8.计算下列各行列式(D,为k阶行列式)

(1) D_n=$\left| \begin{matrix} a & & 1 \\ & \ddots & \\ 1 & & a \end{matrix} \right|$ , 其中对角线上元素都是a , 未写出的元素都是0

(2) D_n=$\left| \begin{array}{r} \begin{matrix} x \\ a \end{matrix} \\ \begin{matrix} \vdots \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a \\ x \end{matrix} \\ \begin{matrix} \vdots \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a \\ a \end{matrix} \\ \begin{matrix} \vdots \\ x \end{matrix} \end{array} \right|$

(3) D_n+1=$\left| \begin{array}{r} \begin{matrix} a^{n} \\ a^{n - 1} \end{matrix} \\ \begin{matrix} \vdots \\ a \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} (a - 1)^{n} \\ (a - 1)^{n - 1} \end{matrix} \\ \begin{matrix} \vdots \\ a - 1 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \\ \cdots \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} (a - n)^{n} \\ (a - n)^{n - 1} \end{matrix} \\ \begin{matrix} \vdots \\ a - n \\ 1 \end{matrix} \end{array} \right|$

(4) D_2n=$\left| \begin{array}{r} \begin{matrix} a_{n} \\ \\ \end{matrix} \\ \begin{matrix} \\ \\ c_{n} \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \\ \ddots \\ \end{matrix} \\ \begin{matrix} \\ ⋰ \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \\ \\ a_{1} \end{matrix} \\ \begin{matrix} c_{1} \\ \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \\ \\ b_{1} \end{matrix} \\ \begin{matrix} d_{1} \\ \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \\ ⋰ \\ \end{matrix} \\ \begin{matrix} \\ \ddots \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} b_{n} \\ \\ \end{matrix} \\ \begin{matrix} \\ \\ d_{n} \end{matrix} \end{array} \right|$ , 其中未写出的元素都是0

(5) D_n=$\left| \begin{array}{r} \begin{matrix} 1 + a_{1} \\ a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ a_{n} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{1} \\ 1 + a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ a_{n} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{1} \\ a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ 1 + a_{n} \end{matrix} \end{array} \right|$

(6) D_n=det(a_ij)=$\left| \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 2 \\ \vdots \\ n - 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 1 \\ \vdots \\ n - 2 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 2 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ \vdots \\ n - 3 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \cdots \\ \\ \cdots \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} n - 1 \\ n - 2 \end{matrix} \\ \begin{matrix} n - 3 \\ \vdots \\ 0 \end{matrix} \end{array} \right|$ , 其中a_ij=|i-j|

(7) D_n=$\left| \begin{array}{r} \begin{matrix} 1 + a_{1} \\ 1 \end{matrix} \\ \begin{matrix} \vdots \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 + a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} \vdots \\ 1 + a_{n} \end{matrix} \end{array} \right|$，其中a₁a₂⋯a_n≠0

解：

(1): 化D为上三角形行列式

$\left| \begin{matrix} a & & 1 \\ & \ddots & \\ 1 & & a \end{matrix} \right|\overset{r_{n} + r_{1}}{\Rightarrow}\left| \begin{matrix} a & & 1 \\ & \ddots & \\ a + 1 & & a + 1 \end{matrix} \right|\overset{c_{1} - c_{n}}{\Rightarrow}\left| \begin{matrix} a - 1 & & 1 \\ & \ddots & \\ & & a + 1 \end{matrix} \right| =$a^n-2(a²-1)

上式中最后那个行列式为上三角形行列式

$\left| \begin{array}{r} \begin{matrix} a \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ a \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} a \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ a \end{matrix} \end{array} \right| =$ $- \left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ a \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ a \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} a \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} a \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right| =$

$- \left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ a \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} a \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} a \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 - a^{2} \end{matrix} \end{array} \right| = - a^{2}\left( 1 - a^{2} \right) = a^{2}\left( a^{2} - 1 \right)$

