( I ) 计算行列式 $|\boldsymbol{A}|$;
(II) 当实数 $a$ 为何值时, 方程组 $\boldsymbol{A x}=\boldsymbol{\beta}$ 有无穷多解, 并求其通解.

$$|\boldsymbol{A}|=1 \cdot\left|\begin{array}{ccc} 1 & a & 0 \\ 0 & 1 & a \\ 0 & 0 & 1 \end{array}\right|+(-1)^{4+1} \cdot a\left|\begin{array}{ccc} a & 0 & 0 \\ 1 & a & 0 \\ 0 & 1 & a \end{array}\right|=1-a^{4} .$$
(II) 当 $|\boldsymbol{A}|=0$ 时, 方程组 $\boldsymbol{A} \boldsymbol{x}=\boldsymbol{\beta}$ 可能有无穷多解, 由（I）可得, $a=1$, 或 $a=-1$.

(1) 当 $a=1$ 时, $(\boldsymbol{A} \vdots \boldsymbol{\beta})=\left[\begin{array}{llll:r}1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 & 0\end{array}\right] \rightarrow\left[\begin{array}{llll:r}1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & -2\end{array}\right]$ 因为 $r(\boldsymbol{A})=3, r(\boldsymbol{A} \vdots \boldsymbol{\beta})=4$, 故方程组无解,即当 $a=1$ 时不合题意, 舍去.
(2) 当 $a=-1$ 时,
$(\boldsymbol{A} \vdots \boldsymbol{\beta})=\left[\begin{array}{cccc:c}1 & -1 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 & -1 \\ 0 & 0 & 1 & -1 & 0 \\ -1 & 0 & 0 & 1 & 0\end{array}\right] \rightarrow\left[\begin{array}{cccc:c}1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 & -1 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0\end{array}\right]$