Linear Algebra and the C Language/a08q

__NOTOC__ Install and compile this file in your working directory. /* ------------------------------------ */ /* Save as : c00b.c */ /* ------------------------------------ */ #include "v_a.h" /* ------------------------------------ */ #define RA R4 #define CA C6 #define Cb C1 /* ------------------------------------ */ #define CB C1 /* B : a basis for the column space of A */ /* ------------------------------------ */ int main(void) { double ab[RA*(CA+Cb)]={ +3, -9, +12, -6, +15, +12, +0, -6, +18, -24, +12, -30, -24, +0, +7, -21, +28, -14, +35, +28, +0, -2, +6, -8, +4, -10, -8, +0 }; double **Ab = ca_A_mR(ab,i_Abr_Ac_bc_mR(RA,CA,Cb)); double **A = c_Ab_A_mR(Ab, i_mR(RA,CA)); double **b = c_Ab_b_mR(Ab, i_mR(RA,Cb)); double **B = i_mR(RA,CB); clrscrn(); printf("Basis for a Column Space by Row Reduction :\n\n"); printf(" A :"); p_mR(A,S6,P1,C10); printf(" b :"); p_mR(b,S6,P1,C10); printf(" Ab :"); p_mR(Ab,S6,P1,C10); stop(); clrscrn(); printf(" The leading 1’s of Ab give the position \n" " of the columns of A which form a basis \n" " for the column space of A \n\n" " A :"); p_mR(A,S7,P3,C10); printf(" gj_PP_mR(Ab,NO) :"); gj_PP_mR(Ab,NO); p_mR(Ab,S7,P3,C10); c_c_mR(A,C1,B,C1); printf(" B : a basis for the column space of A"); p_mR(B,S7,P3,C10); stop(); f_mR(Ab); f_mR(b); f_mR(A); f_mR(B); return 0; } /* ------------------------------------ */ /* ------------------------------------ */ The position of the pivots of Ab gives the position of the columns of A which form a basis for the column space of A. Next, we check whether these columns are linearly independent. Screen output example: Basis for a Column Space by Row Reduction : A : +3.0 -9.0 +12.0 -6.0 +15.0 +12.0 -6.0 +18.0 -24.0 +12.0 -30.0 -24.0 +7.0 -21.0 +28.0 -14.0 +35.0 +28.0 -2.0 +6.0 -8.0 +4.0 -10.0 -8.0 b : +0.0 +0.0 +0.0 +0.0 Ab : +3.0 -9.0 +12.0 -6.0 +15.0 +12.0 +0.0 -6.0 +18.0 -24.0 +12.0 -30.0 -24.0 +0.0 +7.0 -21.0 +28.0 -14.0 +35.0 +28.0 +0.0 -2.0 +6.0 -8.0 +4.0 -10.0 -8.0 +0.0 Press return to continue. The leading 1’s of Ab give the position of the columns of A which form a basis for the column space of A A : +3.000 -9.000 +12.000 -6.000 +15.000 +12.000 -6.000 +18.000 -24.000 +12.000 -30.000 -24.000 +7.000 -21.000 +28.000 -14.000 +35.000 +28.000 -2.000 +6.000 -8.000 +4.000 -10.000 -8.000 gj_PP_mR(Ab,NO) : +1.000 -3.000 +4.000 -2.000 +5.000 +4.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 +0.000 B : a basis for the column space of A +3.000 -6.000 +7.000 -2.000 Press return to continue.