Solution
= Solution
Since $Ap_1=(5,2)^T$, an <A-conjugate direction> $p_2=(a,b)^T$ must satisfy $5a+2b=0$. One nonzero choice is
$$
\boxed{p_2=(2,-5)^T.}
$$
Direct <matrix multiplication> gives $Ap_2=(3,-3)^T$, and $p_1^TAp_2=0$. All nonzero scalar multiples of this <vector> also qualify.