larryleifeng-linear-algebra-lecture
/
线性代数中的反例.lyx

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\begin_body

\begin_layout Title
线性代数中的反例
\end_layout

\begin_layout Section
行列式
\end_layout

\begin_layout Subsection
逆序数
\end_layout

\begin_layout Example
把
\begin_inset Formula $x_{1}x_{2}\cdots x_{n}$
\end_inset

 变成
\begin_inset Formula $123\cdots n$
\end_inset

.
 但这不一定是最少次数的对换.
\end_layout

\begin_layout Example
例
\begin_inset Formula $\tau\left(4132\right)=4$
\end_inset

, 但
\begin_inset Formula $4132\xrightarrow{(2,4)}2134\xrightarrow{(1,2)}1234$
\end_inset

 即两次对换, 就把排列
\begin_inset Formula $4132$
\end_inset

 化为排列
\begin_inset Formula $1234$
\end_inset

.

\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
设
\begin_inset Formula $n$
\end_inset

 元排列
\begin_inset Formula $\cdots i\cdots j\cdots$
\end_inset

 的反序数为
\begin_inset Formula $k$
\end_inset

, 那么
\begin_inset Formula $n$
\end_inset

 元 排列
\begin_inset Formula $\cdots j\cdots i\cdots$
\end_inset

 的反序数不一定为
\begin_inset Formula $k+1$
\end_inset

 或
\begin_inset Formula $k-1$
\end_inset

.
\end_layout

\begin_layout Example
例 取排列
\begin_inset Formula $123$
\end_inset

,
\begin_inset Formula $\tau\left(123\right)=0$
\end_inset

, 对换
\begin_inset Formula $(1,3)$
\end_inset

 得
\begin_inset Formula $321$
\end_inset

,
\begin_inset Formula $\tau(321)=3$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Subsection
\begin_inset Formula $n$
\end_inset

 阶行列式
\end_layout

\begin_layout Example
若
\begin_inset Formula $n$
\end_inset

 阶行列式
\begin_inset Formula $D$
\end_inset

 不等于
\begin_inset Formula $0$
\end_inset

, 那么
\begin_inset Formula $D$
\end_inset

 的
\begin_inset Formula $n-1$
\end_inset

 阶子式不全为
\begin_inset Formula $0$
\end_inset

, 反之不真.

\end_layout

\begin_layout Example
例
\begin_inset Formula $D=\begin{vmatrix}1 & 2\\
3 & 6
\end{vmatrix}$
\end_inset

 的一阶子式全不为
\begin_inset Formula $0$
\end_inset

, 但
\begin_inset Formula $D=0$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
\begin_inset Formula $n$
\end_inset

 阶行列式有一行 (列) 元素均为
\begin_inset Formula $0$
\end_inset

, 则行列式为
\begin_inset Formula $n$
\end_inset

, 反之, 行列式为
\begin_inset Formula $0$
\end_inset

, 那么行列式不一定有一行 (列) 为
\begin_inset Formula $0$
\end_inset

.
\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
\begin{vmatrix}1 & 2 & 3\\
-1 & 1 & 3\\
4 & 5 & 6
\end{vmatrix}=0.
\]

\end_inset


\end_layout

\begin_layout Subsection
克莱姆法则
\end_layout

\begin_layout Example
我们知道, 当方程个数等于未知量个数且系数行列式不等于 0 , 则由克莱姆法则知线性方程组必有唯一解, 但反之不真.
\end_layout

\begin_layout Example
例 线性方程组
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}=1\\
2x_{1}+3x_{2}=4\\
3x_{1}+4x_{2}=5
\end{cases}
\]

\end_inset

的方程个数多于末知量个数, 但却有唯一解.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
如果一个方程组方程的个数比未知量的个数多, 不一定有解.

\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}=1\\
x_{1}+x_{2}=2\\
x_{1}+x_{2}=3
\end{cases}
\]

\end_inset

无解.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
当未知量个数多于方程个数时, 齐次线性方程组必有无穷多解.
 但对非齐次线性方程组来说, 此结论不成立.

