Let m and N be two 3x3 matrices such that MN=NM. Further if `M!=N^2` and `M^2=N^4` then which of the following are correct.

Options

A. determinant of `(M^(2)+Mn^(2))` is 0
B. there is a `3xx3` non-zero matrix U such that `(M^(2)+MN^(2))U` is the zero matrix
C. determinant of `(M^(2)+MN^(2)) ge 1`
D. for a `3xx3` matrix `U`, is the zero matrix

Explanation

To solve the problem, we need to analyze the given conditions about the matrices \( M \) and \( N \):

1. **Given Conditions:**
- \( MN = NM \) (Matrices commute)
- \( M \neq N^2 \)
- \( M^2 = N^4 \)

2. **Rearranging the Equation:**
From the equation \( M^2 = N^4 \), we can rewrite it as:
\[ M^2 - N^4 = 0 \]
This can be factored using the difference of squares:
\[ M^2 - (N^2)^2 = 0 \]
which gives us:
\[ (M - N^2)(M + N^2) = 0 \]

3. **Analyzing the Factors:**
We know that \( M \neq N^2 \), which implies that:
\[ M - N^2 \neq 0 \]
Therefore, for the product to be zero, we must have:
\[ M + N^2 = 0 \]
This implies:
\[ M = -N^2 \]

4. **Determinant Analysis:**
Now, we can analyze the determinant of \( M^2 + MN^2 \):
\[ M^2 + MN^2 = M^2 + M(N^2) = M(M + N^2) \]
Since \( M + N^2 = 0 \), we have:
\[ M(M + N^2) = M \cdot 0 = 0 \]
Thus:
\[ \text{det}(M^2 + MN^2) = 0 \]

5. **Conclusion About Options:**
- **Option 1:** \( \text{det}(M^2 + MN^2) = 0 \) is correct.
- **Option 2:** Since \( \text{det}(M^2 + MN^2) = 0 \), multiplying by any non-zero matrix \( U \) will still yield zero, hence \( \text{det}(M^2 + MN^2)U = 0 \) is also correct.
- **Option 3:** \( \text{det}(M^2 + MN^2) \geq 0 \) is incorrect since it equals zero.
- **Option 4:** If \( M^2 + N^2 + MN^2 = 0 \), then \( U \) must be the zero matrix, which is incorrect since \( U \) can be non-zero.

Thus, the correct options are **Option 1 and Option 2**.