题目 12
$$
D=\left|\begin{array}{cccc}2 a & -1 & 0 & 0 \\ a^{2} & 2 a & -1 & 0 \\ 0 & a^{2} & 2 a & -1 \\ 0 & 0 & a^{2} & 2 a\end{array}\right|=
$$
解析 12
$$
D=\left|\begin{array}{cccc}2 a & -1 & 0 & 0 \\ a^{2} & 2 a & -1 & 0 \\ 0 & a^{2} & 2 a & -1 \\ 0 & 0 & a^{2} & 2 a\end{array}\right|=
$$
$$
2 a\left|\begin{array}{ccc}2 a & -1 & 0 \\ a^{2} & 2 a & -1 \\ 0 & a^{2} & 2 a\end{array}\right|+\left|\begin{array}{ccc}a^{2} & -1 & 0 \\ 0 & 2 a & -1 \\ 0 & a^{2} & 2 a\end{array}\right|=
$$
$$
2 a\left(2 a\left|\begin{array}{cc}2 a & -1 \\ a^{2} & 2 a\end{array}\right|+\left|\begin{array}{cc}a^{2} & -1 \\ 0 & 2 a\end{array}\right|\right)+
$$
$$
a^{2}\left|\begin{array}{cc}2 a & -1 \\ a^{2} & 2 a\end{array}\right| + \left|\begin{array}{cc}0 & -1 \\ 0 & 2 a\end{array}\right|=
$$
$$
2 a\left[\begin{array}{ll}2 a\left(4 a^{2}+a^{2}\right)+2 a^{3}\end{array}\right]+a^{2}\left(4 a^{2}+a^{2}\right)+0 =
$$
$$
2 a\left(10 a^{3}+2 a^{3}\right)+5 a^{4} =
$$
$$
24 a^{4}+5 a^{4}=29 a^{4}.
$$