\end_layout

\begin_layout Example
例 线性方程组
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}+x_{3}=1\\
2x_{1}+2x_{2}+2x_{3}=3
\end{cases}
\]

\end_inset

的方程个数小于未知量个数, 但它却无解.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
齐次线性方程组方程的个数
\begin_inset Formula $m$
\end_inset

 小于末知量的个数
\begin_inset Formula $n$
\end_inset

, 那么有非零解.
 反之, 方程组有非零解, 不一定有
\begin_inset Formula $m<n$
\end_inset

.
\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}+3x_{3}=0\\
x_{1}+4x_{2}+5x_{3}=0\\
x_{1}-2x_{2}+x_{3}=0\\
2x_{1}-x_{2}-4x_{3}=0
\end{cases}
\]

\end_inset

有非零解, 但
\begin_inset Formula $m=4>n=3$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
空间四个平面
\begin_inset Formula $a_{i}x+b_{i}y+c_{i}z+d_{i}=0$
\end_inset

, (
\begin_inset Formula $i=1,2,3,4$
\end_inset

) 相交于一点时, 有
\begin_inset Formula
\[
\Delta=\begin{vmatrix}a_{1} & b_{1} & c_{1} & d_{1}\\
a_{2} & b_{2} & c_{2} & d_{2}\\
a_{3} & b_{3} & c_{3} & d_{3}\\
a_{4} & b_{4} & c_{4} & d_{4}
\end{vmatrix}=0,
\]

\end_inset

但反之不真.
\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
\begin{cases}
x+y+z+1=0\\
2x+2y+2z+2=0\\
x-y+z+3=0\\
2x-2y+2z+6=0
\end{cases}
\]

\end_inset

前两个方程表示同一个平面, 后两个方程表示另一个平面, 由于两个平面的位置关系 (法线不平行), 它们相交于一条直线.
 而
\begin_inset Formula
\[
\begin{aligned}\Delta & =\begin{vmatrix}1 & 1 & 1 & 1\\
2 & 2 & 2 & 2\\
1 & -1 & 1 & 3\\
2 & -2 & 2 & 6
\end{vmatrix}=0,\end{aligned}
\]

\end_inset

但不相交于一点.
\end_layout

\begin_layout Section
矩阵
\end_layout

\begin_layout Example
矩阵的乘法不满足交换律.

\end_layout

\begin_layout Example
(i) 当
\begin_inset Formula $m\neq s$
\end_inset

 时,
\begin_inset Formula $A_{mn}B_{ns}$
\end_inset

 有意义, 但
\begin_inset Formula $B_{ns}A_{mn}$
\end_inset

 没有意义.
\end_layout

\begin_layout Example
例
\begin_inset Formula $A_{23}B_{34}$
\end_inset

 有意义, 而
\begin_inset Formula $B_{34}A_{23}$
\end_inset

 无意义.
\end_layout

\begin_layout Example
(ii)
\begin_inset Formula $A_{mn}B_{nm}$
\end_inset

 和
\begin_inset Formula $B_{nm}A_{mn}$
\end_inset

 都有意义, 当
\begin_inset Formula $m\neq n$
\end_inset

 时, 第一个乘积是
\begin_inset Formula $m$
\end_inset

 阶矩阵, 而第二乘积是
\begin_inset Formula $n$
\end_inset

 阶矩阵, 它们不相等.
\end_layout

\begin_layout Example
例
\begin_inset Formula $A_{23}B_{32}$
\end_inset

 是
\begin_inset Formula $2$
\end_inset

 阶的,
\begin_inset Formula $B_{32}A_{23}$
\end_inset

 是
\begin_inset Formula $3$
\end_inset

 阶的.

\end_layout

\begin_layout Example
(iii)
\begin_inset Formula $A_{nn}B_{nn}$
\end_inset

 和
\begin_inset Formula $B_{nn},A_{nn}$
\end_inset

 虽然都是
\begin_inset Formula $n$
\end_inset

 阶矩阵, 但它们也未必相等.
 例
\begin_inset Formula
\[
\begin{aligned}A=\begin{bmatrix}1 & 2\\
2 & 1
\end{bmatrix}, & \quad B=\begin{bmatrix}2 & -3\\
3 & 1
\end{bmatrix},\\
AB=\begin{bmatrix}8 & -1\\
7 & -5
\end{bmatrix}, & \quad BA=\begin{bmatrix}-4 & 1\\
5 & 7
\end{bmatrix}.
\end{aligned}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
存在零因子, 即
\begin_inset Formula $A\neq0$
\end_inset

,
\begin_inset Formula $B\neq0$
\end_inset

, 但
\begin_inset Formula $AB=0$
\end_inset

.

