行列式题目

发布网友 发布时间：2022-04-25 07:29

我来回答

共1个回答

热心网友 时间：2023-11-06 20:51

先求系数矩阵行列式D=
1 1 1
a b c
bc ca ab
=
1 0 0
a b-a c-a
bc ca-bc ab-bc
=
(b-a)(c-a)*
1 1
-c -b
=
(b-a)(c-a)(c-b)

然后求行列式D1
=
a+b+c 1 1
a^2+b^2+c^2 b c
3abc ca ab
=
a+b+c 0 1
a^2+b^2+c^2 b-c c
3abc ca-ab ab

=(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3abc a ab
=a(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3bc 1 b

=a(c-b)*
a+b 0 1
a^2+b^2 -1 c
2bc 1 b
=a(c-b)*
a+b 0 1
a^2+b^2+2bc 0 b+c
2bc 1 b
=-a(c-b)*
a+b 1
a^2+b^2+2bc b+c
=-a(c-b)*[(a+b)(b+c)-(a^2+b^2+2bc)]
=-a(c-b)*[ab+ac-(a^2+bc)]
=-a(c-b)*[(a-c)(b-a)]
=-a(c-b)(b-a)(c-a)

因此x=D1/D=a
类似地，求出
y=D2/D=b
z=D3/D=c

热心网友 时间：2023-11-06 20:51

先求系数矩阵行列式D=
1 1 1
a b c
bc ca ab
=
1 0 0
a b-a c-a
bc ca-bc ab-bc
=
(b-a)(c-a)*
1 1
-c -b
=
(b-a)(c-a)(c-b)

然后求行列式D1
=
a+b+c 1 1
a^2+b^2+c^2 b c
3abc ca ab
=
a+b+c 0 1
a^2+b^2+c^2 b-c c
3abc ca-ab ab

=(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3abc a ab
=a(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3bc 1 b

=a(c-b)*
a+b 0 1
a^2+b^2 -1 c
2bc 1 b
=a(c-b)*
a+b 0 1
a^2+b^2+2bc 0 b+c
2bc 1 b
=-a(c-b)*
a+b 1
a^2+b^2+2bc b+c
=-a(c-b)*[(a+b)(b+c)-(a^2+b^2+2bc)]
=-a(c-b)*[ab+ac-(a^2+bc)]
=-a(c-b)*[(a-c)(b-a)]
=-a(c-b)(b-a)(c-a)

因此x=D1/D=a
类似地，求出
y=D2/D=b
z=D3/D=c

热心网友 时间：2023-11-06 20:51

先求系数矩阵行列式D=
1 1 1
a b c
bc ca ab
=
1 0 0
a b-a c-a
bc ca-bc ab-bc
=
(b-a)(c-a)*
1 1
-c -b
=
(b-a)(c-a)(c-b)

然后求行列式D1
=
a+b+c 1 1
a^2+b^2+c^2 b c
3abc ca ab
=
a+b+c 0 1
a^2+b^2+c^2 b-c c
3abc ca-ab ab

=(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3abc a ab
=a(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3bc 1 b

=a(c-b)*
a+b 0 1
a^2+b^2 -1 c
2bc 1 b
=a(c-b)*
a+b 0 1
a^2+b^2+2bc 0 b+c
2bc 1 b
=-a(c-b)*
a+b 1
a^2+b^2+2bc b+c
=-a(c-b)*[(a+b)(b+c)-(a^2+b^2+2bc)]
=-a(c-b)*[ab+ac-(a^2+bc)]
=-a(c-b)*[(a-c)(b-a)]
=-a(c-b)(b-a)(c-a)

因此x=D1/D=a
类似地，求出
y=D2/D=b
z=D3/D=c

热心网友 时间：2023-11-06 20:51

先求系数矩阵行列式D=
1 1 1
a b c
bc ca ab
=
1 0 0
a b-a c-a
bc ca-bc ab-bc
=
(b-a)(c-a)*
1 1
-c -b
=
(b-a)(c-a)(c-b)

然后求行列式D1
=
a+b+c 1 1
a^2+b^2+c^2 b c
3abc ca ab
=
a+b+c 0 1
a^2+b^2+c^2 b-c c
3abc ca-ab ab

=(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3abc a ab
=a(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3bc 1 b

=a(c-b)*
a+b 0 1
a^2+b^2 -1 c
2bc 1 b
=a(c-b)*
a+b 0 1
a^2+b^2+2bc 0 b+c
2bc 1 b
=-a(c-b)*
a+b 1
a^2+b^2+2bc b+c
=-a(c-b)*[(a+b)(b+c)-(a^2+b^2+2bc)]
=-a(c-b)*[ab+ac-(a^2+bc)]
=-a(c-b)*[(a-c)(b-a)]
=-a(c-b)(b-a)(c-a)

因此x=D1/D=a
类似地，求出
y=D2/D=b
z=D3/D=c

热心网友 时间：2023-11-06 20:51

先求系数矩阵行列式D=
1 1 1
a b c
bc ca ab
=
1 0 0
a b-a c-a
bc ca-bc ab-bc
=
(b-a)(c-a)*
1 1
-c -b
=
(b-a)(c-a)(c-b)

然后求行列式D1
=
a+b+c 1 1
a^2+b^2+c^2 b c
3abc ca ab
=
a+b+c 0 1
a^2+b^2+c^2 b-c c
3abc ca-ab ab

=(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3abc a ab
=a(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3bc 1 b

=a(c-b)*
a+b 0 1
a^2+b^2 -1 c
2bc 1 b
=a(c-b)*
a+b 0 1
a^2+b^2+2bc 0 b+c
2bc 1 b
=-a(c-b)*
a+b 1
a^2+b^2+2bc b+c
=-a(c-b)*[(a+b)(b+c)-(a^2+b^2+2bc)]
=-a(c-b)*[ab+ac-(a^2+bc)]
=-a(c-b)*[(a-c)(b-a)]
=-a(c-b)(b-a)(c-a)

因此x=D1/D=a
类似地，求出
y=D2/D=b
z=D3/D=c

热心网友 时间：2023-11-06 20:51

先求系数矩阵行列式D=
1 1 1
a b c
bc ca ab
=
1 0 0
a b-a c-a
bc ca-bc ab-bc
=
(b-a)(c-a)*
1 1
-c -b
=
(b-a)(c-a)(c-b)

然后求行列式D1
=
a+b+c 1 1
a^2+b^2+c^2 b c
3abc ca ab
=
a+b+c 0 1
a^2+b^2+c^2 b-c c
3abc ca-ab ab

=(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3abc a ab
=a(c-b)*
a+b+c 0 1
a^2+b^2+c^2 -1 c
3bc 1 b

=a(c-b)*
a+b 0 1
a^2+b^2 -1 c
2bc 1 b
=a(c-b)*
a+b 0 1
a^2+b^2+2bc 0 b+c
2bc 1 b
=-a(c-b)*
a+b 1
a^2+b^2+2bc b+c
=-a(c-b)*[(a+b)(b+c)-(a^2+b^2+2bc)]
=-a(c-b)*[ab+ac-(a^2+bc)]
=-a(c-b)*[(a-c)(b-a)]
=-a(c-b)(b-a)(c-a)

因此x=D1/D=a
类似地，求出
y=D2/D=b
z=D3/D=c