If `A=[(0,c,-b),(-c,0,a),(b,-a,0)],B=[(a^(2),ab,ac),(ab,b^(2),bc),(ac,bc,c^(2))]` then AB=

Options

A. A
B. B
C. It is not necessary that either A=O or B=O
D. O

Answer

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

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

### Step 1: Calculate the first row of \( AB \)

To find the first row of the product \( AB \), we multiply the first row of \( A \) by each column of \( B \).

1. **First Column of \( B \)**:
\[ (0 \cdot a^2) + (c \cdot ab) + (-b \cdot ac) = 0 + abc - abc = 0 \]

2. **Second Column of \( B \)**:
\[ (0 \cdot ab) + (c \cdot b^2) + (-b \cdot bc) = 0 + cb^2 - b^2c = 0 \]

3. **Third Column of \( B \)**:
\[ (0 \cdot ac) + (c \cdot bc) + (-b \cdot c^2) = 0 + c^2b - bc^2 = 0 \]

Thus, the first row of \( AB \) is:
\[ \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} \]

### Step 2: Calculate the second row of \( AB \)

Now we calculate the second row of \( AB \) by multiplying the second row of \( A \) by each column of \( B \).

1. **First Column of \( B \)**:
\[ (-c \cdot a^2) + (0 \cdot ab) + (a \cdot ac) = -ca^2 + 0 + a^2c = 0 \]

2. **Second Column of \( B \)**:
\[ (-c \cdot ab) + (0 \cdot b^2) + (a \cdot bc) = -cab + 0 + abc = 0 \]

3. **Third Column of \( B \)**:
\[ (-c \cdot ac) + (0 \cdot bc) + (a \cdot c^2) = -c^2a + 0 + ac^2 = 0 \]

Thus, the second row of \( AB \) is:
\[ \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} \]

### Step 3: Calculate the third row of \( AB \)

Finally, we calculate the third row of \( AB \) by multiplying the third row of \( A \) by each column of \( B \).

1. **First Column of \( B \)**:
\[ (b \cdot a^2) + (-a \cdot ab) + (0 \cdot ac) = ba^2 - a^2b + 0 = 0 \]

2. **Second Column of \( B \)**:
\[ (b \cdot ab) + (-a \cdot b^2) + (0 \cdot bc) = ab^2 - ab^2 + 0 = 0 \]

3. **Third Column of \( B \)**:
\[ (b \cdot ac) + (-a \cdot bc) + (0 \cdot c^2) = abc - abc + 0 = 0 \]

Thus, the third row of \( AB \) is:
\[ \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} \]

### Final Result

Combining all the rows, we find that:
\[ AB = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

This is the zero matrix.