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

Options

A. a) A
B. b) `-A`
C. c) a null matrix
D. d) a unit matrix

Answer

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

### Step 1: Write down the matrix \( A \)
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{pmatrix} \]

### Step 2: Set up the multiplication \( A^2 = A \cdot A \)
\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{pmatrix} \]

### Step 3: Calculate the elements of \( A^2 \)

- **First row, first column:**
\[ 1 \cdot 1 + 0 \cdot 0 + 0 \cdot a = 1 \]

- **First row, second column:**
\[ 1 \cdot 0 + 0 \cdot 1 + 0 \cdot b = 0 \]

- **First row, third column:**
\[ 1 \cdot 0 + 0 \cdot 0 + 0 \cdot -1 = 0 \]

- **Second row, first column:**
\[ 0 \cdot 1 + 1 \cdot 0 + 0 \cdot a = 0 \]

- **Second row, second column:**
\[ 0 \cdot 0 + 1 \cdot 1 + 0 \cdot b = 1 \]

- **Second row, third column:**
\[ 0 \cdot 0 + 1 \cdot 0 + 0 \cdot -1 = 0 \]

- **Third row, first column:**
\[ a \cdot 1 + b \cdot 0 + (-1) \cdot a = a - a = 0 \]

- **Third row, second column:**
\[ a \cdot 0 + b \cdot 1 + (-1) \cdot b = b - b = 0 \]

- **Third row, third column:**
\[ a \cdot 0 + b \cdot 0 + (-1) \cdot -1 = 1 \]

### Step 4: Combine the results to form the resulting matrix
Putting all the calculated elements together, we get:
\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Conclusion
Thus, \( A^2 \) is equal to the identity matrix \( I \):
\[ A^2 = I \]