简单的3x3矩阵逆码（C ++）
计算3x3矩阵逆的最简单方法是什么？
我只是在寻找一个简短的代码片段，它可以解决非奇异矩阵，可能使用Cramer的规则。它不需要高度优化。我更喜欢简单而不是速度。我宁愿不在其他库中链接。
13 个答案:
答案 0 :(得分：41)
这是batty的答案版本，但这会计算正确的逆。 batty的版本计算逆的转置。
// computes the inverse of a matrix m
double det = m(0, 0) * (m(1, 1) * m(2, 2) - m(2, 1) * m(1, 2)) -
m(0, 1) * (m(1, 0) * m(2, 2) - m(1, 2) * m(2, 0)) +
m(0, 2) * (m(1, 0) * m(2, 1) - m(1, 1) * m(2, 0));
double invdet = 1 / det;
Matrix33d minv; // inverse of matrix m
minv(0, 0) = (m(1, 1) * m(2, 2) - m(2, 1) * m(1, 2)) * invdet;
minv(0, 1) = (m(0, 2) * m(2, 1) - m(0, 1) * m(2, 2)) * invdet;
minv(0, 2) = (m(0, 1) * m(1, 2) - m(0, 2) * m(1, 1)) * invdet;
minv(1, 0) = (m(1, 2) * m(2, 0) - m(1, 0) * m(2, 2)) * invdet;
minv(1, 1) = (m(0, 0) * m(2, 2) - m(0, 2) * m(2, 0)) * invdet;
minv(1, 2) = (m(1, 0) * m(0, 2) - m(0, 0) * m(1, 2)) * invdet;
minv(2, 0) = (m(1, 0) * m(2, 1) - m(2, 0) * m(1, 1)) * invdet;
minv(2, 1) = (m(2, 0) * m(0, 1) - m(0, 0) * m(2, 1)) * invdet;
minv(2, 2) = (m(0, 0) * m(1, 1) - m(1, 0) * m(0, 1)) * invdet;
答案 1 :(得分：31)
为什么不尝试自己编写代码呢？把它作为挑战。 ：）
对于3×3矩阵
（来源：wolfram.com）
矩阵逆是
（来源：wolfram.com）
我假设您知道矩阵| A |的决定因素是
图像（c）Wolfram|Alpha和 mathworld.wolfram（06-11-09， 22.06）
答案 2 :(得分：26)
这段代码计算矩阵A：
的转置逆double determinant = +A(0,0)*(A(1,1)*A(2,2)-A(2,1)*A(1,2))
-A(0,1)*(A(1,0)*A(2,2)-A(1,2)*A(2,0))
+A(0,2)*(A(1,0)*A(2,1)-A(1,1)*A(2,0));
double invdet = 1/determinant;
result(0,0) = (A(1,1)*A(2,2)-A(2,1)*A(1,2))*invdet;
result(1,0) = -(A(0,1)*A(2,2)-A(0,2)*A(2,1))*invdet;
result(2,0) = (A(0,1)*A(1,2)-A(0,2)*A(1,1))*invdet;
result(0,1) = -(A(1,0)*A(2,2)-A(1,2)*A(2,0))*invdet;
result(1,1) = (A(0,0)*A(2,2)-A(0,2)*A(2,0))*invdet;
result(2,1) = -(A(0,0)*A(1,2)-A(1,0)*A(0,2))*invdet;
result(0,2) = (A(1,0)*A(2,1)-A(2,0)*A(1,1))*invdet;
result(1,2) = -(A(0,0)*A(2,1)-A(2,0)*A(0,1))*invdet;
result(2,2) = (A(0,0)*A(1,1)-A(1,0)*A(0,1))*invdet;
虽然问题规定了非奇异矩阵，但您可能仍想检查行列式是否等于零（或非常接近零）并以某种方式标记它以确保安全。
答案 3 :(得分：9)
对我们未知（雅虎）海报的所有应有的尊重，我会看到这样的代码并且只是在里面死一点。调试字母汤非常难以调试。任何地方的任何一个错字都可能真的毁了你的一整天。可悲的是，这个特殊的例子缺少带有下划线的变量。当我们有a_b-c_d * e_f-g_h时，它会更有趣。特别是在使用_和 - 具有相同像素长度的字体时。
根据他的建议接受Suvesh Pratapa，我注意到：
Given 3x3 matrix:
y0x0 y0x1 y0x2
y1x0 y1x1 y1x2
y2x0 y2x1 y2x2
Declared as double matrix [/*Y=*/3] [/*X=*/3];
（A）当拿一个3x3阵列的小调时，我们有4个感兴趣的值。较低的X / Y索引始终为0或1.较高的X / Y索引始终为1或2.始终！因此：
double determinantOfMinor( int theRowHeightY,
int theColumnWidthX,
const double theMatrix [/*Y=*/3] [/*X=*/3] )
{
int x1 = theColumnWidthX == 0 ? 1 : 0; /* always either 0 or 1 */
int x2 = theColumnWidthX == 2 ? 1 : 2; /* always either 1 or 2 */
int y1 = theRowHeightY == 0 ? 1 : 0; /* always either 0 or 1 */
int y2 = theRowHeightY == 2 ? 1 : 2; /* always either 1 or 2 */
