行列式方程式

2015-06-03 3:29 am
1.
|x+2 3 4|
| 2 x+3 4| =0 解行列式方程式
|2 3 x+4|



2.若a-b+c=0 求行列式
| b a -2b |
| -c -b 2a | 之值
| -a c 2c |



3.解行列式方程式
|x-1 x x+1 |
|x x+1 x-1 | =0
|x+1 X-1 x |
更新1:

求1.3有無其他方法 數字大容易算錯@@ 謝謝

回答 (2)

2015-06-03 4:59 am
✔ 最佳答案
1.
令a=x+4
|a-2 3 4|
| 2 a-1 4| =0
|2 3 a l

==>a^3-3a^2-24a+80=0

==>(a+5)(a-4)^2=0

==>a=-5或a=4(重根)

-5=x+4或 4=x+4
==>x=-9或x=0

3.
展開
==>x(x-1)(x+1)+x(x-1)(x+1)+x(x-1)(x+1)
-(x+1)^3-x^3-(x-1)^3=0

==>-9x=0

==>x=0




2015-06-02 21:06:37 補充:
2.
令a=1,b=2,c=1
l2 1 4|
|-1 -2 2 l
|-1 1 2 l

=-8-2-4-8+2-4

=-24

2015-06-02 21:08:41 補充:
1.
a^3-3a^2-24a+80=0

用因式定理求根再長除法除

2015-06-02 21:17:14 補充:
2.更正

令a=1,b=2,c=1
l2 1 -2|
|-1 -2 1l *2
|-1 1 1 l

=2*(-4-1+2+4+1-2)

=0

2015-06-02 22:46:47 補充:
1.用高斯消去法

|x+2 3 4|
| -x x 0|
|2 3 x+4|

=
|x 0 -x|
| -x x 0|
|2 3 x+4|

=
|1 0 -1|
| -1 1 0|
|2 3 x+4|*x^2

=x^2*(x+9)=0
==>x=0或x=-9

2015-06-02 22:53:18 補充:
3.用高斯消去法

|-1 -1 2|
|x x+1 x-1 | =0
|x+1 X-1 x |

=
|-1 -1 2|
|x x+1 x-1 | =0
|1 -2 1 |

=-(x-1)-(x-1)-4x-2(x+1)+x-2(x-1)=0

==>-9x=0
==>x=0
2015-06-03 11:08 pm
2.若a-b+c=0 求行列式
| b a -2b |
| -c -b 2a | 之值
| -a c 2c |


R1 → R1+R2-R3 得 -(a-b+c); a-b+c; 2(a-b+c), 全為 0.
故該行列式值為 0.


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