If A and B are matrices given below:
`A=[(0,c,-b),(-c,0,a),(b,-a,0)]` and `B=[(a^(2),ab,ac),(ab,b^(2),bc),(ac,bc,c^(2))]` then AB is a

Options

A. A
B. O
C. on odd multiple of `pi//2`
D. I

Answer

To find the product of matrices \( A \) and \( B \), we will perform matrix multiplication step by step.

Given matrices:
\[ A = \begin{pmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{pmatrix}, \quad B = \begin{pmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{pmatrix} \]

### Step 1: Matrix Multiplication Setup
The product \( AB \) will be calculated using the formula for matrix multiplication, where the element in the \( i^{th} \) row and \( j^{th} \) column of the resulting matrix is computed as the dot product of the \( i^{th} \) row of \( A \) and the \( j^{th} \) column of \( B \).

### Step 2: Calculate Each Element of the Resulting Matrix
1. **First Row, First Column**:
\[ (0)(a^2) + (c)(ab) + (-b)(ac) = 0 + abc - abc = 0 \]

2. **First Row, Second Column**:
\[ (0)(ab) + (c)(b^2) + (-b)(bc) = 0 + cb^2 - b^2c = 0 \]

3. **First Row, Third Column**:
\[ (0)(ac) + (c)(bc) + (-b)(c^2) = 0 + c^2b - bc^2 = 0 \]

4. **Second Row, First Column**:
\[ (-c)(a^2) + (0)(ab) + (a)(ac) = -ca^2 + 0 + a^2c = 0 \]

5. **Second Row, Second Column**:
\[ (-c)(ab) + (0)(b^2) + (a)(bc) = -cab + 0 + abc = 0 \]

6. **Second Row, Third Column**:
\[ (-c)(ac) + (0)(bc) + (a)(c^2) = -ac^2 + 0 + ac^2 = 0 \]

7. **Third Row, First Column**:
\[ (b)(a^2) + (-a)(ab) + (0)(ac) = ba^2 - a^2b + 0 = 0 \]

8. **Third Row, Second Column**:
\[ (b)(ab) + (-a)(b^2) + (0)(bc) = ab^2 - ab^2 + 0 = 0 \]

9. **Third Row, Third Column**:
\[ (b)(ac) + (-a)(bc) + (0)(c^2) = abc - abc + 0 = 0 \]

### Step 3: Form the Resulting Matrix
Combining all the computed elements, we find:
\[ AB = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

### Conclusion
The product \( AB \) is the zero matrix.