If `A=[{:(1,0,0),(1,0,1),(0,1,0):}]`, then

Options

A. `A^(3)-A^(2)=A-I`
B. `Det(A^(2010)-I)=0`
C. `A^(50)=[{:(1,0,0),(25,1,0),(25,0,1):}]`
D. `A^(50)=[{:(1,1,0),(25,1,0),(25,0,1):}]`

Explanation

To solve the problem, we need to analyze the matrix \( A \) given by:

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

We will find \( A^2 \), \( A^3 \), and then compute \( A^3 - A^2 \).

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

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

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

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

- First row:
- \( (1 \cdot 1 + 0 \cdot 1 + 0 \cdot 0) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 0 + 0 \cdot 1) = 0 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) = 0 \)

- Second row:
- \( (1 \cdot 1 + 0 \cdot 1 + 1 \cdot 0) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 0 + 1 \cdot 1) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 1 \cdot 0) = 0 \)

- Third row:
- \( (0 \cdot 1 + 1 \cdot 1 + 0 \cdot 0) = 1 \)
- \( (0 \cdot 0 + 1 \cdot 0 + 0 \cdot 1) = 0 \)
- \( (0 \cdot 0 + 1 \cdot 1 + 0 \cdot 0) = 1 \)

Thus, we find:

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

### Step 2: Calculate \( A^3 \)

Next, we compute \( A^3 \) by multiplying \( A^2 \) by \( A \):

\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \]

Calculating each element of \( A^3 \):

- First row:
- \( (1 \cdot 1 + 0 \cdot 1 + 0 \cdot 0) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 0 + 0 \cdot 1) = 0 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) = 0 \)

- Second row:
- \( (1 \cdot 1 + 1 \cdot 1 + 1 \cdot 0) = 2 \)
- \( (1 \cdot 0 + 1 \cdot 0 + 1 \cdot 1) = 1 \)
- \( (1 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) = 1 \)

- Third row:
- \( (1 \cdot 1 + 0 \cdot 1 + 1 \cdot 0) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 0 + 1 \cdot 1) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 1 \cdot 0) = 0 \)

Thus, we find:

\[ A^3 = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix} \]

### Step 3: Calculate \( A^3 - A^2 \)

Now, we compute \( A^3 - A^2 \):

\[ A^3 - A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix} \]

Calculating each element:

- First row:
- \( 1 - 1 = 0 \)
- \( 0 - 0 = 0 \)
- \( 0 - 0 = 0 \)

- Second row:
- \( 2 - 1 = 1 \)
- \( 1 - 1 = 0 \)
- \( 1 - 1 = 0 \)

- Third row:
- \( 1 - 1 = 0 \)
- \( 1 - 0 = 1 \)
- \( 0 - 1 = -1 \)

Thus, we find:

\[ A^3 - A^2 = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & -1 \end{pmatrix} \]

### Conclusion

The final result for \( A^3 - A^2 \) is:

\[ \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & -1 \end{pmatrix} \]