Linear Algebra and the C Language/a06t

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00e.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ void fun(int r) { double s = r_I(9); int c = C2; double **A = r_mR( i_mR(r,r),9.); double **B = mulC_mR(c_mR(A, i_mR(r,r)),s,c); clrscrn(); printf(" A:"); p_mR(A, S3,P0,C6); printf(" The column[%d] of A is muliply by %+.3f:\n",c,s); printf(" B:"); p_mR(B, S3,P0,C6); printf(" If B is the matrix that results when \n"); printf(" a single column 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 column 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: +3 +3 -6 +6 -4 -4 -1 +7 +9 The column[3] of A is muliply by +6.000: B: +3 +3 -36 +6 -4 -24 -1 +7 +54 If B is the matrix that results when a single column of A is muliplied by a scalar s, then det(B) = s det(A) det_R(B) = -2412 +6 det_R(A) = -2412 det_R(A) = -402 Press return to continue Press X return to stop