\begin_inset Formula
\[
\begin{aligned}A=\begin{bmatrix}0 & 1\\
0 & 2
\end{bmatrix}, & \quad B=\begin{bmatrix}0 & 1\\
0 & 0
\end{bmatrix},\\
AB=\begin{bmatrix}0 & 1\\
0 & 2
\end{bmatrix} & \begin{bmatrix}0 & 1\\
0 & 0
\end{bmatrix}=\begin{bmatrix}0 & 0\\
0 & 0
\end{bmatrix}.
\end{aligned}
\]

\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
矩阵乘法的消去律不成立, 即
\begin_inset Formula $A\neq0$
\end_inset

,
\begin_inset Formula $AB=AC$
\end_inset

, 未必有
\begin_inset Formula $B=C$
\end_inset

.

\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
A=\begin{bmatrix}1 & 1\\
-1 & 1
\end{bmatrix},\quad B=\begin{bmatrix}1 & -1\\
-1 & 1
\end{bmatrix},\quad C=\begin{bmatrix}-1 & 1\\
1 & -1
\end{bmatrix},
\]

\end_inset

那么
\begin_inset Formula
\[
AB=\begin{bmatrix}1 & 1\\
-1 & 1
\end{bmatrix}\begin{bmatrix}1 & -1\\
-1 & 1
\end{bmatrix}=\begin{bmatrix}0 & 0\\
0 & 0
\end{bmatrix},\quad AC=\begin{bmatrix}1 & 1\\
-1 & 1
\end{bmatrix}\begin{bmatrix}-1 & 1\\
1 & -1
\end{bmatrix}=\begin{bmatrix}0 & 0\\
0 & 0
\end{bmatrix},
\]

\end_inset

于是
\begin_inset Formula $AB=AC$
\end_inset

, 且
\begin_inset Formula $A\neq0$
\end_inset

, 但
\begin_inset Formula $B\neq C$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
由于矩阵乘法不满足交换律, 所以等式:
\begin_inset Formula
\[
(AB)^{m}=A^{m}B^{m}
\]

\end_inset

一般不成立.

\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
\begin{array}{ll}
A=\begin{bmatrix}2 & 1\\
3 & 2
\end{bmatrix}, & B=\begin{bmatrix}1 & -1\\
1 & 1
\end{bmatrix},\\
AB=\begin{bmatrix}3 & -1\\
5 & -1
\end{bmatrix}, & (AB)^{2}=\begin{bmatrix}4 & -2\\
10 & -4
\end{bmatrix},\\
A^{2}=\begin{bmatrix}7 & 4\\
12 & 7
\end{bmatrix}, & B^{2}=\begin{bmatrix}0 & -2\\
2 & 0
\end{bmatrix},\\
A^{2}B^{2}=\begin{bmatrix}8 & -14\\
14 & -24
\end{bmatrix},
\end{array}
\]

\end_inset

所以此时
\begin_inset Formula $(AB)^{2}\neq A^{2}B^{2}$
\end_inset

.
 但是当
\begin_inset Formula $AB=BA$
\end_inset

 时, 就有
\begin_inset Formula $(AB)^{m}=A^{m}B^{m}$
\end_inset

.

\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
式子
\begin_inset Formula $(A+B)^{2}=A^{2}+2AB+B^{2}$
\end_inset

 与
\begin_inset Formula $A^{2}-B^{2}=(A+B)(A-B)$
\end_inset

 一般不成立.
\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
A=\begin{bmatrix}1 & 1\\
0 & 1
\end{bmatrix},\quad B=\begin{bmatrix}2 & 1\\
1 & 1
\end{bmatrix},
\]

\end_inset


\begin_inset Formula
\[
(A+B)^{2}=\begin{bmatrix}11 & 10\\
5 & 6
\end{bmatrix},\ A^{2}+2AB+B^{2}=\begin{bmatrix}12 & 9\\
5 & 5
\end{bmatrix},
\]

\end_inset

所以
\begin_inset Formula $(A+B)^{2}\neq A^{2}+2AB+B^{2}$
\end_inset

,
\begin_inset Formula
\[
A^{2}-B^{2}=\begin{bmatrix}-4 & -1\\
-3 & -1
\end{bmatrix},\ (A+B)(A-B)=\begin{bmatrix}-5 & 0\\
-3 & 0
\end{bmatrix},
\]

\end_inset

所以
\begin_inset Formula $A^{2}-B^{2}\neq(A+B)(A-B)$
\end_inset

, 但当
\begin_inset Formula $AB=BA$
\end_inset

 时, 上述等式成立.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
设
\begin_inset Formula $A$
\end_inset

 是一个
\begin_inset Formula $n$
\end_inset

 阶实矩阵, 若
\begin_inset Formula $A^{2}=0$
\end_inset

, 则
\begin_inset Formula $A=0$
\end_inset

, 若
\begin_inset Formula $A$
\end_inset

是复矩阵，有
\begin_inset Formula $A^{2}=0$
\end_inset

, 不一定有
\begin_inset Formula $A=0$
\end_inset

.

