If `A=[(0,x),(y,0)]` and `A^(3)+A=O` then sum of possible values of xy is

Options

A. 0
B. `-1`
C. 1
D. 2

Explanation

To solve the problem, we need to analyze the matrix \( A = \begin{pmatrix} 0 & x \\ y & 0 \end{pmatrix} \) and the equation \( A^3 + A = O \), where \( O \) is the zero matrix.

### Step-by-Step Solution:

1. **Rewrite the Equation**:
We start with the equation:
\[ A^3 + A = O \]
This can be factored as:
\[ A(A^2 + I) = O \]
where \( I \) is the identity matrix.

**Hint**: Remember that if a product of matrices equals the zero matrix, at least one of the matrices must be singular (have a determinant of zero).

2. **Determinant Condition**:
From the factored equation, we have:
\[ \text{det}(A) \cdot \text{det}(A^2 + I) = 0 \]
This gives us two cases to consider:
- Case 1: \( \text{det}(A) = 0 \)
- Case 2: \( \text{det}(A^2 + I) = 0 \)

**Hint**: Calculate the determinant of matrix \( A \).

3. **Case 1: Determinant of A**:
The determinant of \( A \) is calculated as:
\[ \text{det}(A) = 0 \cdot 0 - x \cdot y = -xy \]
Setting this equal to zero gives:
\[ -xy = 0 \implies xy = 0 \]
This means either \( x = 0 \) or \( y = 0 \).

**Hint**: Think about the implications of \( xy = 0 \) on the values of \( x \) and \( y \).

4. **Case 2: Determinant of \( A^2 + I \)**:
First, we need to compute \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 0 & x \\ y & 0 \end{pmatrix} \begin{pmatrix} 0 & x \\ y & 0 \end{pmatrix} = \begin{pmatrix} xy & 0 \\ 0 & xy \end{pmatrix} \]
Now, adding the identity matrix \( I \):
\[ A^2 + I = \begin{pmatrix} xy & 0 \\ 0 & xy \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} xy + 1 & 0 \\ 0 & xy + 1 \end{pmatrix} \]
The determinant of \( A^2 + I \) is:
\[ \text{det}(A^2 + I) = (xy + 1)(xy + 1) = (xy + 1)^2 \]
Setting this equal to zero gives:
\[ (xy + 1)^2 = 0 \implies xy + 1 = 0 \implies xy = -1 \]

**Hint**: Consider the implications of this determinant condition on the values of \( xy \).

5. **Possible Values of \( xy \)**:
From both cases, we have:
- From Case 1: \( xy = 0 \)
- From Case 2: \( xy = -1 \)

6. **Sum of Possible Values**:
We need to find the sum of the possible values of \( xy \):
\[ 0 + (-1) = -1 \]

### Final Answer:
The sum of possible values of \( xy \) is \( \boxed{-1} \).