(2)本题中D_n是教材例8中行列式的一般形式

解: 利用各列的元素之和相同

把从第二行起的各行统统加到第一行再提取公因式

$$\left| \begin{array}{r} \begin{matrix} x \\ a \end{matrix} \\ \begin{matrix} \vdots \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a \\ x \end{matrix} \\ \begin{matrix} \vdots \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a \\ a \end{matrix} \\ \begin{matrix} \vdots \\ x \end{matrix} \end{array} \right|\overset{r_{1} + r_{2} + \cdots + r_{n}}{\Rightarrow}\left| \begin{array}{r} \begin{matrix} x + (n - 1)a \\ a \end{matrix} \\ \begin{matrix} \vdots \\ a \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} x + (n - 1)a \\ x \end{matrix} \\ \begin{matrix} \vdots \\ a \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} x + (n - 1)a \\ a \end{matrix} \\ \begin{matrix} \vdots \\ x \end{matrix} \end{array} \right|$$

=[x+(n-1)a]$\left| \begin{array}{r} \begin{matrix} 1 \\ a \end{matrix} \\ \begin{matrix} \vdots \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ x \end{matrix} \\ \begin{matrix} \vdots \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a \end{matrix} \\ \begin{matrix} \vdots \\ x \end{matrix} \end{array} \right|\overset{\begin{matrix} r_{i} - ar_{1} \\ i = 2\ ,\ \cdots\ ,\ n \end{matrix}}{\Rightarrow}\lbrack x + (n - 1)a\rbrack\left| \begin{array}{r} \begin{matrix} 1 \\ \end{matrix} \\ \begin{matrix} \\ \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ x - a \end{matrix} \\ \begin{matrix} \\ \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \end{matrix} \\ \begin{matrix} \\ \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ \end{matrix} \\ \begin{matrix} \\ x - a \end{matrix} \end{array} \right|$

=(x-a)^n-1[x+(n-1)a]

$$\left| \begin{array}{r} \begin{matrix} x \\ a \end{matrix} \\ \begin{matrix} a \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a \\ x \end{matrix} \\ \begin{matrix} a \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a \\ a \end{matrix} \\ \begin{matrix} x \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a \\ a \end{matrix} \\ \begin{matrix} a \\ x \end{matrix} \end{array} \right| = \left| \begin{array}{r} \begin{matrix} x + 3a \\ a \end{matrix} \\ \begin{matrix} a \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} x + 3a \\ x \end{matrix} \\ \begin{matrix} a \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} x + 3a \\ a \end{matrix} \\ \begin{matrix} x \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} x + 3a \\ a \end{matrix} \\ \begin{matrix} a \\ x \end{matrix} \end{array} \right| = (x + 3a)\left| \begin{array}{r} \begin{matrix} 1 \\ a \end{matrix} \\ \begin{matrix} a \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ x \end{matrix} \\ \begin{matrix} a \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a \end{matrix} \\ \begin{matrix} x \\ a \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a \end{matrix} \\ \begin{matrix} a \\ x \end{matrix} \end{array} \right|$$

=(x+3a)$\left| \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ x - a \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} x - a \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ x - a \end{matrix} \end{array} \right| = (x - a)^{3}(x + 3a)$

(3)解: 把所给行列式上下翻转,即为范德蒙德行列式

若再将它左右翻转 , 由于上下翻转与左右翻转所用交换次数相等

故行列式经上下翻转再左右翻转(相当于转180° , )其值不变

于是按范德蒙德行列式的结果 , 可得

$$D_{n + 1} = \left| \begin{array}{r} \begin{matrix} 1 \\ a - n \end{matrix} \\ \begin{matrix} \vdots \\ (a - n)^{n} \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ a - n + 1 \end{matrix} \\ \begin{matrix} \vdots \\ (a - n + 1)^{n} \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ a \end{matrix} \\ \begin{matrix} \vdots \\ a^{n} \end{matrix} \end{array} \right| = \prod_{1 \leqslant j < i \leqslant n + 1}^{}\left( x_{i} - x_{j} \right)$$