\end_layout

\begin_layout Example
例
\begin_inset Formula $A=\begin{bmatrix}1 & \ui\\
\ui & 1
\end{bmatrix}$
\end_inset

, 有
\begin_inset Formula $A^{2}=\begin{bmatrix}0 & 0\\
0 & 0
\end{bmatrix}$
\end_inset

, 但
\begin_inset Formula $A\neq O$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
我们知道, 若
\begin_inset Formula $A,B,C$
\end_inset

 都是
\begin_inset Formula $n$
\end_inset

 阶矩阵, 且
\begin_inset Formula $ABC=E$
\end_inset

, 则
\begin_inset Formula $BCA=E$
\end_inset

,
\begin_inset Formula $CAB=E$
\end_inset

 总成立.
 但
\begin_inset Formula $BAC=E$
\end_inset

,
\begin_inset Formula $ACB=E$
\end_inset

,
\begin_inset Formula $CBA=E$
\end_inset

 却不一定成立.

\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
A=\begin{bmatrix}1 & 1\\
0 & 1
\end{bmatrix},\quad B=\begin{bmatrix}1 & 1\\
1 & 0
\end{bmatrix},\quad C=\begin{bmatrix}0 & 1\\
1 & -2
\end{bmatrix},
\]

\end_inset

有
\begin_inset Formula $ABC=E$
\end_inset

, 但
\begin_inset Formula $BAC=\begin{bmatrix}2 & -3\\
1 & -1
\end{bmatrix}\neq E$
\end_inset

,
\begin_inset Formula $ACB=\begin{bmatrix}0 & 1\\
-1 & 1
\end{bmatrix}\neq E$
\end_inset

,
\begin_inset Formula $CBA=\begin{bmatrix}1 & 1\\
-1 & 0
\end{bmatrix}\neq E$
\end_inset

, 当
\begin_inset Formula $AB=BA$
\end_inset

 时, 上述等式成立.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
我们知道, 单位矩阵
\begin_inset Formula $E$
\end_inset

 与任意
\begin_inset Formula $n$
\end_inset

 阶矩阵
\begin_inset Formula $A$
\end_inset

, 左乘或右乘的乘积仍然是
\begin_inset Formula $A$
\end_inset

 自身, 即
\begin_inset Formula
\[
EA=AE=A.
\]

\end_inset

但是, 对某个别矩阵左乘或右乘不变的不一定就是单位矩阵.

\end_layout

\begin_layout Example
例
\begin_inset Formula $\begin{bmatrix}3 & 1\\
2 & 2
\end{bmatrix}\begin{bmatrix}1 & -1\\
-2 & 2
\end{bmatrix}=\begin{bmatrix}1 & -1\\
-2 & 2
\end{bmatrix}\begin{bmatrix}3 & 1\\
2 & 2
\end{bmatrix}$
\end_inset

, 但
\begin_inset Formula $\begin{bmatrix}3 & 1\\
2 & 2
\end{bmatrix}$
\end_inset

 不是二阶单位矩阵.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
矩阵乘积的行列式等于矩阵的行列式的乘积, 即
\begin_inset Formula $|AB|=|A|\cdot|B|$
\end_inset

.
\end_layout

\begin_layout Example
但对于矩阵和的行列式一般不等于矩阵的行列式之和;
\begin_inset Formula $kA$
\end_inset

 的行列式一般也不等于
\begin_inset Formula $k\cdot|A|$
\end_inset

, 即
\begin_inset Formula $|A+B|\neq|A|+|B|$
\end_inset

,
\begin_inset Formula $|kA|\neq k\cdot|A|$
\end_inset

.

\end_layout

\begin_layout Example
例
\end_layout

\begin_layout Example
(i).

\begin_inset Formula $A=\begin{bmatrix}1 & 0\\
0 & 0
\end{bmatrix}$
\end_inset

,
\begin_inset Formula $B=\begin{bmatrix}0 & 0\\
0 & 1
\end{bmatrix}$
\end_inset

,
\begin_inset Formula $A+B=\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix}$
\end_inset

, 而
\begin_inset Formula $|A+B|=1$
\end_inset

, 又
\begin_inset Formula $|A|+|B|=0+0=0$
\end_inset

.
\end_layout

\begin_layout Example
(ii).

