If `A=[{:(0,1),(-1,0)],B=[{:(i,0),(0,-i):}],C=[{:(0,-i),(-i,0):}]`, then which one of the following in not correct ?

Options

A. `A^(2)=B^(2)`
B. `B^(2)=C^(2)`
C. AB = C
D. AB = BA

Explanation

To solve the problem, we need to analyze the matrices \( A \), \( B \), and \( C \) given in the question and check the correctness of the statements related to these matrices.

Given:
\[ A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \quad B = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, \quad C = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \]

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

To find \( A^2 \), we perform the multiplication \( A \times A \):

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

Calculating the elements:

- First row, first column: \( 0 \cdot 0 + 1 \cdot (-1) = -1 \)
- First row, second column: \( 0 \cdot 1 + 1 \cdot 0 = 0 \)
- Second row, first column: \( -1 \cdot 0 + 0 \cdot (-1) = 0 \)
- Second row, second column: \( -1 \cdot 1 + 0 \cdot 0 = -1 \)

Thus, we have:

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

### Step 2: Calculate \( B^2 \)

Now, we calculate \( B^2 \):

\[ B^2 = B \cdot B = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \cdot \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( i \cdot i + 0 \cdot 0 = i^2 = -1 \)
- First row, second column: \( i \cdot 0 + 0 \cdot (-i) = 0 \)
- Second row, first column: \( 0 \cdot i + (-i) \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 0 + (-i) \cdot (-i) = (-i)^2 = -1 \)

Thus, we have:

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

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

Next, we calculate \( C^2 \):

\[ C^2 = C \cdot C = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( 0 \cdot 0 + (-i)(-i) = i^2 = -1 \)
- First row, second column: \( 0 \cdot (-i) + (-i) \cdot 0 = 0 \)
- Second row, first column: \( (-i) \cdot 0 + 0 \cdot (-i) = 0 \)
- Second row, second column: \( (-i)(-i) + 0 \cdot 0 = (-i)^2 = -1 \)

Thus, we have:

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

### Conclusion

Now we can summarize the results:

- \( A^2 = -I \)
- \( B^2 = -I \)
- \( C^2 = -I \)

Since all the squares of the matrices \( A \), \( B \), and \( C \) yield the same result, we can conclude that none of the statements regarding \( A^2 \), \( B^2 \), or \( C^2 \) being different from \( -I \) is correct.

### Final Answer

Thus, the statement that is not correct is the one that claims any of the matrices' squares is different from \( -I \).