以4阶行列式为例

$$D_{4} = \left| \begin{array}{r} \begin{matrix} a^{3} \\ a^{2} \end{matrix} \\ \begin{matrix} a \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} (a - 1)^{3} \\ (a - 1)^{2} \end{matrix} \\ \begin{matrix} a - 1 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} (a - 2)^{3} \\ (a - 2)^{2} \end{matrix} \\ \begin{matrix} a - 2 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} (a - 3)^{3} \\ (a - 3)^{2} \end{matrix} \\ \begin{matrix} a - 3 \\ 1 \end{matrix} \end{array} \right| = \left| \begin{array}{r} \begin{matrix} 1 \\ a \end{matrix} \\ \begin{matrix} a^{2} \\ a^{3} \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ a - 1 \end{matrix} \\ \begin{matrix} (a - 1)^{2} \\ (a - 1)^{3} \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ a - 2 \end{matrix} \\ \begin{matrix} (a - 2)^{2} \\ (a - 2)^{3} \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ a - 3 \end{matrix} \\ \begin{matrix} (a - 3)^{2} \\ (a - 3)^{3} \end{matrix} \end{array} \right|$$

$$由公式det(V) = \prod_{1 \leqslant i \leqslant j \leqslant n}^{}{x_{j} - x_{i}}\ \ \ \ 和n = 4$$

得x₁=a x₂=a-1 x₃=a-2 x₄=a-3

行列式为det(V)=(x₂-x₁)(x₃-x₁)(x₄-x₁)(x₃-x₂)(x₄-x₂)(x₄-x₃)

计算每一项：

1. x₂-x₁=(a-1)-a=-1

2. x₃-x₁=(a-2)-a=-2

3. x₄-x₁=(a-3)-a=-3

4. x₃-x₂=(a-2)-(a-1)=-1

5. x₄-x₂=(a-3)-(a-1)=-2

6. x₄-x₃=(a-3)-(a-2)=-1

因此det(V)=(-1)×(-2)×(-3)×(-1)×(-2)×(-1)=12

(4)解:本题与例11相仿 , 解法也大致相同 , 用递推法

D_2n$\overset{\begin{matrix} r_{2} \leftrightarrow r_{2n} \\ c_{2} \leftrightarrow c_{2n} \end{matrix}}{\Rightarrow}\left| \begin{matrix} a_{n} \\ c_{n} \\ \end{matrix}\ \ \ \begin{matrix} \\ \\ 0 \end{matrix}\ \ \ \begin{matrix} b_{n} \\ d_{n} \\ \end{matrix}\ \ \ \begin{matrix} 0 \\ \\ D_{2(n - 1)} \end{matrix} \right|\overset{由例10}{\Rightarrow}$(a_nd_n-b_nc_n)D_2(n-1)

即有递推公式D_2n=(a_nd_n-b_nc_n)D_2(n-1)

另一方面 , 归纳基础为D₂=$\left| \begin{matrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{matrix} \right|$=a₁d₁-b₁c₁

利用这些结果 , 递推得

$$D_{2n} = \left( a_{n}d_{n} - b_{n}c_{n} \right)\cdots\left( a_{1}d_{1} - b_{1}c_{1} \right) = \prod_{k = 1}^{n}\left( a_{k}d_{k} - b_{k}c_{k} \right)$$

(5)解: 把所有的行(第一行除外)都加到第一行,并提取第一行的公因子

得D_n=$\left| \begin{array}{r} \begin{matrix} 1 + a_{1} \\ a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ a_{n} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{1} \\ 1 + a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ a_{n} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{1} \\ a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ 1 + a_{n} \end{matrix} \end{array} \right|$=(1+a₁+a₂+⋯+a_n)$\left| \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ a_{n} \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 + a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ a_{n} \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ 1 + a_{n} \end{matrix} \end{array} \right|$

