设线性方程组为
{
a
11
x
1
+
a
11
x
2
+
⋯
+
a
1
n
x
n
=
b
1
a
21
x
1
+
a
22
x
2
+
⋯
+
a
2
n
x
n
=
b
2
.
.
.
.
.
.
a
n
1
x
1
+
a
n
2
x
2
+
⋯
+
a
n
n
x
n
=
b
n
\begin{cases} a_{11}x_1+a_{11}x_2+\cdots+a_{1n}x_n=b_1\\ a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n=b_2\\ ……\\ a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n=b_n\\ \end{cases}
⎩
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎧
a
1
1
x
1
+
a
1
1
x
2
+
⋯
+
a
1
n
x
n
=
b
1
a
2
1
x
1
+
a
2
2
x
2
+
⋯
+
a
2
n
x
n
=
b
2
.
.
.
.
.
.
a
n
1
x
1
+
a
n
2
x
2
+
⋯
+
a
n
n
x
n
=
b
n
其系数行列式为
D
=
∣
a
11
a
12
⋯
a
1
n
a
21
a
22
⋯
a
2
n
⋮
⋮
⋱
⋮
a
n
1
a
n
2
⋯
a
n
n
∣
≠
0
D=\begin{vmatrix} a_{11}&a_{12}&\cdots&a_{1n} \\ a_{21}&a_{22}&\cdots&a_{2n} \\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\cdots&a_{nn} \end{vmatrix} \not=0
D
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
a
1
1
a
2
1
⋮
a
n
1
a
1
2
a
2
2
⋮
a
n
2
⋯
⋯
⋱
⋯
a
1
n
a
2
n
⋮
a
n
n
∣
∣
∣
∣
∣
∣
∣
∣
∣
=
0
则该线性方程组有且仅有唯一解：
x
1
=
D
1
D
,
x
2
=
D
2
D
,
⋯
,
x
n
=
D
n
D
x_1=\frac{D_1}{D},x_2=\frac{D_2}{D},\cdots,x_n=\frac{D_n}{D}
x
1
=
D
D
1
,
x
2
=
D
D
2
,
⋯
,
x
n
=
D
D
n
其中
D
j
(
j
=
1
,
2
,
…
,
n
)
D_j(j=1,2,\ldots,n)
D
j
(
j
=
1
,
2
,
…
,
n
)
是把系数行列式
D
D
D
中的第
j
j
j
列的元素用常数项
b
1
,
b
2
,
…
,
b
n
b_1,b_2,\ldots,b_n
b
1
,
b
2
,
…
,
b
n
代替后得到的
n
n
n
阶行列式，即
D
j
=
∣
a
11
⋯
a
1
,
j
−
1
b
1
a
1
,
j
+
1
⋯
a
1
n
a
21
⋯
a
2
,
j
−
1
b
1
a
1
,
j
+
1
⋯
a
2
n
⋮
⋮
⋮
⋮
a
n
1
⋯
a
n
,
j
−
1
b
1
a
1
,
j
+
1
⋯
a
n
n
∣
D_j = \begin{vmatrix} a_{11}&\cdots&a_{1,j-1}&b_1&a_{1,j+1}&\cdots&a_{1n} \\ a_{21}&\cdots&a_{2,j-1}&b_1&a_{1,j+1}&\cdots&a_{2n} \\ \vdots&\vdots&\vdots&\vdots\\ a_{n1}&\cdots&a_{n,j-1}&b_1&a_{1,j+1}&\cdots&a_{nn} \end{vmatrix}
D
j
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
a
1
1
a
2
1
⋮
a
n
1
⋯
⋯
⋮
⋯
a
1
,
j
−
1
a
2
,
j
−
1
⋮
a
n
,
j
−
1
b
1
b
1
⋮
b
1
a
1
,
j
+
1
a
1
,
j
+
1
a
1
,
j
+
1
⋯
⋯
⋯
a
1
n
a
2
n
a
n
n
∣
∣
∣
∣
∣
∣
∣
∣
∣
举例
{
2
x
1
+
3
x
2
−
5
x
3
=
3
x
1
−
2
x
2
+
x
3
=
0
3
x
1
+
x
2
+
3
x
3
=
7
\begin{cases} 2x_1+3x_2-5x_3=3\\ x_1-2x_2+x_3=0\\ 3x_1+x_2+3x_3=7\\ \end{cases}
⎩
⎪
⎨
⎪
⎧
2
x
1
+
3
x
2
−
5
x
3
=
3
x
1
−
2
x
2
+
x
3
=
0
3
x
1
+
x
2
+
3
x
3
=
7
解：
D
=
∣
2
3
−
5
1
−
2
1
3
1
3
∣
=
∣
2
7
−
7
1
0
0
3
7
0
∣
=
−
∣
7
−
7
7
0
∣
=
−
49
≠
0
D= \begin{vmatrix} 2&3&-5\\ 1 & -2 & 1\\ 3 & 1 &3\\ \end{vmatrix}= \begin{vmatrix} 2 & 7 & -7\\ 1 & 0 & 0\\ 3 & 7 & 0\\ \end{vmatrix}=- \begin{vmatrix} 7&-7\\ 7 &0\\ \end{vmatrix}=-49\not=0
D
=
∣
∣
∣
∣
∣
∣
2
1
3
3
−
2
1
−
5
1
3
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
2
1
3
7
0
7
−
7
0
0
∣
∣
∣
∣
∣
∣
=
−
∣
∣
∣
∣
7
7
−
7
0
∣
∣
∣
∣
=
−
4
9
=
0
D
1
=
∣
3
3
−
5
0
−
2
1
7
1
3
∣
=
∣
2
7
−
7
1
0
0
3
7
0
∣
=
−
∣
3
−
7
7
7
∣
=
−
70
D_1= \begin{vmatrix} 3&3&-5\\ 0 & -2 & 1\\ 7 & 1 &3\\ \end{vmatrix}= \begin{vmatrix} 2 & 7 & -7\\ 1 & 0 & 0\\ 3 & 7 & 0\\ \end{vmatrix}=- \begin{vmatrix} 3 & -7\\ 7 & 7\\ \end{vmatrix}=-70
D
1
=
∣
∣
∣
∣
∣
∣
3
0
7
3
−
2
1
−
5
1
3
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
2
1
3
7
0
7
−
7
0
0
∣
∣
∣
∣
∣
∣
=
−
∣
∣
∣
∣
3
7
−
7
7
∣
∣
∣
∣
=
−
7
0
x
1
=
D
1
D
=
−
70
−
49
=
10
7
x_1=\frac{D_1}{D}=\frac{-70}{-49}=\frac{10}{7}
x
1
=
D
D
1
=
−
4
9
−
7
0
=
7
1
0
其余两个解同理。