\begin_inset Formula
\[
A=\begin{bmatrix}1 & 0\\
0 & 1
\end{bmatrix},\ k=2,
\]

\end_inset

因此有
\begin_inset Formula $|kA|=4$
\end_inset

,
\begin_inset Formula $k|A|=2$
\end_inset

.
\end_layout

\begin_layout Section
矩阵与线性方程组
\end_layout

\begin_layout Subsection
增广矩阵的秩与线性方程组的解的关系
\end_layout

\begin_layout Example
设
\begin_inset Formula $A=\left(a_{ij}\right)$
\end_inset

 是
\begin_inset Formula $n$
\end_inset

 阶矩阵, 令
\begin_inset ERT
status open

\begin_layout Plain Layout

$$B=
\backslash
begin{bNiceArray}{ccc|c}[margin]
\end_layout

\begin_layout Plain Layout


\backslash
Block{3-3}{A} & & & b_1
\backslash

\backslash

\end_layout

\begin_layout Plain Layout

&
\backslash
hspace*{1cm} & &
\backslash
Vdots
\backslash

\backslash

\end_layout

\begin_layout Plain Layout

& & & b_n
\backslash

\backslash

\end_layout

\begin_layout Plain Layout


\backslash
hline
\end_layout

\begin_layout Plain Layout

b_1 &
\backslash
Cdots & b_n & 0
\end_layout

\begin_layout Plain Layout


\backslash
end{bNiceArray}=
\backslash
begin{bmatrix}
\end_layout

\begin_layout Plain Layout

a_{11} &
\backslash
cdots & a_{1n} & b_{1}
\backslash

\backslash

\end_layout

\begin_layout Plain Layout


\backslash
vdots &  &
\backslash
vdots &
\backslash
vdots
\backslash

\backslash

\end_layout

\begin_layout Plain Layout

a_{n1} &
\backslash
cdots & a_{nn} & b_{n}
\backslash

\backslash

\end_layout

\begin_layout Plain Layout

b_{1} &
\backslash
cdots & b_{n} & 0
\end_layout

\begin_layout Plain Layout


\backslash
end{bmatrix},$$
\end_layout

\end_inset

若秩
\begin_inset Formula $B=$
\end_inset

 秩
\begin_inset Formula $A$
\end_inset

, 可以证明方程组
\begin_inset Formula
\[
\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1n}x_{n}=b_{1}\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=b_{2}\\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n}x_{2}+\cdots+a_{n}x_{n}=b_{n}
\end{cases}
\]

\end_inset

有解, 但反之不真.
\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
\begin{cases}
x_{1}+2x_{2}=3\\
2x_{1}+x_{2}=1
\end{cases}
\]

\end_inset


\begin_inset Formula
\[
\begin{aligned}A=\begin{bmatrix}1 & 2\\
2 & 1
\end{bmatrix}, & \overline{A}=\begin{bmatrix}1 & 2 & 3\\
2 & 1 & 1
\end{bmatrix},\end{aligned}
\]

\end_inset

因为秩
\begin_inset Formula $\overline{A}=$
\end_inset

 秩
\begin_inset Formula $A=2$
\end_inset

, 方程组有解, 但是
\begin_inset Formula
\[
\begin{vmatrix}1 & 2 & 3\\
2 & 1 & 1\\
3 & 1 & 0
\end{vmatrix}=2\neq0.
\]

\end_inset

因此
\begin_inset Formula
\[
B=\begin{bmatrix}1 & 2 & 3\\
2 & 1 & 1\\
3 & 1 & 0
\end{bmatrix}
\]

\end_inset

的秩是
\begin_inset Formula $3$
\end_inset

, 所以, 秩
\begin_inset Formula $B\ne$
\end_inset

 秩
\begin_inset Formula $A$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
若方程组
\begin_inset Formula
\[
\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1,n-1}x_{n-1}=a_{1n},\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2,n-1}x_{n-1}=a_{2n},\\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{n1}x_{1}+a_{n2}x_{2}+\cdots+a_{n,n-1}x_{n-1}=a_{nn},
\end{cases}
\]

\end_inset

有解, 则
\begin_inset Formula
\[
D=\begin{vmatrix}a_{11} & \cdots & a_{1n}\\
\vdots & \ddots & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{vmatrix}=0.
\]

\end_inset

但反之不真.

