To determine which of the given matrices is NOT the square of a 3x3 matrix with real entries, we can use properties of such square matrices.

Let $B$ be a given 3x3 matrix and suppose there exists a 3x3 matrix $A$ with real entries such that $A^2 = B$.

Property 1: Determinant The determinant of a matrix with real entries is a real number. Taking the determinant of the equation $A^2 = B$, we get: $\det(B) = \det(A^2) = (\det(A))^2$ Since $\det(A)$ is a real number, $(\det(A))^2$ must be non-negative. Therefore, a necessary condition for $B$ to be the square of a real matrix is that $\det(B) \ge 0$. Any matrix with a negative determinant cannot be the square of a real matrix.

Property 2: Eigenvalues A stronger condition relates to the eigenvalues of $B$. If $B=A^2$ for a real matrix $A$, then any real negative eigenvalue of $B$ must have an even algebraic multiplicity. This is because the eigenvalues of $A$ are either real or come in complex conjugate pairs. A real negative eigenvalue of $B$ can only arise from squaring a purely imaginary eigenvalue of $A$. Since non-real eigenvalues of a real matrix $A$ must come in conjugate pairs (e.g., $ic$ and $-ic$), they contribute to the multiplicity of the negative eigenvalue of $B$ in pairs. Thus, the total multiplicity must be even.

Let's analyze each option:

A: $B_A = \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & { - 1} \\ \end{matrix} } \right]$

1. Using the determinant: $\det(B_A) = 1 \times 1 \times (-1) = -1$. Since the determinant is negative, $B_A$ cannot be the square of a real matrix.
2. Using eigenvalues: The eigenvalues of $B_A$ are the diagonal entries: $1, 1, -1$. The negative eigenvalue is $\lambda = -1$, and its multiplicity is 1, which is an odd number. This also shows that $B_A$ cannot be the square of a real matrix. Thus, option A is a correct answer.

B: $B_B = \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & { - 1} & 0 \\ 0 & 0 & { - 1} \\ \end{matrix} } \right]$

1. Using the determinant: $\det(B_B) = 1 \times (-1) \times (-1) = 1$. Since the determinant is non-negative, this matrix could be the square of a real matrix.
2. Using eigenvalues: The eigenvalues are $1, -1, -1$. The negative eigenvalue $\lambda = -1$ has multiplicity 2, which is an even number. This condition is satisfied.
3. Finding a matrix A: We need to check if such a real matrix $A$ exists. Let's try to construct one. We need $A^2=B_B$. Consider a block diagonal matrix $A = \left[ {\begin{matrix} a & 0 & 0 \\ 0 & b & c \\ 0 & d & e \\ \end{matrix} } \right]$. Then $A^2 = \left[ {\begin{matrix} a^2 & 0 & 0 \\ 0 & b^2+cd & bc+ce \\ 0 & bd+de & cd+e^2 \\ \end{matrix} } \right]$. We need $a^2=1$, so let $a=1$. We need the 2x2 submatrix $A' = \left[ {\begin{matrix} b & c \\ d & e \\ \end{matrix} } \right]$ to satisfy $(A')^2 = \left[ {\begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} } \right] = -I_2$. A known real matrix whose square is $-I_2$ is the rotation matrix for 90 degrees, scaled. Let $A' = \left[ {\begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix} } \right]$. $(A')^2 = \left[ {\begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix} } \right] \left[ {\begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix} } \right] = \left[ {\begin{matrix} -1 & 0 \\ 0 & -1 \\ \end{matrix} } \right]$. So, we can take $A = \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{matrix} } \right]$. This is a real matrix, and $A^2 = B_B$. Therefore, $B_B$ IS the square of a real matrix. Thus, option B is not a correct answer to the question.

C: $B_C = \left[ {\begin{matrix} { - 1} & 0 & 0 \\ 0 & { - 1} & 0 \\ 0 & 0 & { - 1} \\ \end{matrix} } \right]$

1. Using the determinant: $\det(B_C) = (-1) \times (-1) \times (-1) = -1$. Since the determinant is negative, $B_C$ cannot be the square of a real matrix.
2. Using eigenvalues: The eigenvalues are $-1, -1, -1$. The negative eigenvalue $\lambda = -1$ has multiplicity 3, which is an odd number. This confirms that $B_C$ cannot be the square of a real matrix. Thus, option C is a correct answer.

D: $B_D = \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} } \right]$ This is the identity matrix $I_3$. $\det(B_D) = 1 > 0$. It can be a square. We can choose $A = I_3$, which is a real matrix. Then $A^2 = I_3^2 = I_3 = B_D$. So, $B_D$ IS the square of a real matrix. Thus, option D is not a correct answer.

Conclusion: The matrices that are NOT the square of a 3x3 matrix with real entries are those in options A and C.