已知线性方程组
$$
\text { ( I ) }\left\{\begin{array}{c}
a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1,2 n} x_{2 n}=0, \\
a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2,2 n} x_{2 n}=0, \\
\cdots \cdots \\
a_{n 1} x_{1}+a_{n 2} x_{2}+\cdots+a_{n, 2 n} x_{2 n}=0
\end{array}\right.
$$
的一个基础解系为 $\left(b_{11}, b_{12}, \cdots, b_{1,2 n}\right)^{\mathrm{T}},\left(b_{21}, b_{22}, \cdots, b_{2,2 n}\right)^{\mathrm{T}}, \cdots,\left(b_{n 1}, b_{n 2}, \cdots, b_{n, 2 n}\right)^{\mathrm{T}}$. 试写出线性方程组 || 的
$$
\left\{\begin{array}{c}
b_{11} y_{1}+b_{12} y_{2}+\cdots+b_{1,2 n} y_{2 n}=0 \\
b_{21} y_{1}+b_{22} y_{2}+\cdots+b_{2,2 n} y_{2 n}=0 \\
\cdots \cdots \\
b_{n 1} y_{1}+b_{n 2} y_{2}+\cdots+b_{n, 2 n} y_{2 n}=0
\end{array}\right.
$$
的通解, 并说明理由.