Proving a matrix has non trivial solution

Question

A & B are 3x3 matrices and $$\ A^{T} = -A $$

I need to prove that

$\ (A^{2}B)x = 0 $ has a non trivial solution.

so I understand that

$\ |A| = |A^{T}| = |-A| = -1^{n}|A| = 0 $ because n=3. and from that I conclude that A is not invertible and if $\ A^T = -A $ it must have tr(A) = 0 but i'm stuck and would love get a hint.

thanks.

Answer 1

$$det(A) = 0 \Rightarrow det(A^2B)= 0 $$

As $det(A^2B) = 0 \Rightarrow rank(A^2B) < 3 \Rightarrow$ the columns of $A^2B$ are linearly dependent $\Rightarrow$ there is a non-zero vector $x = (x_1\; x_2\; x_3)^T$ such that $A^2B x = 0$.

Answer 2

Since $\det(A)=0$,$$\det(A^2B)=\det(A)^2\det(B)=0,$$and therefore, $(A^2B).x=0$ has a non-trivial solution.