Program C++ Metode Gauss Untuk Aljabar Linier Ordo 4x5 X, Y, Z, U

Disini saya akan memberikan source code C++ dengan metode Gauss 4x5 X, Y, Z, U. Sehingga yang digunakan ini adalah statis bukan dinamis, jumlah angka yang sudah ditentukan.

Berikut adalah source code dengan metode Gauss:

#include "stdio.h" int main(){ int b1k1 = 1; int b1k2 = -1; int b1k3 = -1; int b1k4 = 1; int b1k5 = 0; int b2k1 = 2; int b2k2 = 0; int b2k3 = 2; int b2k4 = 0; int b2k5 = 8; int b3k1 = 0; int b3k2 = -1; int b3k3 = -2; int b3k4 = 0; int b3k5 = -8; int b4k1 = 3; int b4k2 = -3; int b4k3 = -2; int b4k4 = 4; int b4k5 = 7; printf("%i | %i | %i | %i [X] = %i\n", b1k1, b1k2, b1k3, b1k4, b1k5); printf("%i | %i | %i | %i [Y] = %i\n", b2k1, b2k2, b2k3, b2k4, b2k5); printf("%i | %i | %i | %i [Z] = %i\n", b3k1, b3k2, b3k3, b3k4, b3k5); printf("%i | %i | %i | %i [U] = %i\n\n", b4k1, b4k2, b4k3, b4k4, b4k5); printf("[%i | %i | %i | %i | %i]\n", b1k1, b1k2, b1k3, b1k4, b1k5); printf("[%i | %i | %i | %i | %i]\n", b2k1, b2k2, b2k3, b2k4, b2k5); printf("[%i | %i | %i | %i | %i]\n", b3k1, b3k2, b3k3, b3k4, b3k5); printf("[%i | %i | %i | %i | %i]\n\n", b4k1, b4k2, b4k3, b4k4, b4k5); int b4k1b1 = b4k1 - (3*b1k1); int b4k2b1 = b4k2 - (3*b1k2); int b4k3b1 = b4k3 - (3*b1k3); int b4k4b1 = b4k4 - (3*b1k4); printf("B4 - 3.B1\n"); printf("[%i | %i | %i | %i | %i]\n", b1k1, b1k2, b1k3, b1k4, b1k5); printf("[%i | %i | %i | %i | %i]\n", b2k1, b2k2, b2k3, b2k4, b2k5); printf("[%i | %i | %i | %i | %i]\n", b3k1, b3k2, b3k3, b3k4, b3k5); printf("[%i | %i | %i | %i | %i]\n\n", b4k1b1, b4k2b1, b4k3b1, b4k4b1, b4k5); int b2k1b1 = b2k1 - (2*b1k1); int b2k2b1 = b2k2 - (2*b1k2); int b2k3b1 = b2k3 - (2*b1k3); int b2k4b1 = b2k4 - (2*b1k4); printf("B2 - 2.B1\n"); printf("[%i | %i | %i | %i | %i]\n", b1k1, b1k2, b1k3, b1k4, b1k5); printf("[%i | %i | %i | %i | %i]\n", b2k1b1, b2k2b1, b2k3b1, b2k4b1, b2k5); printf("[%i | %i | %i | %i | %i]\n", b3k1, b3k2, b3k3, b3k4, b3k5); printf("[%i | %i | %i | %i | %i]\n\n", b4k1b1, b4k2b1, b4k3b1, b4k4b1, b4k5); int b3k2b2 = b3k2 + (2/b2k2b1); int b3k3b2 = b3k3 + (b2k3b1/2); int b3k4b2 = b3k4 + (2/b2k4b1); int b3k5b2 = b3k5 + (b2k5/2); printf("B3 + 1/2.B2\n"); printf("[%i | %i | %i | %i | %i]\n", b1k1, b1k2, b1k3, b1k4, b1k5); printf("[%i | %i | %i | %i | %i]\n", b2k1b1, b2k2b1, b2k3b1, b2k4b1, b2k5); printf("[%i | %i | %i | %i | %i]\n", b3k1, b3k2b2, b3k3b2, b3k4b2, b3k5b2); printf("[%i | %i | %i | %i | %i]\n\n", b4k1b1, b4k2b1, b4k3b1, b4k4b1, b4k5); int b4k4b2 = b4k4b1 + b3k4b2; int b4k5b2 = b4k5 + b3k5b2; printf("B4 + B2\n"); printf("[%i | %i | %i | %i | %i]\n", b1k1, b1k2, b1k3, b1k4, b1k5); printf("[%i | %i | %i | %i | %i]\n", b2k1b1, b2k2b1, b2k3b1, b2k4b1, b2k5); printf("[%i | %i | %i | %i | %i]\n", b3k1, b3k2b2, b3k3b2, b3k4b2, b3k5b2); printf("[%i | %i | %i | %i | %i]\n\n", b4k1b1, b4k2b1, b4k3b1, b4k4b2, b4k5b2); int Z = 3; int U = -4/-1; int Z1 = 4*3; int U1 = 2*4; int ZU1 = Z1-U1; int ZUTot = 8-ZU1; int Y = ZUTot/2; int YZUTot = -Y-Z+U; int X = 0+(YZUTot)*-1; printf("\n Z = %i \n", Z); printf("\n -U = -4"); printf("\n U = %i \n", U); printf("\n 2Y + 4Z - 2U = 8 "); printf("\n 2Y + %i - %i = 8 ", Z1, U1 ); printf("\n 2Y + %i = 8 ", ZU1 ); printf("\n 2Y = %i ", ZUTot ); printf("\n Y = %i \n", Y ); printf("\n X - Y - Z + U = 0 "); printf("\n X - %i - %i + %i = 0 ", Y, Z, U); printf("\n X %i = 0 ", YZUTot ); printf("\n X = %i", X ); }

Silahkan masukkan source code diatas keaplikas DEV C++, kemudian Compile dan Run klik F11.

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