return ( theMatrix [y1] [x1] * theMatrix [y2] [x2] )
- ( theMatrix [y1] [x2] * theMatrix [y2] [x1] );
}
（B）行列式现在是:(注意减号！）
double determinant( const double theMatrix [/*Y=*/3] [/*X=*/3] )
{
return ( theMatrix [0] [0] * determinantOfMinor( 0, 0, theMatrix ) )
- ( theMatrix [0] [1] * determinantOfMinor( 0, 1, theMatrix ) )
+ ( theMatrix [0] [2] * determinantOfMinor( 0, 2, theMatrix ) );
}
（C）现在反过来了：
bool inverse( const double theMatrix [/*Y=*/3] [/*X=*/3],
double theOutput [/*Y=*/3] [/*X=*/3] )
{
double det = determinant( theMatrix );
/* Arbitrary for now. This should be something nicer... */
if ( ABS(det) < 1e-2 )
{
memset( theOutput, 0, sizeof theOutput );
return false;
}
double oneOverDeterminant = 1.0 / det;
for ( int y = 0; y < 3; y ++ )
for ( int x = 0; x < 3; x ++ )
{
/* Rule is inverse = 1/det * minor of the TRANSPOSE matrix. *
* Note (y,x) becomes (x,y) INTENTIONALLY here! */
theOutput [y] [x]
= determinantOfMinor( x, y, theMatrix ) * oneOverDeterminant;
/* (y0,x1) (y1,x0) (y1,x2) and (y2,x1) all need to be negated. */
if( 1 == ((x + y) % 2) )
theOutput [y] [x] = - theOutput [y] [x];
}
return true;
}
用更低质量的测试代码完成它：
void printMatrix( const double theMatrix [/*Y=*/3] [/*X=*/3] )
{
for ( int y = 0; y < 3; y ++ )
{
cout << "[ ";
for ( int x = 0; x < 3; x ++ )
cout << theMatrix [y] [x] << " ";
cout << "]" << endl;
}
cout << endl;
}
void matrixMultiply( const double theMatrixA [/*Y=*/3] [/*X=*/3],
const double theMatrixB [/*Y=*/3] [/*X=*/3],
double theOutput [/*Y=*/3] [/*X=*/3] )
{
for ( int y = 0; y < 3; y ++ )
for ( int x = 0; x < 3; x ++ )
{
theOutput [y] [x] = 0;
for ( int i = 0; i < 3; i ++ )
theOutput [y] [x] += theMatrixA [y] [i] * theMatrixB [i] [x];
}
}
int
main(int argc, char **argv)
{
if ( argc > 1 )
SRANDOM( atoi( argv[1] ) );
double m[3][3] = { { RANDOM_D(0,1e3), RANDOM_D(0,1e3), RANDOM_D(0,1e3) },
{ RANDOM_D(0,1e3), RANDOM_D(0,1e3), RANDOM_D(0,1e3) },
{ RANDOM_D(0,1e3), RANDOM_D(0,1e3), RANDOM_D(0,1e3) } };
double o[3][3], mm[3][3];
if ( argc <= 2 )
cout << fixed << setprecision(3);
printMatrix(m);
cout << endl << endl;
SHOW( determinant(m) );
cout << endl << endl;
BOUT( inverse(m, o) );
printMatrix(m);
printMatrix(o);
cout << endl << endl;
matrixMultiply (m, o, mm );
printMatrix(m);
printMatrix(o);
printMatrix(mm);
cout << endl << endl;
}
有感：
您可能还想检测非常大的决定因素，因为舍入误差会影响您的准确性！
答案 4 :(得分：3)
大多数OpenGL工具包都提供了一个相当不错的（我认为）头文件，其中包含大多数2x2,3x3和4x4矩阵操作的宏。不是标准，但我已经在不同的地方看到过。
你可以在这里查看。最后你会发现2x2,3x3和4x4的倒数。
答案 5 :(得分：3)
如果您认真对待边缘案件，请不要自己尝试这样做。因此，虽然他们的许多天真/简单方法在理论上是精确的，但它们对于几乎奇异的矩阵可能具有令人讨厌的数值行为。特别是你可以得到取消/舍入错误，导致你得到任意不好的结果。
