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

Options

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

Explanation

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 matrices to be multiplied

We have:

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

We will multiply \( A \) by itself:

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

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

We will calculate each element of the resulting matrix \( A^2 \) by taking the dot product of the rows of the first matrix with the columns of the second matrix.

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 the results into the final 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, the result of \( A^2 \) is the identity matrix:

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