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

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

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

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

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

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

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

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