“正确”的方法是使用行和列旋转进行高斯消除，这样您总是除以最大的剩余数值。 （这对于NxN矩阵也是稳定的。）。请注意，单独行旋转并不能捕获所有不良情况。
然而IMO正确而快速地实现这一点并不值得花时间 - 使用经过良好测试的库并且只有一堆标题库。
答案 6 :(得分：3)
我刚刚创建了一个QMatrix类。它使用内置向量＆gt;容器。 QMatrix.h 它使用Jordan-Gauss方法计算方阵的倒数。
您可以按如下方式使用它：
#include "QMatrix.h"
#include <iostream>
int main(){
QMatrix<double> A(3,3,true);
QMatrix<double> Result = A.inverse()*A; //should give the idendity matrix
std::cout<<A.inverse()<<std::endl;
std::cout<<Result<<std::endl; // for checking
return 0;
}
反函数实现如下：
给定一个包含以下字段的类：
template<class T> class QMatrix{
public:
int rows, cols;
std::vector<std::vector<T> > A;
inverse（）函数：
template<class T>
QMatrix<T> QMatrix<T>:: inverse(){
Identity<T> Id(rows); //the Identity Matrix as a subclass of QMatrix.
QMatrix<T> Result = *this; // making a copy and transforming it to the Identity matrix
T epsilon = 0.000001;
for(int i=0;i<rows;++i){
//check if Result(i,i)==0, if true, switch the row with another
for(int j=i;j<rows;++j){
if(std::abs(Result(j,j))<epsilon) { //uses Overloading()(int int) to extract element from Result Matrix
Result.replace_rows(i,j+1); //switches rows i with j+1
}
else break;
}
// main part, making a triangular matrix
Id(i)=Id(i)*(1.0/Result(i,i));
Result(i)=Result(i)*(1.0/Result(i,i)); // Using overloading ()(int) to get a row form the matrix
for(int j=i+1;j<rows;++j){
T temp = Result(j,i);
Result(j) = Result(j) - Result(i)*temp;
Id(j) = Id(j) - Id(i)*temp; //doing the same operations to the identity matrix
Result(j,i)=0; //not necessary, but looks nicer than 10^-15
}
}
// solving a triangular matrix
for(int i=rows-1;i>0;--i){
for(int j=i-1;j>=0;--j){
T temp = Result(j,i);
Id(j) = Id(j) - temp*Id(i);
Result(j)=Result(j)-temp*Result(i);
}
}
return Id;
}
答案 7 :(得分：1)
我还推荐Ilmbase，它是OpenEXR的一部分。这是一组很好的模板2,3,4矢量和矩阵例程。
答案 8 :(得分：1)
# include <conio.h>
# include<iostream.h>
const int size = 9;
int main()
{
char ch;
do
{
clrscr();
int i, j, x, y, z, det, a[size], b[size];
cout << " **** MATRIX OF 3x3 ORDER ****"
<< endl
<< endl
<< endl;
for (i = 0; i <= size; i++)
a[i]=0;
for (i = 0; i < size; i++)
{
cout << "Enter "
<< i + 1
<< " element of matrix=";
cin >> a[i];
cout << endl
<<endl;
}
clrscr();
cout << "your entered matrix is "
<< endl
<<endl;
for (i = 0; i < size; i += 3)
cout << a[i]
<< " "
<< a[i+1]
<< " "
<< a[i+2]
<< endl
<<endl;
cout << "Transpose of given matrix is"
<< endl
<< endl;
for (i = 0; i < 3; i++)
cout << a[i]
<< " "
<< a[i+3]
<< " "
<< a[i+6]
<< endl
<< endl;
cout << "Determinent of given matrix is = ";
x = a[0] * (a[4] * a[8] -a [5] * a[7]);
y = a[1] * (a[3] * a[8] -a [5] * a[6]);
z = a[2] * (a[3] * a[7] -a [4] * a[6]);
det = x - y + z;
cout << det
<< endl
