88 lines
1.9 KiB
Go
88 lines
1.9 KiB
Go
// InverseA
|
||
/*
|
||
------------------------------------------------------
|
||
作者 : Black Ghost
|
||
日期 : 2018-11-20
|
||
版本 : 0.0.0
|
||
------------------------------------------------------
|
||
求矩阵逆的列主元消去法
|
||
理论：
|
||
参考 李信真, 车刚明, 欧阳洁, 等. 计算方法. 西北工业大学
|
||
出版社, 2000, pp 51.
|
||
------------------------------------------------------
|
||
输入 :
|
||
a 矩阵
|
||
输出 :
|
||
sol 解值
|
||
err 解出标志：false-未解出或达到步数上限；
|
||
true-全部解出
|
||
------------------------------------------------------
|
||
*/
|
||
|
||
package goNum
|
||
|
||
// InverseA 求矩阵逆的列主元消去法
|
||
func InverseA(a [][]float64) ([][]float64, bool) {
|
||
/*
|
||
求矩阵逆的列主元消去法
|
||
输入 :
|
||
a 矩阵
|
||
输出 :
|
||
sol 解值
|
||
err 解出标志：false-未解出或达到步数上限；
|
||
true-全部解出
|
||
*/
|
||
|
||
var err bool = false
|
||
n := len(a)
|
||
temp0, _ := E_Mat(n)
|
||
b := temp0
|
||
sol := b
|
||
temp1 := make([]float64, n)
|
||
|
||
//判断是否方阵
|
||
if len(a) != len(a[0]) {
|
||
return sol, err
|
||
}
|
||
|
||
//主元消去
|
||
for i := 0; i < n; i++ {
|
||
//求第i列的主元素并调整行顺序
|
||
acol := make([]float64, n-i)
|
||
for icol := i; icol < n; icol++ {
|
||
acol[icol-i] = a[icol][i]
|
||
}
|
||
_, ii, _ := MaxAbs(acol)
|
||
if ii+i != i {
|
||
temp1 = a[ii+i]
|
||
a[ii+i] = a[i]
|
||
a[i] = temp1
|
||
temp1 = b[ii+i]
|
||
b[ii+i] = b[i]
|
||
b[i] = temp1
|
||
}
|
||
|
||
//列消去
|
||
//本行主元置一
|
||
mul := a[i][i]
|
||
for j := 0; j < n; j++ {
|
||
a[i][j] = a[i][j] / mul
|
||
b[i][j] = b[i][j] / mul
|
||
}
|
||
//其它列置零
|
||
for j := 0; j < n; j++ {
|
||
if j != i {
|
||
mul = a[j][i] / a[i][i]
|
||
for k := 0; k < n; k++ {
|
||
a[j][k] = a[j][k] - a[i][k]*mul
|
||
b[j][k] = b[j][k] - b[i][k]*mul
|
||
}
|
||
}
|
||
}
|
||
}
|
||
|
||
sol = b
|
||
err = true
|
||
return sol, err
|
||
}
|