$\overset{\begin{matrix} c_{2} - c_{1} \\ \cdots\cdots \\ c_{n} - c_{1} \end{matrix}}{\Rightarrow}$(1+a₁+a₂+⋯+a_n)$\left| \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ a_{n} \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} \vdots \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} \vdots \\ 1 \end{matrix} \end{array} \right|$=1+a₁+a₂+⋯+a_n

D₄=$\left| \begin{array}{r} \begin{matrix} 1 + a_{1} \\ a_{2} \end{matrix} \\ \begin{matrix} a_{3} \\ a_{4} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{1} \\ 1 + a_{2} \end{matrix} \\ \begin{matrix} a_{3} \\ a_{4} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{1} \\ a_{2} \end{matrix} \\ \begin{matrix} 1 + a_{3} \\ a_{4} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} a_{1} \\ a_{2} \end{matrix} \\ \begin{matrix} a_{3} \\ 1 + a_{4} \end{matrix} \end{array} \right|$

=$\left( 1 + a_{1} + a_{2} + a_{3} + a_{4} \right)\left| \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} a_{3} \\ a_{4} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 + a_{2} \end{matrix} \\ \begin{matrix} a_{3} \\ a_{4} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} 1 + a_{3} \\ a_{4} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} a_{3} \\ 1 + a_{4} \end{matrix} \end{array} \right|$

=$\left( 1 + a_{1} + a_{2} + a_{3} + a_{4} \right)\left| \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} a_{3} \\ a_{4} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right|$=$1 + a_{1} + a_{2} + a_{3} + a_{4}$

(6)解:

D_n=$\left| \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 2 \\ \vdots \\ n - 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 1 \\ \vdots \\ n - 2 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 2 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ \vdots \\ n - 3 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \cdots \\ \\ \cdots \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} n - 1 \\ n - 2 \end{matrix} \\ \begin{matrix} n - 3 \\ \vdots \\ 0 \end{matrix} \end{array} \right|\overset{\begin{array}{r} \begin{matrix} r_{n} - r_{n - 1} \\ r_{n - 1} - r_{n - 2} \end{matrix} \\ \begin{matrix} \cdots\cdots \\ r_{2} - r_{1} \end{matrix} \end{array}}{\Rightarrow}\left| \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ \vdots \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ - 1 \end{matrix} \\ \begin{matrix} 1 \\ \vdots \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \cdots \\ \\ \cdots \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} n - 2 \\ - 1 \end{matrix} \\ \begin{matrix} - 1 \\ \vdots \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} n - 1 \\ - 1 \end{matrix} \\ \begin{matrix} - 1 \\ \vdots \\ - 1 \end{matrix} \end{array} \right|$

$\overset{\begin{array}{r} \begin{matrix} c_{1} + c_{n} \\ c_{2} + c_{n} \end{matrix} \\ \begin{matrix} \cdots\cdots \\ c_{n - 1} + c_{n} \end{matrix} \end{array}}{\Rightarrow}\left| \begin{array}{r} \begin{matrix} n - 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ \vdots \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} n \\ - 2 \end{matrix} \\ \begin{matrix} 0 \\ \vdots \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \cdots \\ \\ \cdots \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 2n - 3 \\ - 2 \end{matrix} \\ \begin{matrix} - 2 \\ \vdots \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} n - 1 \\ - 1 \end{matrix} \\ \begin{matrix} - 1 \\ \vdots \\ - 1 \end{matrix} \end{array} \right|$=(-1)^n-1(n-1)2^n-2

对于n=3

D₃=$\left| \begin{matrix} a_{11} \\ a_{21} \\ a_{31} \end{matrix}\ \ \ \ \begin{matrix} a_{12} \\ a_{22} \\ a_{32} \end{matrix}\ \ \ \ \begin{matrix} a_{13} \\ a_{23} \\ a_{33} \end{matrix} \right| = \left| \begin{matrix} |1 - 1| \\ |2 - 1| \\ |3 - 1| \end{matrix}\ \ \ \ \begin{matrix} |1 - 2| \\ |2 - 2| \\ |3 - 2| \end{matrix}\ \ \ \ \begin{matrix} |1 - 3| \\ |2 - 3| \\ |3 - 3| \end{matrix} \right| = \left| \begin{matrix} 0 \\ 1 \\ 2 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 0 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} 2 \\ 1 \\ 0 \end{matrix} \right|$