<< endl
<< endl
<< endl;
if (det == 0)
{
cout << "As Determinent=0 so it is singular matrix and its inverse cannot exist"
<< endl
<< endl;
goto quit;
}
b[0] = a[4] * a[8] - a[5] * a[7];
b[1] = a[5] * a[6] - a[3] * a[8];
b[2] = a[3] * a[7] - a[4] * a[6];
b[3] = a[2] * a[7] - a[1] * a[8];
b[4] = a[0] * a[8] - a[2] * a[6];
b[5] = a[1] * a[6] - a[0] * a[7];
b[6] = a[1] * a[5] - a[2] * a[4];
b[7] = a[2] * a[3] - a[0] * a[5];
b[8] = a[0] * a[4] - a[1] * a[3];
cout << "Adjoint of given matrix is"
<< endl
<< endl;
for (i = 0; i < 3; i++)
{
cout << b[i]
<< " "
<< b[i+3]
<< " "
<< b[i+6]
<< endl
<<endl;
}
cout << endl
<<endl;
cout << "Inverse of given matrix is "
<< endl
<< endl
<< endl;
for (i = 0; i < 3; i++)
{
cout << b[i]
<< "/"
<< det
<< " "
<< b[i+3]
<< "/"
<< det
<< " "
<< b[i+6]
<< "/"
<< det
<< endl
<<endl;
}
quit:
cout << endl
<< endl;
cout << "Do You want to continue this again press (y/yes,n/no)";
cin >> ch;
cout << endl
<< endl;
} /* end do */
while (ch == 'y');
getch ();
return 0;
}
答案 9 :(得分：0)
#include <iostream>
using namespace std;
int main()
{
double A11, A12, A13;
double A21, A22, A23;
double A31, A32, A33;
double B11, B12, B13;
double B21, B22, B23;
double B31, B32, B33;
cout << "Enter all number from left to right, from top to bottom, and press enter after every number: ";
cin >> A11;
cin >> A12;
cin >> A13;
cin >> A21;
cin >> A22;
cin >> A23;
cin >> A31;
cin >> A32;
cin >> A33;
B11 = 1 / ((A22 * A33) - (A23 * A32));
B12 = 1 / ((A13 * A32) - (A12 * A33));
B13 = 1 / ((A12 * A23) - (A13 * A22));
B21 = 1 / ((A23 * A31) - (A21 * A33));
B22 = 1 / ((A11 * A33) - (A13 * A31));
B23 = 1 / ((A13 * A21) - (A11 * A23));
B31 = 1 / ((A21 * A32) - (A22 * A31));
B32 = 1 / ((A12 * A31) - (A11 * A32));
B33 = 1 / ((A11 * A22) - (A12 * A21));
cout << B11 << "\t" << B12 << "\t" << B13 << endl;
cout << B21 << "\t" << B22 << "\t" << B23 << endl;
cout << B31 << "\t" << B32 << "\t" << B33 << endl;
return 0;
}
答案 10 :(得分：0)
//Title: Matrix Header File
//Writer: Say OL
//This is a beginner code not an expert one
//No responsibilty for any errors
//Use for your own risk
using namespace std;
int row,col,Row,Col;
double Coefficient;
//Input Matrix
void Input(double Matrix[9][9],int Row,int Col)
{
for(row=1;row<=Row;row++)
for(col=1;col<=Col;col++)
{
cout<<"e["<<row<<"]["<<col<<"]=";
cin>>Matrix[row][col];
}
}
//Output Matrix
void Output(double Matrix[9][9],int Row,int Col)
{
for(row=1;row<=Row;row++)
{
for(col=1;col<=Col;col++)
cout<<Matrix[row][col]<<"\t";
cout<<endl;
}
}
//Copy Pointer to Matrix
void CopyPointer(double (*Pointer)[9],double Matrix[9][9],int Row,int Col)
{