\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
\begin{cases}
x_{1}+2x_{2}=1\\
2x_{1}+4x_{2}=1\\
3x_{1}+6x_{2}=2
\end{cases}
\]

\end_inset


\begin_inset Formula
\[
\begin{aligned}D & =\begin{vmatrix}1 & 2 & 1\\
2 & 4 & 1\\
3 & 6 & 2
\end{vmatrix}=0,\end{aligned}
\]

\end_inset

但方程组无解, 因为秩
\begin_inset Formula $A=1$
\end_inset

, 秩
\begin_inset Formula $\overline{A}=2$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
设
\begin_inset Formula $D$
\end_inset

 为方程组
\begin_inset Formula
\begin{equation}
\begin{cases}
a_{11}x_{1}+a_{12}x_{2}+\cdots+a_{1}x_{n}=b_{1}\\
a_{21}x_{1}+a_{22}x_{2}+\cdots+a_{2n}x_{n}=b_{2}\\
\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\\
a_{n}x_{1}+a_{n}x_{2}+\cdots+a_{n}x_{n}=b_{n}
\end{cases}\label{eq:2}
\end{equation}

\end_inset

的系数行列式,
\begin_inset Formula $D_{i}$
\end_inset

 是把
\begin_inset Formula $D$
\end_inset

 中第
\begin_inset Formula $i$
\end_inset

 列换成常数项
\begin_inset Formula $b_{1},b_{2},\cdots,b_{n}$
\end_inset

, 所得到的行列式.
 当
\begin_inset Formula $D=D_{1}=D_{2}=\cdots=D_{n}=0$
\end_inset

 时, (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:2"
plural "false"
caps "false"
noprefix "false"

\end_inset

) 不一定有无穷多解.
\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}+x_{3}=1\\
2x_{1}+2x_{2}+2x_{3}=1\\
3x_{1}+3x_{2}+3x_{3}=0
\end{cases}
\]

\end_inset

显然有
\begin_inset Formula $D=D_{1}=D_{2}=D_{3}=0$
\end_inset

, 但无解, 因为 秩
\begin_inset Formula $A=1$
\end_inset

, 秩
\begin_inset Formula $\overline{A}=2$
\end_inset

, 秩
\begin_inset Formula $A\neq$
\end_inset

 秩
\begin_inset Formula $\overline{A}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Subsection
同解方程与非齐次方程组的导出组*
\end_layout

\begin_layout Example
设非齐次线性方程组 (I) 与 (II) 的导出组
\begin_inset Foot
status collapsed

\begin_layout Plain Layout
导出组是该非齐次线性方程组的导出齐次线性方程组或相应的齐次线性方程组的缩写, 指的是将非齐次线性方程组右端的常数项换为零, 得到的齐次线性方程组,
 即为齐次线性方程组的通解.
\end_layout

\end_inset

分别为 (I') 与 (II'), 若 (I) 有解, 而 (I) 与 (II) 同解, 则 (I') 与 (II') 同解.
\end_layout

\begin_layout Example
但条件 (I) 有解去掉, (I) 与 (II) 同解, 不一定有 (I') 与 (II') 同解.

\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}=1\\
2x_{1}+2x_{2}=3
\end{cases}\text{ 与 }\begin{cases}
\frac{1}{2}x_{1}+x_{2}=1\\
x_{1}+2x_{2}=1
\end{cases}
\]

\end_inset


\family roman
\series medium
\shape up
\size normal
\emph off
\bar no
\strikeout off
\xout off
\uuline off
\uwave off
\noun off
\color none
同解.

\family default
\series default
\shape default
\size default
\emph default
\bar default
\strikeout default
\xout default
\uuline default
\uwave default
\noun default
\color inherit
但其导出组不同解.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Example
设非齐次线性方程组 (I) 与 (II) 的导出组分别为 (I') 与 (II'); 若 (I') 与 (II') 同解, (I) 与 (II)
 不一定同解.

\end_layout

\begin_layout Example
例
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}=1\\
2x_{2}+x_{2}=1
\end{cases}
\]

\end_inset

与
\begin_inset Formula
\[
\begin{cases}
x_{1}+x_{2}=1\\
2x_{1}+x_{2}=3
\end{cases}
\]

\end_inset

的导出组同解, 但两个原方程组不同解.
\end_layout

\end_body
\end_document