20 For a given integer $a_{1}(\neq-1)$, find a sequence of real numbers $\left\{a_{n}\right\}\left(a_{i} \neq 0\right.$, $i=1,2, \cdots, 5)$, such that: if $x_{1}, x_{2}, \cdots, x_{5}, y_{1}, y_{2}, \cdots, y_{5}$ satisfy
$$
\left\{\begin{array}{l}
\left\{\begin{array}{l}
b_{11} x_{1}+b_{12} x_{2}+\cdots+b_{15} x_{5}=x_{1}, \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
b_{51} x_{1}+b_{52} x_{2}+\cdots+b_{55} x_{5}=x_{5} ;
\end{array}\right. \\
\left\{\begin{array}{l}
b_{11} y_{1}+b_{12} y_{2}+\cdots+b_{15} y_{5}=2 y_{1}, \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
b_{51} y_{1}+b_{52} y_{2}+\cdots+b_{55} y_{5}=2 y_{5},
\end{array}\right.
\end{array}\right.
$$

and

then $x_{1} y_{1}+x_{2} y_{2}+\cdots+x_{5} y_{5}=0$, where $b_{i j}=\prod_{1 \leq k \leq i}\left(1+j a_{k}\right)$.
$$
y_{2}, \cdots, y_{5} \text { get }\left\{\begin{array}{l}
b_{11} x_{1} y_{1}+b_{12} x_{2} y_{1}+\cdots+b_{15} x_{5} y_{1}=x_{1} y_{1}, \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
b_{51} x_{1} y_{5}+b_{52} x_{2} y_{5}+\cdots+b_{55} x_{5} y_{5}=x_{5} y_{5},
\end{array}\right.
$$
Multiply each by $y_{1}$,
Solution: From $\left\{\begin{array}{l}b_{11} x_{1}+b_{12} x_{2}+\cdots+b_{15} x_{5}=x_{1}, \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ b_{51} x_{1}+b_{52} x_{2}+\cdots+b_{55} x_{5}=x_{5}\end{array}\right.$

Summing up, we get
$$
\begin{aligned}
& \left(b_{11} y_{1}+\cdots+b_{51} y_{5}\right) x_{1}+\cdots+\left(b_{15} y_{1}+\cdots+b_{55} y_{5}\right) x_{5} \\
= & x_{1} y_{1}+\cdots+x_{5} y_{5}
\end{aligned}
$$

If $b_{i j}=b_{j i}$, then $x_{1} y_{1}+\cdots+x_{5} y_{5}=2\left(x_{1} y_{1}+\cdots+x_{5} y_{5}\right)$, solving gives $x_{1} y_{1}+\cdots+x_{5} y_{5}=0$. Therefore, when $b_{i j}=b_{j i}$, the conclusion holds.

By definition, $1+n a_{1}=b_{1} n=b_{n 1}=b_{(n-1) 1}\left(1+a_{n}\right)=b_{1(n-1)}(1+$ $\left.a_{n}\right)=\left[1+(n-1) a_{1}\right]\left(1+a_{n}\right)$, we get $a_{n}=\frac{a_{1}}{1+(n-1) a_{1}}$.
Now verify that $a_{n}=\frac{a_{1}}{1+(n-1) a_{1}}$ when $b_{i j}=b_{j i}$.
Assume $i>j$, then
$$
\begin{aligned}
b_{i j} & =\prod_{1 \leq k \leq i}\left(1+j a_{k}\right)=\prod_{k=1}^{i} \frac{1+(j+k-1) a_{1}}{1+(k-1) a_{1}} \\
& =\frac{\prod_{k=1-i+j}^{j}\left[1+(i+k-1) a_{1}\right]}{\prod_{k=1}^{i}\left[1+(k-1) a_{1}\right]} \\
& =\frac{\prod_{k=1}^{j}\left[1+(i+k-1) a_{1}\right]}{\prod_{k=1}^{j}\left[1+(k-1) a_{1}\right]} \cdot \frac{\prod_{k=1-i+j}^{j}\left[1+(i+k-1) a_{1}\right]}{\prod_{k=j+1}^{j+i}\left[1+(k-1) a_{1}\right]} \\
& =\prod_{k=1}^{j} \frac{1+(i+k-1) a_{1}}{1+(k-1) a_{1}}=b_{i j},
\end{aligned}
$$

Thus, the required real sequence $\left\{a_{n}\right\}=\left\{\frac{a_{1}}{1+(n-1) a_{1}}\right\}$.
If $a_{3}=\cdots=a_{5}=0$, it can be deduced that $x_{1}=\cdots=x_{5}=0, y_{1}=\cdots=y_{5}=0$, hence the conclusion holds.