for(row=1;row<=Row;row++)
for(col=1;col<=Col;col++)
Matrix[row][col]=Pointer[row][col];
}
//Copy Matrix to Matrix
void CopyMatrix(double MatrixInput[9][9],double MatrixTarget[9][9],int Row,int Col)
{
for(row=1;row<=Row;row++)
for(col=1;col<=Col;col++)
MatrixTarget[row][col]=MatrixInput[row][col];
}
//Transpose of Matrix
double MatrixTran[9][9];
double (*(Transpose)(double MatrixInput[9][9],int Row,int Col))[9]
{
for(row=1;row<=Row;row++)
for(col=1;col<=Col;col++)
MatrixTran[col][row]=MatrixInput[row][col];
return MatrixTran;
}
//Matrix Addition
double MatrixAdd[9][9];
double (*(Addition)(double MatrixA[9][9],double MatrixB[9][9],int Row,int Col))[9]
{
for(row=1;row<=Row;row++)
for(col=1;col<=Col;col++)
MatrixAdd[row][col]=MatrixA[row][col]+MatrixB[row][col];
return MatrixAdd;
}
//Matrix Subtraction
double MatrixSub[9][9];
double (*(Subtraction)(double MatrixA[9][9],double MatrixB[9][9],int Row,int Col))[9]
{
for(row=1;row<=Row;row++)
for(col=1;col<=Col;col++)
MatrixSub[row][col]=MatrixA[row][col]-MatrixB[row][col];
return MatrixSub;
}
//Matrix Multiplication
int mRow,nCol,pCol,kcol;
double MatrixMult[9][9];
double (*(Multiplication)(double MatrixA[9][9],double MatrixB[9][9],int mRow,int nCol,int pCol))[9]
{
for(row=1;row<=mRow;row++)
for(col=1;col<=pCol;col++)
{
MatrixMult[row][col]=0.0;
for(kcol=1;kcol<=nCol;kcol++)
MatrixMult[row][col]+=MatrixA[row][kcol]*MatrixB[kcol][col];
}
return MatrixMult;
}
//Interchange Two Rows
double RowTemp[9][9];
double MatrixInter[9][9];
double (*(InterchangeRow)(double MatrixInput[9][9],int Row,int Col,int iRow,int jRow))[9]
{
CopyMatrix(MatrixInput,MatrixInter,Row,Col);
for(col=1;col<=Col;col++)
{
RowTemp[iRow][col]=MatrixInter[iRow][col];
MatrixInter[iRow][col]=MatrixInter[jRow][col];
MatrixInter[jRow][col]=RowTemp[iRow][col];
}
return MatrixInter;
}
//Pivote Downward
double MatrixDown[9][9];
double (*(PivoteDown)(double MatrixInput[9][9],int Row,int Col,int tRow,int tCol))[9]
{
CopyMatrix(MatrixInput,MatrixDown,Row,Col);
Coefficient=MatrixDown[tRow][tCol];
if(Coefficient!=1.0)
for(col=1;col<=Col;col++)
MatrixDown[tRow][col]/=Coefficient;
if(tRow<Row)
for(row=tRow+1;row<=Row;row++)
{
Coefficient=MatrixDown[row][tCol];
for(col=1;col<=Col;col++)
MatrixDown[row][col]-=Coefficient*MatrixDown[tRow][col];
}
return MatrixDown;
}
//Pivote Upward
double MatrixUp[9][9];
double (*(PivoteUp)(double MatrixInput[9][9],int Row,int Col,int tRow,int tCol))[9]
{
CopyMatrix(MatrixInput,MatrixUp,Row,Col);
Coefficient=MatrixUp[tRow][tCol];
if(Coefficient!=1.0)
for(col=1;col<=Col;col++)
MatrixUp[tRow][col]/=Coefficient;
if(tRow>1)
for(row=tRow-1;row>=1;row--)
{
