解四元一次方程组,急用方程如下0.8624X+0.825Y+0.8535Z+0.8161H=0.82293.15X+0.42Y+0.98Z+1.3H=1.23770.06X+0.0288Y+0.032Z+0.8029H=0.5474X+Y+Z+H=1一定有解.xyzh都大于0

Question

解四元一次方程组,急用
方程如下
0.8624X+0.825Y+0.8535Z+0.8161H=0.8229
3.15X+0.42Y+0.98Z+1.3H=1.2377
0.06X+0.0288Y+0.032Z+0.8029H=0.5474
X+Y+Z+H=1
一定有解.xyzh都大于0

Answer 1

用matlab求解
a=[0.8624 0.825 0.8535 0.8161;3.15 0.42 0.98 1.3;0.06 0.0288 0.032 0.8029;1 1 1 1]
b=[0.8229;1.2377;0.5474;1]
b\a
a =
0.8624 0.8250 0.8535 0.8161
3.1500 0.4200 0.9800 1.3000
0.0600 0.0288 0.0320 0.8029
1.0000 1.0000 1.0000 1.0000
b =
0.8229
1.2377
0.5474
1.0000
ans =
1.6078 0.6311 0.8359 1.0602