If `A=[(a^(2),ab,ac),(ab,b^(2),bc),(ac,bc,c^(2))] "and B"= [(0,c,-b),(-c,0,a),(b,-a,0)]` then the product AB is equal to

Options

A. O
B. A
C. B
D. I

Answer

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

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

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 product is obtained by taking the dot product of the \( i^{th} \) row of \( A \) and the \( j^{th} \) column of \( B \).

### Step 1: Calculate \( (AB)_{11} \)

\[ (AB)_{11} = a^2 \cdot 0 + ab \cdot (-c) + ac \cdot b = 0 - abc + abc = 0 \]

### Step 2: Calculate \( (AB)_{12} \)

\[ (AB)_{12} = a^2 \cdot c + ab \cdot 0 + ac \cdot (-a) = a^2c + 0 - a^2c = 0 \]

### Step 3: Calculate \( (AB)_{13} \)

\[ (AB)_{13} = a^2 \cdot (-b) + ab \cdot a + ac \cdot 0 = -a^2b + a^2b + 0 = 0 \]

### Step 4: Calculate \( (AB)_{21} \)

\[ (AB)_{21} = ab \cdot 0 + b^2 \cdot (-c) + bc \cdot b = 0 - b^2c + b^2c = 0 \]

### Step 5: Calculate \( (AB)_{22} \)

\[ (AB)_{22} = ab \cdot c + b^2 \cdot 0 + bc \cdot (-a) = abc + 0 - abc = 0 \]

### Step 6: Calculate \( (AB)_{23} \)

\[ (AB)_{23} = ab \cdot (-b) + b^2 \cdot a + bc \cdot 0 = -ab^2 + ab^2 + 0 = 0 \]

### Step 7: Calculate \( (AB)_{31} \)

\[ (AB)_{31} = ac \cdot 0 + bc \cdot (-c) + c^2 \cdot b = 0 - bc^2 + bc^2 = 0 \]

### Step 8: Calculate \( (AB)_{32} \)

\[ (AB)_{32} = ac \cdot c + bc \cdot 0 + c^2 \cdot (-a) = ac^2 + 0 - ac^2 = 0 \]

### Step 9: Calculate \( (AB)_{33} \)

\[ (AB)_{33} = ac \cdot (-b) + bc \cdot a + c^2 \cdot 0 = -acb + acb + 0 = 0 \]

### Final Result

Combining all the calculated elements, we find that:

\[ AB = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]

This is a null matrix.