Coefficient=MatrixUp[row][tCol];
for(col=1;col<=Col;col++)
MatrixUp[row][col]-=Coefficient*MatrixUp[tRow][col];
}
return MatrixUp;
}
//Pivote in Determinant
double MatrixPiv[9][9];
double (*(Pivote)(double MatrixInput[9][9],int Dim,int pTarget))[9]
{
CopyMatrix(MatrixInput,MatrixPiv,Dim,Dim);
for(row=pTarget+1;row<=Dim;row++)
{
Coefficient=MatrixPiv[row][pTarget]/MatrixPiv[pTarget][pTarget];
for(col=1;col<=Dim;col++)
{
MatrixPiv[row][col]-=Coefficient*MatrixPiv[pTarget][col];
}
}
return MatrixPiv;
}
//Determinant of Square Matrix
int dCounter,dRow;
double Det;
double MatrixDet[9][9];
double Determinant(double MatrixInput[9][9],int Dim)
{
CopyMatrix(MatrixInput,MatrixDet,Dim,Dim);
Det=1.0;
if(Dim>1)
{
for(dRow=1;dRow<Dim;dRow++)
{
dCounter=dRow;
while((MatrixDet[dRow][dRow]==0.0)&(dCounter<=Dim))
{
dCounter++;
Det*=-1.0;
CopyPointer(InterchangeRow(MatrixDet,Dim,Dim,dRow,dCounter),MatrixDet,Dim,Dim);
}
if(MatrixDet[dRow][dRow]==0)
{
Det=0.0;
break;
}
else
{
Det*=MatrixDet[dRow][dRow];
CopyPointer(Pivote(MatrixDet,Dim,dRow),MatrixDet,Dim,Dim);
}
}
Det*=MatrixDet[Dim][Dim];
}
else Det=MatrixDet[1][1];
return Det;
}
//Matrix Identity
double MatrixIdent[9][9];
double (*(Identity)(int Dim))[9]
{
for(row=1;row<=Dim;row++)
for(col=1;col<=Dim;col++)
if(row==col)
MatrixIdent[row][col]=1.0;
else
MatrixIdent[row][col]=0.0;
return MatrixIdent;
}
//Join Matrix to be Augmented Matrix
double MatrixJoin[9][9];
double (*(JoinMatrix)(double MatrixA[9][9],double MatrixB[9][9],int Row,int ColA,int ColB))[9]
{
Col=ColA+ColB;
for(row=1;row<=Row;row++)
for(col=1;col<=Col;col++)
if(col<=ColA)
MatrixJoin[row][col]=MatrixA[row][col];
else
MatrixJoin[row][col]=MatrixB[row][col-ColA];
return MatrixJoin;
}
//Inverse of Matrix
double (*Pointer)[9];
double IdentMatrix[9][9];
int Counter;
double MatrixAug[9][9];
double MatrixInv[9][9];
double (*(Inverse)(double MatrixInput[9][9],int Dim))[9]
{
Row=Dim;
Col=Dim+Dim;
Pointer=Identity(Dim);
CopyPointer(Pointer,IdentMatrix,Dim,Dim);
Pointer=JoinMatrix(MatrixInput,IdentMatrix,Dim,Dim,Dim);
CopyPointer(Pointer,MatrixAug,Row,Col);
for(Counter=1;Counter<=Dim;Counter++)
{
Pointer=PivoteDown(MatrixAug,Row,Col,Counter,Counter);
CopyPointer(Pointer,MatrixAug,Row,Col);
}
for(Counter=Dim;Counter>1;Counter--)
{
Pointer=PivoteUp(MatrixAug,Row,Col,Counter,Counter);
CopyPointer(Pointer,MatrixAug,Row,Col);
}
for(row=1;row<=Dim;row++)
for(col=1;col<=Dim;col++)
MatrixInv[row][col]=MatrixAug[row][col+Dim];
return MatrixInv;
}
//Gauss-Jordan Elemination
double MatrixGJ[9][9];
double VectorGJ[9][9];
double (*(GaussJordan)(double MatrixInput[9][9],double VectorInput[9][9],int Dim))[9]
