Linear Algebra and the C Language/a06u

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00f.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ void fun(int rc) { double s = r_I(9); int r = R2; double **A = r_mR( i_mR(rc,rc),9.); double **B = mulR_mR(c_mR(A, i_mR(rc,rc)),s,r) ; clrscrn(); printf(" A:"); p_mR(A, S3,P0,C6); printf(" The row[%d] of A is muliply by %+.0f\n",r,s); printf(" B:"); p_mR(B, S3,P0,C6); printf(" If B is the matrix that results when \n"); printf(" a single row of A is muliplied by \n"); printf(" a scalar s, then det(B) = s det(A)\n\n"); printf(" det_R(B) = %+.0f\n", det_R(B)); printf(" %+.0f det_R(A) = %+.0f\n",s,s*det_R(A)); printf(" det_R(A) = %+.0f\n",det_R(A)); f_mR(A); f_mR(B); } /* ------------------------------------ */ int main(void) { time_t t; srand(time(&t)); do { fun(rp_I(RC3)+RC1); } while(stop_w()); return 0; } /* ------------------------------------ */ /* ------------------------------------ */ We can verify numerically that if we multiply a row of A by s, the determinant of A must be multiplied by s to obtain the determinant of the modified matrix A Screen output example: A: -6 +2 +2 +5 -8 -9 +4 +2 +7 -1 -6 +8 +4 +1 +5 -9 The row[4] of A is muliply by +8 B: -6 +2 +2 +5 -8 -9 +4 +2 +7 -1 -6 +8 +32 +8 +40 -72 If B is the matrix that results when a single row of A is muliplied by a scalar s, then det(B) = s det(A) det_R(B) = -36168 +8 det_R(A) = -36168 det_R(A) = -4521 Press return to continue Press X return to stop