復(fù)數(shù)矩陣求逆運(yùn)算

1、1實(shí)、虛部矩陣均可逆的復(fù):數(shù)矩陣求 逆法諛C是個(gè)八階史數(shù)矩陣，將C寫成C = A 4 加的形式，其中實(shí)部矩陣4,虛部矩陣是越階可逆實(shí) 矩陣.巧由于I =1，朋八T 1 + 8人/尸=(A +出-酒)口4 A)而f ,小 + jmL =(B 1 - iA-+ A) T1 =(ttX - - X) (Bti1) (BXAB +4一)%”二(H-H iA)BlTA8, + A-)B =(1 - iAti)(A 4 ZM廣 =(A * OAO)1 - iA B(A bA-1 U)-1故此，且數(shù)矩陣C的逆矩陣為c1 = (a + 出廠 = (a + 8a .”廠 iAB(A + A)”fl)所以按照以上

2、介紹的方法，用C語言實(shí)現(xiàn)過程如下： #include ttinclude float getA(float *arcs, int n)按第一行展開計(jì)算|A| (if (n=l)( return arcs00;)int i, j, k;float ans = 0;float *temp = (float *)malloc(sizeof(float *) * n); for(i = 0; i n; i+)temp Li = (float *)malloc(sizeof(float *) * n);)for(i=0;in;i+)(for(j=0;jn-l;j+)(for(k_0;k=i)?k+l：k

3、;float t = getA(temp, n-l);if(i%2=0)(ans +二 arcs0i*t;else(ans -= arcs0i*t;)return ans;)void getAStart (float *arcs, int n,float *ans)(if (n=l)ans00 = 1;return;)int i, j, k, t;float num = getA(arcs, n);float *tcmp = (float *)malloc(sizeof(float *) * n);for(i = 0; i n; i+)tempi = (float *)malloc(sizeo

4、f(float *) * n);)for (i=0;in;i+)(for (j=0;jn;j+)(for (k=0;kn-l;k+)for (t=0;t=i? k+l:k t=j? t+1: t;/找出除第i行第j列的(n-1)矩陣計(jì)算arcsij的余子式計(jì)算arcsij的余子式ansj i = getA(temp, n-l) / num;ansEj i二-ansj i;計(jì)算arcs i j的代數(shù)余子式)void Matrix_Add(float *a, float *b, float *c, int n) 矩陣相加int i, j;for(i = 0; i n; i+) for(j = 0;

5、 j n; j+) )c = a * bvoid Matrix Mul(float *a, float *b, float *c, int n) int i, j, k;for (i = 0;i3;i+)(for(j =0;j3;j+)= 0; 清零for (k = 0;k3;k+) (ci j += ai k*bk j;)void Inverse(float *a, float *b, int n) int i, j;float *c = (float mal1oc(sizeof(float *) * n); for(i = 0; i n; i+)ci = (float )malloc(si

6、zeof(float) * n);)float *d = (float *)malloc(sizeof(float *) * n); for(i _ 0; i n; i+) di = (float )malloc(sizeof(float) * n);)getAStart (a, n, c);求實(shí)部矩陣的逆：C = ATprintf(實(shí)部矩陣的逆：C = ATn)； for(i=0;in;i+)for(j=0;jn;j+)printf (z/%0. 5f，z,cij);)printf(n)； printf CAn);Matrix Mui (c, b, d, n) ; /D = AT * Bpr

7、intf (實(shí)部矩陣的逆與虛部矩陣相乘:D = A-1 * Bn); for(i=0;in;i+)( for(j=0;jn;j+) (printf (，z%0. 5f) printf (n)； printf (/，n，/);Matrix_Mul (b, d, c, n) ; C = B * AT * B printfCC = B * AT * Bn); for (i=0;in;i+) (for(j=0;jn;j+) (printf (z/%0. 5f z，, cij);) printf (n)； printf(n)；Matrix_Add(a, c, c, n) ; /C = A + B * A

8、-1 * B printfCC = A + B * A-l * Bn);for(i=0;in;i+) (for(j=0;jn;j+) printf (zz%0. 5f z/, cij);) printf(n)； printf(n)；getAStart (c, n, a) ;/A = (A + B * AT * B)T,復(fù)數(shù)矩陣的求逆后的實(shí)部printf (復(fù)數(shù)矩陣的求逆后的實(shí)部：A -(A + B * B)Tn)；for(i=0;in;i+)(for(j=0;jn;j+)printf(n)； printf (n)；Matrix_Mul(d, a, b, n) ; /B = (AT * B) * (A + B * AT * B)T,復(fù)數(shù)矩陣的求逆后的虛部 printf (復(fù)數(shù)矩陣的求逆后的虛部:B = (A2T * B) * (A + B * AT * B)Tn);for (i=0;in;i+)(for(j=0;jn;j+)printf (z/%0. 5f)printf (rT); printf(n)；)void main ()(int i;float *a = (float *)malloc(sizeof(float *) *2);for(i = 0; i 2; i+)ai二(float *)malloc(sizeof(float) * 2);)float *b = (float