If `A=[{:(0," "c,-b),(-c," "0," "a),(b,-a," "0):}]" and "B=[{:(a^(2),ab,ac),(ab,b^(2),bc),(ac,bc,c^(2)):}]`, show that AB is a zero matrix.

Options

A. 0
B. 1
C. 2I
D. none of these

Answer

To show that the product of matrices \( A \) and \( B \) is a zero matrix, we will perform matrix multiplication step by step.

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

We need to compute \( AB \) and show that it results in a zero matrix.

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

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

2. **Second element (Row 1, Column 2)**:
\[ 0 \cdot ab + c \cdot b^2 + (-b) \cdot bc = 0 + cb^2 - b^2c = 0 \]

3. **Third element (Row 1, Column 3)**:
\[ 0 \cdot ac + c \cdot bc + (-b) \cdot c^2 = 0 + c^2b - bc^2 = 0 \]

Thus, the first row of \( AB \) is:
\[ (0, 0, 0) \]

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

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

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

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

Thus, the second row of \( AB \) is:
\[ (0, 0, 0) \]

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

1. **First element (Row 3, Column 1)**:
\[ b \cdot a^2 + (-a) \cdot ab + 0 \cdot ac = ba^2 - ab^2 + 0 = 0 \]

2. **Second element (Row 3, Column 2)**:
\[ b \cdot ab + (-a) \cdot b^2 + 0 \cdot bc = ab^2 - ab^2 + 0 = 0 \]

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

Thus, the third row of \( AB \) is:
\[ (0, 0, 0) \]

### Conclusion
Combining all rows, we find that:
\[ AB = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]
which is indeed a zero matrix.