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1、习题三（第1、2、3、6、9、11、15、16、17、18、19、20、22题）1.设L为单位下三角阵，试写出解方程组的算法。解:■1，211兀2〃2，31■■lyi■■1■■兀3■■=■■■Jnl•仃2•-i3…1■■dnx=〃1Y7-1YjijXj+习=dj,i=2,3,…昇2戶1x=d]

2、l,〃•易Llc3■■■r6久届=5nDijXj=Ci，—,〃-2,・・・,1••J=lXi-（Ci~工hjXj）rii'戶i+1/乘除法运算次数=工（〃1=一i+1)=1+2n=n(n+1)3.解:3.2-1.50.50.901.62.5-1.01.551.04.1-1.52.08_3.2-1.50.50.9003.25-1.251.104.56875—1.656251.798753.2-1.50.50.90_03.25-1.251.1000.100960.25240『3・1・40577耳》厂2_0・5厂］q・0.31

3、》5「］、6.解:0.3571r2rx°2.000()0.6000Gri'1.8214530.4286>r41.788221.000()厂5一，航…：x2=(1.1+1.25x2.5)十3.25=1.3「(09兀］-0+1.5xl.3-0.5x2.5)//3.2_=0.52.00001.00005.5200'0.35712.00000.64294.31440.60002.00000.40003.26610.42862.00000.57142.42871.00002.00002」1502.00001.00005.520001

4、.821450.64293.32880001.788220.40002.16957兀30001.904130.57141.90870°-2524%.10096=2-51.904131.69992L11260等价二角方程组Q・OOO()M()+1.0000M]1.82145M]+0.6429M?1.78822^2+0.4000M31.90413M3+0.5714M41.69992M45.5200=3.32880=2.16957=1.90870=1.11260冋代得M4=l・1126%69992=°&4501牡=(1・9087

5、0-0.5714x0.654501)/^〔§=0.80599“_(2.1695-0.4000x0.80599)/_1/1.78822=LU3297M-(3.32880一0.6429x1.03297)/_〕M]-/1.82145_L4O^°册(5.5200-1.0000x1.46296)/”门父“%J/2.0000=Z02852・•・Mo=2.0285,M]=1.4630,=1.0330,M3=0.8060,M4=0.65459.W：2134031041040321310410022T233—r2>0422T042n3814

6、等价三角方程组r3兀］+4x2=322Tx2+4x3=8_14~11X3=2,A*?=0，—111.解：Gauss消去法1592A=592430843910.1000xlO12251J-0.592xlO30.5920x0.4308x0.439xlO30.2251xlO40.1000x10*0r2一592厂］0.5920xlO30.439xlO30.4308xlO4-(0.5920xlO3)20.225lxlO4-0.592x0.439x106_O.IOOOxlO10.5920xio30.439xlO3一00.4308xlO

7、4-0.3505x1060.225IxlO4-0.2560xlO6__O.IOOOxlO10.592xIO30.439xlO3_0-0.3462xlO6-0.2574xl06_回代y=(-0.2574xlO6)-(-0.3462xl06)=0.7435x10°x=[0.4390xIO3-(0.592xlO3)x(0.7435x10°)]十(0.1000xIO1)=(0.4390—0.4402)xIO3=-0.1200xIO1列主元Gauss消去"59243082251'AT1592439__0.5920xIO30.4308

8、x1040.2251xlO4'-O.IOOOxlO10.5920xlO30.4390xlO310.5390x103"0.5920xlO30.0000x10°0.4308xlO40.2251xIO40.4390xlO30.2251x10“0.5390xlO30.5920xlO300.4308xlO40.2