考研线性代数：行列式部分初级专项练习题（2024 年）

题目 13

$$
D=\left|\begin{array}{llll}a & 1 & 0 & 0 \\ 0 & a & 1 & 0 \\ 0 & 0 & a & 1 \\ 1 & x & x^{2} & x^{3}\end{array}\right|=
$$

解析 13

$$
D=\left|\begin{array}{llll}a & 1 & 0 & 0 \\ 0 & a & 1 & 0 \\ 0 & 0 & a & 1 \\ 1 & x & x^{2} & x^{3}\end{array}\right|=
$$

$$
a\left|\begin{array}{lll}a & 1 & 0 \\ 0 & a & 1 \\ x & x^{2} & x^{3}\end{array}\right|-\left|\begin{array}{ccc}0 & 1 & 0 \\ 0 & a & 1 \\ 1 & x^{2} & x^{3}\end{array}\right|=
$$

$$
a\left(a\left|\begin{array}{cc}a & 1 \\ x^{2} & x^{3}\end{array}\right|-\left|\begin{array}{ll}0 & 1 \\ x & x^{3}\end{array}\right|\right)+\left|\begin{array}{ll}0 & 1 \\ 1 & x^{3}\end{array}\right|=
$$

$$
a\left[a\left(a x^{3}-x^{2}\right)-(-x)\right]+(-1)=
$$

$$
a\left(a^{2} x^{3}-a x^{2}+x\right)+1=
$$

$$
a^{3} x^{3}-a^{2} x^{2}+a x-1.
$$

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