If A and B are skew-symmetric matrices, then which one of the following is NOT true?

The correct option is: A⁴ + B⁵ is symmetric.

Let us analyze why this statement is NOT true step-by-step.
Recall that a square matrix $M$ is:
1. Symmetric if $M^T=M$
2. Skew-symmetric if $M^T=\−M$

We are given that $A$ and $B$ are skew-symmetric matrices. Therefore:
$A^T=\−A$
$B^T=\−B$

Let us first look at the transpose of any power of a matrix, ${\left(M^k\right)}^T={\left(M^T\right)}^k$.
For a skew-symmetric matrix $M$:
${\left(M^k\right)}^T={(\−M)}^k={(\−1)}^k{M}^k$
Thus:
- If $k$ is an odd integer, ${\left(M^k\right)}^T=\−M^k$ (so $M^k$ is skew-symmetric).
- If $k$ is an even integer, ${\left(M^k\right)}^T=M^k$ (so $M^k$ is symmetric).

Let us test the given options to find which statement is NOT true:

1. For option 1: $A^3+B^5$
Since 3 and 5 are odd, $A^3$ and $B^5$ are both skew-symmetric:
${(A^3+B^5)}^T={\left(A^3\right)}^T+{\left(B^5\right)}^T=\−A^3\−B^5=\−(A^3+B^5)$.
Thus, $A^3+B^5$ is skew-symmetric. This statement is true.

2. For option 2: $A^{19}$
Since 19 is odd, ${\left(A^{19}\right)}^T=\−A^{19}$.
Thus, $A^{19}$ is skew-symmetric. This statement is true.

3. For option 3: $B^{14}$
Since 14 is even, ${\left(B^{14}\right)}^T=B^{14}$.
Thus, $B^{14}$ is symmetric. This statement is true.

4. For option 4: $A^4+B^5$
Let us find its transpose:
${(A^4+B^5)}^T={\left(A^4\right)}^T+{\left(B^5\right)}^T$
Since 4 is even, ${\left(A^4\right)}^T=A^4$ (symmetric).
Since 5 is odd, ${\left(B^5\right)}^T=\−B^5$ (skew-symmetric).
Therefore:
${(A^4+B^5)}^T=A^4\−B^5$
Since $A^4\−B^5\≠A^4+B^5$ (in general), the matrix $A^4+B^5$ is not symmetric. Hence, this statement is NOT true.