Which of the following is (are) NOT the square of a `3xx3` matrix with real entries ?

Options

A. `[{:(,1,0,0),(,0,1,0),(,0,0,-1):}]`
B. `[{:(,1,0,0),(,0,-1,0),(,0,0,-1):}]`
C. `[{:(,-1,0,0),(,0,-1,0),(,0,0,-1):}]`
D. `[{:(,1,0,0),(,0,-1,0),(,0,0,1):}]`

Answer

To determine which of the given matrices is (are) NOT the square of a \(3 \times 3\) matrix with real entries, we will compute the determinants of each matrix. A key point to remember is that the determinant of the square of a \(3 \times 3\) matrix with real entries cannot be negative.

### Step-by-Step Solution:

1. **Calculate the Determinant of Matrix A:**
- Given Matrix A:
\[ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \]
- The determinant is calculated as:
\[ \text{det}(A) = 1 \cdot (-1) \cdot (-1) - 0 = 1 \]
- Since the determinant is positive, Matrix A can be the square of a \(3 \times 3\) matrix.

2. **Calculate the Determinant of Matrix B:**
- Given Matrix B:
\[ B = \begin{pmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} \]
- The determinant is calculated as:
\[ \text{det}(B) = 1 \cdot (-1) \cdot (-1) + 1 \cdot 1 \cdot 1 + 1 \cdot 1 \cdot 1 - 1 \cdot 1 \cdot 1 - 1 \cdot (-1) \cdot 1 - 1 \cdot 1 \cdot (-1) \]
- Simplifying this gives:
\[ \text{det}(B) = 1 + 1 + 1 - 1 - 1 - 1 = 0 \]
- Since the determinant is zero, Matrix B can also be the square of a \(3 \times 3\) matrix.

3. **Calculate the Determinant of Matrix C:**
- Given Matrix C:
\[ C = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \]
- The determinant is calculated as:
\[ \text{det}(C) = (-1) \cdot (-1) \cdot (-1) - 0 = -1 \]
- Since the determinant is negative, Matrix C cannot be the square of a \(3 \times 3\) matrix.

4. **Calculate the Determinant of Matrix D:**
- Given Matrix D:
\[ D = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{pmatrix} \]
- The determinant is calculated as:
\[ \text{det}(D) = 1 \cdot (1 \cdot 1 - (-1) \cdot 1) - 1 \cdot (1 \cdot 1 - (-1) \cdot 1) + 1 \cdot (1 \cdot (-1) - 1 \cdot 1) \]
- Simplifying this gives:
\[ \text{det}(D) = 1 \cdot (1 + 1) - 1 \cdot (1 + 1) + 1 \cdot (-1 - 1) = 2 - 2 - 2 = -2 \]
- Since the determinant is negative, Matrix D cannot be the square of a \(3 \times 3\) matrix.

### Conclusion:
The matrices that are NOT the square of a \(3 \times 3\) matrix with real entries are:
- Matrix C
- Matrix D

### Summary of Results:
- **Matrix A:** Possible (det = 1)
- **Matrix B:** Possible (det = 0)
- **Matrix C:** Not possible (det = -1)
- **Matrix D:** Not possible (det = -2)