篇一：工程数学线性代数课后答案 同济第五版
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篇二：线性代数第二章答案
第二章 矩阵及其运算
1? 已知线性变换?
??x1?2y1?2y2?y3?x2?3y1?y2?5y3? ??x3?3y1?2y2?3y3
求从变量x1? x2? x3到变量y1? y2? y3的线性变换?
解 由已知?
?x1??221??y1?
?x2???315??y2?? ?x??323??y???2??3??
?y1??221??x1???7?49??y1???故y2??315??x2???63?7??y2?? ?y??323??x??32?4?????3????y3??2??
??y1??7x1?4x2?9x3
?y2?6x1?3x2?7x3? ??y3?3x1?2x2?4x3
2? 已知两个线性变换 ?1
???y1??3z1?z2?x1?2y1?y3
?x2??2y1?3y2?2y3? ?y2?2z1?z3? ???y3??z2?3z3?x3?4y1?y2?5y3
求从z1? z2? z3到x1? x2? x3的线性变换?
解 由已知
?x1??201??y1??201???31
?x2????232??y2????232??20?x??415??y??415??0?1??2?????3??
??613??z1?
??12?49??z2?? ??10?116??z????3?0??z1?1??z2? ?z?3???3?
??x1??6z1?z2?3z3
所以有?x2?12z1?4z2?9z3? ??x3??10z1?z2?16z3
?111??123?
3? 设A??11?1?? B???1?24?? 求3AB?2A及ATB? ?1?11??051?????
?111??123??111?
解 3AB?2A?3?11?1???1?24??2?11?1? ?1?11??051??1?11???????
?058??111???21322?
?3?0?56??2?11?1????2?1720?? ?290??1?11??429?2???????
?111??123??058?
ATB??11?1???1?24???0?56?? ?1?11??051??290???????
4? 计算下列乘积?
?431??7?
(1)?1?23??2???570??1?????
?431??7??4?7?3?2?1?1??35? 解 ?1?23??2???1?7?(?2)?2?3?1???6?? ?570??1??5?7?7?2?0?1??49?????????
?3?
(2)(123)?2???1???
?3?
解 (123)?2??(1?3?2?2?3?1)?(10)??1???
?2?
(3)?1?(?12)? ?3???
?2?(?1)2?2???24??2?
解 ?1?(?12)??1?(?1)1?2????12?? ?3??3?(?1)3?2???36???????
?131??0?12?2140?? (4)???1?31? ? 1?134?????40?2?
?131??0?12??6?78?2140?? 解 ???1?31???20?5?6?? 1?134???????40?2?
?a11a12a13??x1??? (5)(x1x2x3)a12a22a23?x2?? ????aaa?132333??x3?
解
?a11a12a13??x1??? (x1x2x3)a12a22a23?x2? ????aaa?132333??x3?
?x1?
?(a11x1?a12x2?a13x3 a12x1?a22x2?a23x3 a13x1?a23x2?a33x3)?x2? ?x??3?
5? 设A??22?a11x12?a22x2?a33x3?2a12x1x2?2a13x1x3?2a23x2x3? ?1
?12?? B??1?13???0?? 问? 2??
(1)AB?BA吗?
解 AB?BA?
因为AB???3
?44?? BA??1?36???2?? 所以AB?BA? 8??
(2)(A?B)2?A2?2AB?B2吗?
解 (A?B)2?A2?2AB?B2?
因为A?B??
?2?22?? 5??2??2?25???2(A?B)2???2?
但2???814?? ?1429?5????38??68???1A2?2AB?B2???411???812??3?????0???1016?? ?1527?4????所以(A?B)2?A2?2AB?B2?
(3)(A?B)(A?B)?A2?B2吗?
解 (A?B)(A?B)?A2?B2?
因为A?B??
?2?22?? A?B??0?05???2?? 1??22??02???06?? (A?B)(A?B)???25??01??09???????
10???28?? ?38?而A2?B2??????????411??34??17?
故(A?B)(A?B)?A2?B2?
6? 举反列说明下列命题是错误的?（也可参考书上的答案）
(1)若A2?0? 则A?0?
解 取A???0
?0
?1
?01?? 则A2?0? 但A?0? 0??1?? 则A2?A? 但A?0且A?E? 0?? (2)若A2?A? 则A?0或A?E?解 取A??
(3)若AX?AY? 且A?0? 则X?Y ?
解 取
1A???0?0?? X??11?? Y??1??11??00?????1?? 1??
则AX?AY? 且A?0? 但X?Y ?
7? 设A??
解
?10?? 求A2? A3? ? ? ?? Ak? ???1?10?10???10?? A2????1????1??2?1???????10??10???10?? A3?A2A???2?1???1??3?1???????
0?? 1?? ? ? ? ? ? ?? 1Ak???k??
??10?
8? 设A??0?1?? 求Ak ? ?00????
解 首先观察
??10???1A2??0?1??0??00???00?????33?2A3?A2?A??0?3?00???44?3A4?A3?A??0?4?00???55?4A5?A4?A??0?5
?00?
??k?kA??0??0k?k?10???22?1?1???0?22??? ?2?????00??3??3?2?? ?3??6?2?4?3?? ?4??10?3?5?4?? ?5????? ? ?? ? ? ? ? ? ?? ?k0k(k?1)k?2?2k?k?1?k
用数学归纳法证明?
篇三：线性代数第二章答案
《线性代数第二版答案》出自：百味书屋
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