If `A=[(1,0,0),(0,1,0),(a,b,-1)]` then `A^(2)` is eqal to

Options

A. `A`
B. `-A`
C. Null matrix
D. I

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 \( A \times A \).

### Step 1: Write down the multiplication

We need to calculate \( A^2 = A \times A \):

\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{pmatrix} \times \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{pmatrix}. \]

### Step 2: Calculate the elements of the resulting matrix

Now, we will calculate each element of the resulting matrix \( A^2 \):

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

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

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

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

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

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

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

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

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

### Step 3: Combine results into the resulting matrix

Putting all these results together, we get:

\[ A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. \]

### Conclusion

Thus, \( A^2 \) is the identity matrix \( I \):

\[ A^2 = I. \]