If `A=[{:(1,0,0),(0,1,0),(a,b,-1):}]` then `A^(2)` is equal to

Options

A. A
B. `-A`
C. a null matrix
D. a unit matrix

Answer

To find \( A^2 \) where \( A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{pmatrix} \), we will perform matrix multiplication of \( A \) with itself.

### Step-by-step Solution:

1. **Write down the matrices**:
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{pmatrix} \]

2. **Multiply \( A \) by itself**:
\[ A^2 = A \times A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{pmatrix} \times \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{pmatrix} \]

3. **Calculate the elements of the resulting matrix**:

- **First row**:
- First element: \( 1 \cdot 1 + 0 \cdot 0 + 0 \cdot a = 1 \)
- Second element: \( 1 \cdot 0 + 0 \cdot 1 + 0 \cdot b = 0 \)
- Third element: \( 1 \cdot 0 + 0 \cdot 0 + 0 \cdot -1 = 0 \)

- **Second row**:
- First element: \( 0 \cdot 1 + 1 \cdot 0 + 0 \cdot a = 0 \)
- Second element: \( 0 \cdot 0 + 1 \cdot 1 + 0 \cdot b = 1 \)
- Third element: \( 0 \cdot 0 + 1 \cdot 0 + 0 \cdot -1 = 0 \)

- **Third row**:
- First element: \( a \cdot 1 + b \cdot 0 + -1 \cdot a = a - a = 0 \)
- Second element: \( a \cdot 0 + b \cdot 1 + -1 \cdot b = b - b = 0 \)
- Third element: \( a \cdot 0 + b \cdot 0 + -1 \cdot -1 = 1 \)

4. **Combine the results**:
\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

5. **Identify the resulting matrix**:
The resulting matrix is the identity matrix \( I \).

### Final Answer:
\[ A^2 = I \]