Let A and B be any two $$3 \times 3$$ matrices. If A is symmetric and B is skew symmetric, then the matrix AB $$-$$ BA is:

We are given that A is a symmetric matrix and B is a skew symmetric matrix, both of size $$3 \times 3$$. We need to determine the nature of the matrix $$C = AB - BA$$.

Recall the definitions:

• A symmetric matrix satisfies $$A^T = A$$.
• A skew symmetric matrix satisfies $$B^T = -B$$.

To determine if $$C$$ is symmetric, skew symmetric, or neither, we compute its transpose $$C^T$$ and compare it to $$C$$ and $$-C$$.

Start with $$C = AB - BA$$. The transpose of a difference is the difference of transposes, so:

$$C^T = (AB - BA)^T = (AB)^T - (BA)^T.$$

Using the property $$(XY)^T = Y^T X^T$$ for any matrices $$X$$ and $$Y$$, we get:

$$C^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T.$$

Substitute the properties of A and B:

• Since A is symmetric, $$A^T = A$$.
• Since B is skew symmetric, $$B^T = -B$$.

So, replace $$A^T$$ with $$A$$ and $$B^T$$ with $$-B$$:

$$C^T = (-B) A - A (-B).$$

Simplify the expression:

$$C^T = -B A - (-A B) \quad \text{(since } A(-B) = -A B\text{)}.$$

$$C^T = -B A + A B.$$

$$C^T = A B - B A.$$

Note that $$A B - B A$$ is the same as $$AB - BA$$, which is exactly $$C$$. Therefore:

$$C^T = AB - BA = C.$$

Since $$C^T = C$$, the matrix $$C$$ is symmetric.

Now, verify with the options:

• A. skew symmetric: This would require $$C^T = -C$$, but we have $$C^T = C$$, so not skew symmetric.
• B. I or $$-I$$: This is not necessarily true, as $$C$$ depends on A and B and may not be a multiple of the identity matrix.
• C. symmetric: Matches our result $$C^T = C$$.
• D. neither symmetric nor skew symmetric: But we found it is symmetric.

Hence, the correct answer is Option C.