15.37 Let * be an operation on ordered pairs of real numbers, defined by $(a, b) *(c, d)=$ $(a c+b d, a d+b c)$. If $(a, b) *(x, y)=(a, b)$, then $(x, y)$ is
(A) $(0,0)$.
(B) $(0,1)$.
(C) $(1,0)$.
(D) $(1,1)$.
(E) $(1,-1)$.
(Jilin Province Seven Cities and Prefectures Junior High School Mathematics Competition, 1985)
[Solution] Given $(a, b) *(x, y)=(a x+b y, a y+b x)$, then
$$
\left\{\begin{array}{l}
a x+b y=a, \\
a y+b x=b,
\end{array}\right.
$$

If $a^{2} \neq b^{2}$, solving yields $x=1, \quad y=0$.
Therefore, the answer is $(C)$.