{
Row=Dim;
Col=Dim+1;
Pointer=JoinMatrix(MatrixInput,VectorInput,Dim,Dim,1);
CopyPointer(Pointer,MatrixGJ,Row,Col);
for(Counter=1;Counter<=Dim;Counter++)
{
Pointer=PivoteDown(MatrixGJ,Row,Col,Counter,Counter);
CopyPointer(Pointer,MatrixGJ,Row,Col);
}
for(Counter=Dim;Counter>1;Counter--)
{
Pointer=PivoteUp(MatrixGJ,Row,Col,Counter,Counter);
CopyPointer(Pointer,MatrixGJ,Row,Col);
}
for(row=1;row<=Dim;row++)
for(col=1;col<=1;col++)
VectorGJ[row][col]=MatrixGJ[row][col+Dim];
return VectorGJ;
}
//Generalized Gauss-Jordan Elemination
double MatrixGGJ[9][9];
double VectorGGJ[9][9];
double (*(GeneralizedGaussJordan)(double MatrixInput[9][9],double VectorInput[9][9],int Dim,int vCol))[9]
{
Row=Dim;
Col=Dim+vCol;
Pointer=JoinMatrix(MatrixInput,VectorInput,Dim,Dim,vCol);
CopyPointer(Pointer,MatrixGGJ,Row,Col);
for(Counter=1;Counter<=Dim;Counter++)
{
Pointer=PivoteDown(MatrixGGJ,Row,Col,Counter,Counter);
CopyPointer(Pointer,MatrixGGJ,Row,Col);
}
for(Counter=Dim;Counter>1;Counter--)
{
Pointer=PivoteUp(MatrixGGJ,Row,Col,Counter,Counter);
CopyPointer(Pointer,MatrixGGJ,Row,Col);
}
for(row=1;row<=Row;row++)
for(col=1;col<=vCol;col++)
VectorGGJ[row][col]=MatrixGGJ[row][col+Dim];
return VectorGGJ;
}
//Matrix Sparse, Three Diagonal Non-Zero Elements
double MatrixSpa[9][9];
double (*(Sparse)(int Dimension,double FirstElement,double SecondElement,double ThirdElement))[9]
{
MatrixSpa[1][1]=SecondElement;
MatrixSpa[1][2]=ThirdElement;
MatrixSpa[Dimension][Dimension-1]=FirstElement;
MatrixSpa[Dimension][Dimension]=SecondElement;
for(int Counter=2;Counter<Dimension;Counter++)
{
MatrixSpa[Counter][Counter-1]=FirstElement;
MatrixSpa[Counter][Counter]=SecondElement;
MatrixSpa[Counter][Counter+1]=ThirdElement;
}
return MatrixSpa;
}
将上述代码复制并保存为Matrix.h，然后尝试以下代码：
#include<iostream>
#include<conio.h>
#include"Matrix.h"
int Dim;
double Matrix[9][9];
int main()
{
cout<<"Enter your matrix dimension: ";
cin>>Dim;
Input(Matrix,Dim,Dim);
cout<<"Your matrix:"<<endl;
Output(Matrix,Dim,Dim);
cout<<"The inverse:"<<endl;
Output(Inverse(Matrix,Dim),Dim,Dim);
getch();
}
答案 11 :(得分：0)
//Function for inverse of the input square matrix 'J' of dimension 'dim':
vector<vector<double > > inverseVec33(vector<vector<double > > J, int dim)
{
//Matrix of Minors
vector<vector<double > > invJ(dim,vector<double > (dim));
for(int i=0; i<dim; i++)
{
for(int j=0; j<dim; j++)
{
invJ[i][j] = (J[(i+1)%dim][(j+1)%dim]*J[(i+2)%dim][(j+2)%dim] -
J[(i+2)%dim][(j+1)%dim]*J[(i+1)%dim][(j+2)%dim]);
}
}
//determinant of the matrix:
double detJ = 0.0;
for(int j=0; j<dim; j++)
{ detJ += J[0][j]*invJ[0][j];}
//Inverse of the given matrix.
