发布时间 : 星期五 文章数值分析简明教程课后习题答案更新完毕开始阅读b6f70d176c175f0e7cd13785
??(k?1)(k)(k)(k)(k)?x1?x1(k?1)??5x2??5x2?3x3?2?3x3?2??25(k)?(k?)5(k?1)1(k)?(k?)(k) 或 x?x?x?2x??x2?8x3?3 ?2?213222??1(k)14(k)18?(k?1)2(k?1)1(k?1)11?(k?1)x?x?x?x??x2?x3?3123??555255??G??8?1,不收敛。
3、（p.171，题6）加工上述题5的方程组，比如调换方程组的排列顺序，以保证迭代过程
的收敛性。
【解】加工后结果如下：
?3x1?x2?2（1）?
x?2x??12?1?5x1?2x2?x3?4?（2）?x1?5x2?3x3?2
?2x?x?5x??1123?1方程组（1）的雅可比迭代：
1(k)2?(k?1)3x??x2?1??33，GJ??x(k?1)??1x(k)?121?22?方程组（1）的高斯-赛德尔迭代：
??1?1，迭代收敛。 21(k)2?(k?1)3x??x2???133，GG?S??x(k?1)?1x(k)?221?63?方程组（2）的雅可比迭代：
??1?1，迭代收敛。 3?(k?1)2(k)1(k)4?x2?x3??x1555?1(k)3(k)2?(k?1)x??x1?x3?，GJ?2555??(k?1)2(k)1(k)11?x1?x2??x3555?方程组（1）的高斯-赛德尔迭代：
??4?1，迭代收敛。 5
?(k?1)2(k)1(k)4?x2?x3??x1555?2(k)16(k)6?(k?1)x??x2?x3?，GG?S?2252525?6(k)321?(k?1)18(k)x?x?x3?2?3125125125?
??18?1，迭代收敛。 256.1 高斯消元法
1、（p.198，题2）用选列主元高斯消元法求解下列方程组：
?x1?x2?x3??4?（1）?5x1?4x2?3x3??12
?2x?x?x?1123?1?2x1?3x2?5x3?5?（2）?3x1?4x2?7x3?6
?x?3x?3x?523?15?4?1?11?4??5?43?12??1r1?r2????r1?r2??51【解】 （1）?5?43?12???1?11?4???0?5??21111??21111???????21
3?12?28??? 55?111?????5?43?12??2r1?r3?5?43?12??5?43?12?5?r3??5???5?r3???0?12?8???0?12?8???0?12?8?
13179???21111???013?1790??????555??????5?43?12?1r2?r3?5?43?12?13r3?5?43?12?r2?r3??13???25???013?179???013?179???013?179?
2525???0?12?8???001?100??????1313???x?794x?3x3?124?6?3?(?1)?12?6,x1?2??3. 所以： x3??1，x2?31355?2355??3476??2r1?r2?3476??3476????r1?r2??3?11?3r2?（2）?3476???2355???01???0113?
33??1335??1335??1335???1335?????????1?r1?r33??3??0??0?41537123??3476?6??3476?3r3?r2?r3????3??0113???0529? ???0113?3??0529?????
?3476?5r3?3476???3????0529???0529?
?003565??0012??????2x3?9?4x2?7x3?6?4?1?7?2?6?1,x1????4. 所以： x3?2，x2?5351?r2?r35 2、（p.199，题9）计算下列三阶坡度阵的条件数：
??1?1（1）??2?1??31213141?3?1??。 4?1?5??1213141?3?1??，先求A-1。 4?1?5???00??10?
?01?????1?1【解】令：A???2?1??3
??1?1??2?1??312131413141511??100?11?123??2r1?r2?111010???0?12122??1?110001???345???1?12r2??0?1???31?r2?r3121211411??100??1r1?r2?1323??1?6120?01??11?0001?5????121211??100?1?3180r3?1?6120??0??11?00?11?1806????1?100?31?6120?
?41?01?453??12101?100?31?6120?
?130?180180?????1??0??0??1210??1?r3?r2??0??0??11???100??1r3?r1?10?960?60?323??0?36192?180?010?36192?180? ???130?180180??00130?180180???????
1?r2?r129?3630??3630??100?9?，所以 A?1???36192?180?
??010?36192?180???????00130?180180???30?180180???1?A?
最后求得条件数为：cond(A)?A
??11?408?748 6