矩阵A是一个3×3的矩阵，B是一个3×2的矩阵，编程求A×B 已知A是一个3行4列的矩阵。B是一个4行5列的矩阵。编程求A...
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\u95ee\uff1a\u4e3a\u5565\u6570\u7ec4c\u662f3\u884c5\u5217
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\u95ee\uff1a\u8fd9\u91cc\u662f\u600e\u4e48\u53d8\u8fc7\u6765\u7684
\u4e5f\u5c31\u662f\u6211\u8fd8\u662f\u4e0d\u660e\u767d\u6570\u7ec4a*\u6570\u7ec4b\u662f\u4ec0\u4e48\u6837\u7684\u8fdb\u884c\u7684
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\u6ce8\u610f\u697c\u4e3b\u7684\u9898\u4e2d\uff0cm=3,r=4,n=5\uff0ci=0,1,2,3\uff1bj=0,1,2,3,4
\u8fd9\u5c31\u8868\u73b0\u5728\u8fd9\u91cc\uff1a
for(k=0;k<4;k++)
c[i][j]+=a[i][k]*b[k][j];
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#include
void main()
{
float a[50][50],b[50][50],c[50][50];
int n,j,k,i,l,y;
printf("\u8bf7\u8f93\u5165\u4f60\u6240\u9700\u7684a\u77e9\u9635\u884c\u6570:\n");
scanf("%d",&n);
printf("\u8bf7\u8f93\u5165\u4f60\u6240\u9700\u7684a\u77e9\u9635\u5217\u6570:\n");
scanf("%d",&l);
printf("\u8bf7\u8f93\u5165\u4f60\u6240\u9700\u7684b\u77e9\u9635\u5217\u6570:\n");
scanf("%d",&k);
printf("\u8bf7\u8f93\u5165\u77e9\u9635a:\n");
for(i=0;i<n;i++)
{
for(j=0;j<l;j++)
scanf("%f",&a[i][j]);
}
for(i=0;i<n;i++)
{
for(j=0;j<l;j++)
printf("%6.2f",a[i][j]);
printf("\n");
}
printf("\u8bf7\u8f93\u5165\u77e9\u9635b:\n");
for(i=0;i<l;i++)
{
for(j=0;j<k;j++)
scanf("%f",&b[i][j]);
}
for(i=0;i<l;i++)
{
for(j=0;j<k;j++)
printf("%6.2f",b[i][j]);
printf("\n");
}
for(i=0;i<n;i++)
for(j=0;j<k;j++)
{
c[i][j]=0;
for(y=0;y<l;y++)
c[i][j]=c[i][j]+a[i][y]*b[y][j];
}
printf("\u77e9\u9635\u76f8\u4e58\u7684\u77e9\u9635C\u5982\u4e0b:\n");
for(i=0;i<n;i++)
{
for(j=0;j<k;j++)
printf("%8.2f",c[i][j]);
printf("\n");
}
getch();
}
int main(void)
{
int a[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
int b[3][2] = {{1,2,3},{4,5,6}};
int c[3][3];
int i,j,k;
for(i=0;i<3;i++)
{
for(j=0;j<2;j++)
{
//c[i][j]是矩阵c的每一个元素
c[i][j] = 0;
for(k=0;k<3;k++)
{
//矩阵求积的方法，c[i][j]等于矩阵a的i行每一个元素乘以矩阵b的j列每一个元素，求这些积的和
c[i][j] += a[i][k] * b[k][j];
printf("%d ",c[i][j]);
}
printf("\n");
}
}
return 0;
}
#include<stdio.h>
void main()
{
int i,j,k,a[3][3],b[3][2],c[3][2];
printf("输入矩阵A:\n");
for(i=0;i<3;i++)
for(j=0;j<3;j++)
scanf("%d",&a[i][j]);
printf("输入矩阵B:\n");
for(i=0;i<3;i++)
for(j=0;j<2;j++)
scanf("%d",&b[i][j]);
printf("AB相乘后得矩阵C:\n");
for(i=0;i<3;i++){
for(j=0;j<2;j++){
c[i][j]=0;
for(k=0;k<3;k++)
c[i][j]+=a[i][k]*b[k][j];
printf("%d ",c[i][j]);
}
printf("\n");
}
}
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