解待定系数多元一次方程组用矩阵行列式方法思路
已知(X0,Y0),(X1,Y1),(X2,Y2)
y=ax^2+bx+c
求
/ a=?
| b=?
\ c=?
解:
带入状态值得到如下
Y0=aX0^2+bX0+C aX0^2+bX0+C-Y0=0
Y1=aX1^2+bX1+C ==> aX1^2+bX1+C-Y1=0
Y2=aX2^2+bX2+C aX2^2+bX2+C-Y2=0
整理成矩阵行列式
/ X0^2 X0 1 -Y0 \ ① / a \
| X1^2 X1 1 -Y1 | ② =| b |
\ X2^2 X2 1 -Y2 / ③ \ c /
通过消去法化简行列式得到结果
①-②:
X0^2 X0 1 -Y0
-X1^2 -X1 -1 Y1
-------------------------------
X1^2-X0^2 X1-X0 0 Y1-Y0
①-③:
X0^2 X0 1 -Y0
-X2^2 -X2 -1 Y2
-------------------------------
X2^2-X0^2 X2-X0 0 Y2-Y0
重整理列出行列式
/ X0^2 X0 1 -Y0 \ ①
| X1^2-X0^2 X1-X0 0 Y1-Y0 | ②
\ X2^2-X0^2 X2-X0 0 Y2-Y0 / ③
再化简:
②/(X1-X0)
(X1^2-X0^2)/(X1-X0) 1 0 (Y1-Y0)/(X1-X0)
③/(X2-X0)
(X2^2-X0^2)/(X2-X0) 1 0 (Y2-Y0)/(X2-X0)
再整理行列式
/ X0^2 X0 1 -Y0 \ ①
| (X1^2-X0^2)/(X1-X0) 1 0 (Y1-Y0)/(X1-X0) | ②
\ (X2^2-X0^2)/(X2-X0) 1 0 (Y2-Y0)/(X2-X0) / ③
②-③
(X1^2-X0^2)/(X1-X0)-(X2^2-X0^2)/(X2-X0) 0 0 (Y1-Y0)/(X1-X0)-(Y2-Y0)/(X2-X0)
重列
/ X0^2 X0 1 -Y0 \ ①
| (X1^2-X0^2)/(X1-X0) 1 0 (Y1-Y0)/(X1-X0) | ②
\ (X1^2-X0^2)/(X1-X0)-(X2^2-X0^2)/(X2-X0) 0 0 (Y1-Y0)/(X1-X0)-(Y2-Y0)/(X2-X0) / ③
③/(X1^2-X0^2)/(X1-X0)-(X2^2-X0^2)/(X2-X0)
1 0 0 [(Y1-Y0)/(X1-X0)-(Y2-Y0)/(X2-X0)]/[(X1^2-X0^2)/(X1-X0)-(X2^2-X0^2)/(X2-X0)]
重列
/ X0^2 X0 1 -Y0 \ ①
| (X1^2-X0^2)/(X1-X0) 1 0 (Y1-Y0)/(X1-X0) | ②
\ 1 0 0 [(Y1-Y0)/(X1-X0)-(Y2-Y0)/(X2-X0)]/[(X1^2-X0^2)/(X1-X0)-(X2^2-X0^2)/(X2-X0)] / ③
a+[(Y1-Y0)/(X1-X0)-(Y2-Y0)/(X2-X0)]/[(X1^2-X0^2)/(X1-X0)-(X2^2-X0^2)/(X2-X0)]=0
同理化简可以得到b,c
①-②*X0 得到
X0^2*X0 0 1 -Y0-[(Y1-Y0)/(X1-X0)]*X0 ===>① /*把新值赋给①
重列
/ X0^2*X0 0 1 -Y0-[(Y1-Y0)/(X1-X0)]*X0 \ ①
| (X1^2-X0^2)/(X1-X0) 1 0 (Y1-Y0)/(X1-X0) | ②
\ 1 0 0 [(Y1-Y0)/(X1-X0)-(Y2-Y0)/(X2-X0)]/[(X1^2-X0^2)/(X1-X0)-(X2^2-X0^2)/(X2-X0)] / ③
[①*(X1^2-X0^2)/(X1-X0)]-[②*X0^2*
X0],得到的结果消去a,剩下b带系数,可以得到b的值了,
.......开源,没时间写了.........等勇士补充呀