=$- \left| \begin{matrix} 1 \\ 0 \\ 2 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ 1 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 2 \\ 0 \end{matrix} \right|$=$- \left| \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ 1 \\ 1 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 2 \\ - 2 \end{matrix} \right|$=$- \left| \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix}\ \ \ \ \begin{matrix} 1 \\ 2 \\ - 4 \end{matrix} \right|$=4

(7)解: 将原行列式化为上三角形行列式

为此 , 从第2行起 , 各行均减去第1行 , 得与例1.3相仿的行列式

D_n$\left| \begin{array}{r} \begin{matrix} 1 + a_{1} \\ 1 \end{matrix} \\ \begin{matrix} \vdots \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 + a_{2} \end{matrix} \\ \begin{matrix} \vdots \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \cdots \end{matrix} \\ \begin{matrix} \\ \cdots \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} \vdots \\ 1 + a_{n} \end{matrix} \end{array} \right|\overset{\begin{matrix} r_{i} - r_{1} \\ i = 2\ ,\ \cdots\ ,\ n \end{matrix}}{\Rightarrow}\left| \begin{array}{r} \begin{matrix} 1 + a_{1} \\ - a_{1} \end{matrix} \\ \begin{matrix} \vdots \\ - a_{1} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} \\ \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \end{matrix} \\ \begin{matrix} \ddots \\ \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ \end{matrix} \\ \begin{matrix} \\ a_{n} \end{matrix} \end{array} \right|\overset{\begin{matrix} c_{1} + \frac{a_{1}}{a_{i}}c_{i} \\ i = 2\ ,\ \cdots\ ,\ n \end{matrix}}{\Rightarrow}\left| \begin{array}{r} \begin{matrix} b \\ 0 \end{matrix} \\ \begin{matrix} \vdots \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} \\ \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} \cdots \\ \end{matrix} \\ \begin{matrix} \ddots \\ \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ \end{matrix} \\ \begin{matrix} \\ a_{n} \end{matrix} \end{array} \right|$

$$其中b = 1 + a_{1} + a_{1}\sum_{i = 2}^{n}\frac{1}{a_{i}} = a_{1}\left( 1 + \sum_{i = 1}^{n}\frac{1}{a_{i}} \right)$$

$$于是D_{n} = a_{1}\cdots a_{n}\left( 1 + \sum_{i = 1}^{n}\frac{1}{a_{i}} \right)$$

当n=4时

$$\left| \begin{array}{r} \begin{matrix} 1 + a_{1} \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 + a_{2} \end{matrix} \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 1 + a_{3} \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 1 + a_{4} \end{matrix} \end{array} \right| = \left| \begin{array}{r} \begin{matrix} 1 + a_{1} \\ - a_{1} \end{matrix} \\ \begin{matrix} - a_{1} \\ - a_{1} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} a_{3} \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ a_{4} \end{matrix} \end{array} \right|$$

$$= \left| \begin{array}{r} \begin{matrix} 1 + a_{1} + \frac{a_{1}}{a_{2}} \\ 0 \end{matrix} \\ \begin{matrix} - a_{1} \\ - a_{1} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} a_{3} \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ a_{4} \end{matrix} \end{array} \right| = \left| \begin{array}{r} \begin{matrix} 1 + a_{1} + \frac{a_{1}}{a_{2}} + \frac{a_{1}}{a_{3}} \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ - a_{1} \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} a_{3} \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ a_{4} \end{matrix} \end{array} \right|$$

$$= \left| \begin{array}{r} \begin{matrix} 1 + a_{1} + \frac{a_{1}}{a_{2}} + \frac{a_{1}}{a_{3}} + \frac{a_{1}}{a_{4}} \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ a_{2} \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} a_{3} \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ a_{4} \end{matrix} \end{array} \right|$$

