华为 $5 \mathrm{G}$ 通信编码的极化码技术方案基于矩阵的乘法，如:

$$
\left(\begin{array}{ll}
c_{1} & c_{2}
\end{array}\right)=\left(\begin{array}{ll}
a_{1} & a_{2}
\end{array}\right) \times\left(\begin{array}{ll}
b_{11} & b_{12} \\
b_{21} & b_{22}
\end{array}\right) \text {, 其中 }
$$

$c_{1}=a_{1} b_{11}+a_{2} b_{21}, \quad c_{2}=a_{1} b_{12}+a_{2} b_{22}$, 已知定义在 $\mathrm{R}$ 上不恒为 0 的函数 $f(x)$ ，对任意 $a, b \in R$ 有: $\left(\begin{array}{ll}y_{1} & y_{2}\end{array}\right)=(f(a) \quad f(b)) \times\left(\begin{array}{cc}-1 & b+1 \\ a-1 & 1\end{array}\right)$ 且满足 $f(a b)=y_{1}+y_{2}$, 则 $(\quad)$
答案: A,D

【解析】 $\left(\begin{array}{ll}y_{1} & y_{2}\end{array}\right)=(f(a) f(b)) \times\left(\begin{array}{cc}-1 & b+1 \\ a-1 & 1\end{array}\right)$,

$y_{1}=-f(a)+f(b)(a-1), \quad y_{2}=f(a)(b+1)+f(b)$,

$f(a b)=y_{1}+y_{2}$,
$f(a b)=-f(a)+f(b)(a-1)+f(a)(b+1)+f(b)=b f(a)+a f(b)$,

令 $a=b=0$, 则 $f(0)=0 f(0)+0 f(0)=0$,

令 $a=b=1$, 则 $f(1)=f(1)+f(1), f(1)=0$,

令 $a=b=-1$, 则 $f(1)=-f(-1)-f(-1), f(-1)=0$,

令 $a=x, b=-1$, 则 $f(-x)=-f(x)+x f(-1), f(x)+f(-x)=0$.

故答案为: $\mathrm{AD}$

【分析】创新题型, 利用新知识矩阵定义求出 $f(a b)=b f(a)+a f(b)$, 再赋值即可得解.