vector<vector<double > > invJT(dim,vector<double > (dim));
for(int i=0; i<dim; i++)
{
for(int j=0; j<dim; j++)
{
invJT[i][j] = invJ[j][i]/detJ;
}
}
return invJT;
}
void main()
{
//given matrix:
vector<vector<double > > Jac(3,vector<double > (3));
Jac[0][0] = 1; Jac[0][1] = 2; Jac[0][2] = 6;
Jac[1][0] = -3; Jac[1][1] = 4; Jac[1][2] = 3;
Jac[2][0] = 5; Jac[2][1] = 1; Jac[2][2] = -4;`
//Inverse of the matrix Jac:
vector<vector<double > > JacI(3,vector<double > (3));
//call function and store inverse of J as JacI:
JacI = inverseVec33(Jac,3);
}
答案 12 :(得分：-1)
我继续在python中编写它，因为我觉得它比c ++中的这个问题更具可读性。功能顺序是按照this video手动解决此问题的操作顺序。只需导入它并在矩阵上调用“print_invert”。
def print_invert (matrix):
i_matrix = invert_matrix (matrix)
for line in i_matrix:
print (line)
return
def invert_matrix (matrix):
determinant = str (determinant_of_3x3 (matrix))
cofactor = make_cofactor (matrix)
trans_matrix = transpose_cofactor (cofactor)
trans_matrix[:] = [[str (element) +'/'+ determinant for element in row] for row in trans_matrix]
return trans_matrix
def determinant_of_3x3 (matrix):
multiplication = 1
neg_multiplication = 1
total = 0
for start_column in range (3):
for row in range (3):
multiplication *= matrix[row][(start_column+row)%3]
neg_multiplication *= matrix[row][(start_column-row)%3]
total += multiplication - neg_multiplication
multiplication = neg_multiplication = 1
if total == 0:
total = 1
return total
def make_cofactor (matrix):
cofactor = [[0,0,0],[0,0,0],[0,0,0]]
matrix_2x2 = [[0,0],[0,0]]
# For each element in matrix...
for row in range (3):
for column in range (3):
# ...make the 2x2 matrix in this inner loop
matrix_2x2 = make_2x2_from_spot_in_3x3 (row, column, matrix)
cofactor[row][column] = determinant_of_2x2 (matrix_2x2)
return flip_signs (cofactor)
def make_2x2_from_spot_in_3x3 (row, column, matrix):
c_count = 0
r_count = 0
matrix_2x2 = [[0,0],[0,0]]
# ...make the 2x2 matrix in this inner loop
for inner_row in range (3):
for inner_column in range (3):
if row is not inner_row and inner_column is not column:
matrix_2x2[r_count % 2][c_count % 2] = matrix[inner_row][inner_column]
c_count += 1
if row is not inner_row:
r_count += 1
return matrix_2x2
def determinant_of_2x2 (matrix):
total = matrix[0][0] * matrix [1][1]
return total - (matrix [1][0] * matrix [0][1])
def flip_signs (cofactor):
sign_pos = True
# For each element in matrix...
for row in range (3):
for column in range (3):
if sign_pos:
sign_pos = False
else:
cofactor[row][column] *= -1
sign_pos = True
return cofactor
def transpose_cofactor (cofactor):
new_cofactor = [[0,0,0],[0,0,0],[0,0,0]]
for row in range (3):
for column in range (3):
new_cofactor[column][row] = cofactor[row][column]
return new_cofactor
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