=$\left( 1 + a_{1} + \frac{a_{1}}{a_{2}} + \frac{a_{1}}{a_{3}} + \frac{a_{1}}{a_{4}} \right)a_{2}a_{3}a_{4} = a_{1}\left( \frac{1}{a_{1}} + 1 + \frac{1}{a_{2}} + \frac{1}{a_{3}} + \frac{1}{a_{4}} \right)a_{2}a_{3}a_{4}$

$$= a_{1}a_{2}a_{3}a_{4}\left( 1 + \frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + \frac{1}{a_{4}} \right) = a_{1}a_{2}a_{3}a_{4}\left( 1 + \sum_{i = 1}^{4}\frac{1}{a_{i}} \right)$$

=$a_{1}a_{2}a_{3}a_{4} + a_{2}a_{3}a_{4} + a_{1}a_{3}a_{4} + a_{1}a_{2}a_{4} + a_{1}a_{2}a_{3}$

9 , 设D=$\left| \begin{array}{r} \begin{matrix} 3 \\ - 5 \end{matrix} \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 1 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ - 5 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} - 1 \\ 3 \end{matrix} \\ \begin{matrix} 1 \\ 3 \end{matrix} \end{array}\ \ \ \begin{array}{r} \begin{matrix} 2 \\ - 4 \end{matrix} \\ \begin{matrix} - 1 \\ - 3 \end{matrix} \end{array} \right|$ , D的(i , j)元的代数余子式记作A_ij

求A₃₁+3A₃₂-2A₃₃+2A₃₄

解: 根据代数余子式定义：$A_{ij} = ( - 1)^{i + j}M_{ij}$，其中$M_{ij}$为元素$a_{ij}$的余子式。

步骤1：计算$A_{31}$（$i = 3,j = 1$）

1. 划去第3行、第1列，得余子式$M_{31}$：$M_{31} = \left| \begin{matrix} 1 & - 1 & 2 \\ 1 & 3 & - 4 \\ - 5 & 3 & - 3 \end{matrix} \right| = 16$

2. 计算代数余子式：$A_{31} = ( - 1)^{3 + 1}M_{31} = 1 \times 16 = 16$

步骤2：计算$A_{32}$（$i = 3,j = 2$）

1. 划去第3行、第2列，得余子式$M_{32}$：$M_{32} = \left| \begin{matrix} 3 & - 1 & 2 \\ - 5 & 3 & - 4 \\ 1 & 3 & - 3 \end{matrix} \right| = - 8$

2. 计算代数余子式：$A_{32} = ( - 1)^{3 + 2}M_{32} = - 1 \times ( - 8) = 8$

步骤3：计算$A_{33}$（$i = 3,j = 3$）

1. 划去第3行、第3列，得余子式$M_{33}$：$M_{33} = \left| \begin{matrix} 3 & 1 & 2 \\ - 5 & 1 & - 4 \\ 1 & - 5 & - 3 \end{matrix} \right| = - 40$

2. 计算代数余子式：$A_{33} = ( - 1)^{3 + 3}M_{33} = 1 \times ( - 40) = - 40$

步骤4：计算$A_{34}$（$i = 3,j = 4$）

1. 划去第3行、第4列，得余子式$M_{34}$：$M_{34} = \left| \begin{matrix} 3 & 1 & - 1 \\ - 5 & 1 & 3 \\ 1 & - 5 & 3 \end{matrix} \right| = 48$

2. 计算代数余子式：$A_{34} = ( - 1)^{3 + 4}M_{34} = - 1 \times 48 = - 48$

步骤5.

A₃₁+3A₃₂-2A₃₃+2A₃₄=16+3×8-2×(-40)+2×(-48)=24