If `A=[(ab,b^(2)),(-a^(2),-ab)]` then A is

Options

A. Idempotent
B. Involutary
C. Nilpotent
D. Scalar

Answer

To determine the properties of the matrix \( A = \begin{pmatrix} ab & b^2 \\ -a^2 & -ab \end{pmatrix} \), we will check if it is a nilpotent matrix. A matrix \( A \) is nilpotent if there exists a positive integer \( k \) such that \( A^k = 0 \) (the zero matrix).

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply matrix \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} ab & b^2 \\ -a^2 & -ab \end{pmatrix} \cdot \begin{pmatrix} ab & b^2 \\ -a^2 & -ab \end{pmatrix} \]

### Step 2: Perform the multiplication

Calculating each element of \( A^2 \):

- **Element (1,1)**:
\[ (ab)(ab) + (b^2)(-a^2) = a^2b^2 - a^2b^2 = 0 \]

- **Element (1,2)**:
\[ (ab)(b^2) + (b^2)(-ab) = ab^3 - ab^3 = 0 \]

- **Element (2,1)**:
\[ (-a^2)(ab) + (-ab)(-a^2) = -a^3b + a^3b = 0 \]

- **Element (2,2)**:
\[ (-a^2)(b^2) + (-ab)(-ab) = -a^2b^2 + a^2b^2 = 0 \]

### Step 3: Write the resulting matrix

Thus, we have:

\[ A^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Conclusion

Since \( A^2 = 0 \), we conclude that \( A \) is a nilpotent matrix.

### Final Answer

The matrix \( A \) is nilpotent.

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