Question:
If A is a skew-symmetric matrix and n is an odd natural number, write whether An is symmetric or skew-symmetric or neither of the two.
Solution:
If $A$ is a skew-symmetric matrix, then $A^{T}=-A$.
$\left(A^{n}\right)^{T}=\left(A^{T}\right)^{n} \quad[$ For all $n \in N]$
$\Rightarrow\left(A^{n}\right)^{T}=(-A)^{n} \quad\left[\because A^{T}=-A\right]$
$\Rightarrow\left(A^{n}\right)^{T}=(-1)^{n} A^{n}$
$\Rightarrow\left(A^{n}\right)^{T}=A^{n}$, if $n$ is even or $-A^{n}$, if $n$ is odd.
Hence, $A^{n}$ is skew-symmetric when $n$ is an odd natural number.