Let `A(x)=[(0,x-2,x-3),(x+2,0,x-5),(x+3,x+5,0)]`,
then the matrix `A(0)(A(0))^(T)` is a

Options

A. null matrix
B. symmetric matrix
C. skew symmetric matrix
D. non singular matrix

Answer

To solve the problem, we need to find the matrix \( A(0) \) and then compute the product \( A(0) \cdot (A(0))^T \). Let's go through the steps systematically.

### Step 1: Evaluate \( A(0) \)

Given the matrix \( A(x) \):
\[ A(x) = \begin{pmatrix} 0 & x-2 & x-3 \\ x+2 & 0 & x-5 \\ x+3 & x+5 & 0 \end{pmatrix} \]

Substituting \( x = 0 \):
\[ A(0) = \begin{pmatrix} 0 & 0-2 & 0-3 \\ 0+2 & 0 & 0-5 \\ 0+3 & 0+5 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -2 & -3 \\ 2 & 0 & -5 \\ 3 & 5 & 0 \end{pmatrix} \]

### Step 2: Compute \( (A(0))^T \)

Now we need to find the transpose of \( A(0) \):
\[ (A(0))^T = \begin{pmatrix} 0 & 2 & 3 \\ -2 & 0 & 5 \\ -3 & -5 & 0 \end{pmatrix} \]

### Step 3: Compute \( A(0) \cdot (A(0))^T \)

Now, we will calculate the product \( A(0) \cdot (A(0))^T \):
\[ A(0) \cdot (A(0))^T = \begin{pmatrix} 0 & -2 & -3 \\ 2 & 0 & -5 \\ 3 & 5 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 2 & 3 \\ -2 & 0 & 5 \\ -3 & -5 & 0 \end{pmatrix} \]

Calculating each element of the resulting matrix:

- **Element (1,1)**:
\[ 0 \cdot 0 + (-2) \cdot (-2) + (-3) \cdot (-3) = 0 + 4 + 9 = 13 \]

- **Element (1,2)**:
\[ 0 \cdot 2 + (-2) \cdot 0 + (-3) \cdot 5 = 0 + 0 - 15 = -15 \]

- **Element (1,3)**:
\[ 0 \cdot 3 + (-2) \cdot 5 + (-3) \cdot 0 = 0 - 10 + 0 = -10 \]

- **Element (2,1)**:
\[ 2 \cdot 0 + 0 \cdot (-2) + (-5) \cdot (-3) = 0 + 0 + 15 = 15 \]

- **Element (2,2)**:
\[ 2 \cdot 2 + 0 \cdot 0 + (-5) \cdot 5 = 4 + 0 - 25 = -21 \]

- **Element (2,3)**:
\[ 2 \cdot 3 + 0 \cdot 5 + (-5) \cdot 0 = 6 + 0 + 0 = 6 \]

- **Element (3,1)**:
\[ 3 \cdot 0 + 5 \cdot (-2) + 0 \cdot (-3) = 0 - 10 + 0 = -10 \]

- **Element (3,2)**:
\[ 3 \cdot 2 + 5 \cdot 0 + 0 \cdot 5 = 6 + 0 + 0 = 6 \]

- **Element (3,3)**:
\[ 3 \cdot 3 + 5 \cdot 5 + 0 \cdot 0 = 9 + 25 + 0 = 34 \]

Putting it all together, we get:
\[ A(0) \cdot (A(0))^T = \begin{pmatrix} 13 & -15 & -10 \\ 15 & -21 & 6 \\ -10 & 6 & 34 \end{pmatrix} \]

### Final Result

Thus, the matrix \( A(0) \cdot (A(0))^T \) is:
\[ \begin{pmatrix} 13 & -15 & -10 \\ 15 & -21 & 6 \\ -10 & 6 & 34 \end{pmatrix} \]