 2bc - c^2 \)
2. For \( (c+a)^2 - b^2 = c^2 + 2ac + a^2 - b^2 \)
3. The third column becomes \( (a+b)^2 - c^2 = a^2 + 2ab + b^2 - c^2 \)

Thus, we have:

\[ A' = \begin{pmatrix} (b+c)^2 & a^2 - b^2 - 2bc - c^2 & a^2 - b^2 - 2bc - c^2 \\ b^2 & c^2 + 2ac + a^2 - b^2 & 0 \\ c^2 & 0 & a^2 + 2ab + b^2 - c^2 \end{pmatrix} \]

### Step 3: Factor Out Common Terms
Notice that the second and third columns are now similar. We can factor out common terms from the determinant.

### Step 4: Calculate the Determinant
Now, we can calculate the determinant using the properties of determinants. Since we have two identical columns, the determinant will be zero.

Thus, we have:

\[ \text{det}(A) = 0 \]

### Conclusion
The determinant of the matrix \( A \) is:

\[ \text{det}(A) = 0 \]