已知方程组

$$
I:\left\{\begin{array}{c}
a_{11} x_1+a_{12} x_2+\cdots+a_{1,2 n} x_{2 n}=0, \\
a_{21} x_1+a_{22} x_2+\cdots+a_{2,2 n} x_{2 n}=0, \\
\cdots \ldots \ldots \ldots \\
a_{n 1} x_1+a_{n 2} x_2+\cdots+a_{n, 2 n} x_{2 n}=0
\end{array}\right.
$$


的一个基础解系为

$$
\left(\begin{array}{c}
b_{11} \\
b_{12} \\
\vdots \\
b_{1,2 n}
\end{array}\right),\left(\begin{array}{c}
b_{21} \\
b_{22} \\
\vdots \\
b_{2,2 n}
\end{array}\right), \cdots,\left(\begin{array}{c}
b_{n 1} \\
b_{n 2} \\
\vdots \\
b_{n, 2 n}
\end{array}\right),
$$


试写出方程组
$$
\text { II : }\left\{\begin{array}{c}
b_{11} y_1+b_{12} y_2+\cdots+b_{1,2 n} y_{2 n}=0, \\
b_{21} y_1+b_{22} y_2+\cdots+b_{2,2 n} y_{2 n}=0, \\
\cdots \cdots \cdots \cdots \\
b_{n 1} y_1+b_{n 2} y_2+\cdots+b_{n, 2 n} y_{2 n}=0
\end{array}\right.
$$


的通解，并说明理由。