工程数学线性代数课后习题答案

??210??13?11?1所以 A?A*???3??|A|22???167?1??

?a1a?0??2 (4)??(a1a2

??0?a?n??a1?0?a?2 解 A?????0an????1??a1?0?1a??12? A?????1?0???a?n?an 0)

由对角矩阵的性质知

12 解下列矩阵方程

2 (1)??1?5?X??4?6??21?3?????12 解 X???1?5??4?6???3?5??4?6???2?23?????????3???21???12??21??08??21?1?1?13 (2)X?210????432???1?11??????1

?21?1?1?13???210? 解 X????432??1?11????101?1?131????23?2? ???3?432???330?????221?85?2? ?????3??31 (3)???1?4?X?2?2????10???31???1???0?1? 0?

1???11 解 X????1?4??31??2?0?1???12??????12?4?31??10? ?1??11???0?1??12?12??????6 ?1??312?6??1?0???111?0????1??2??0??4?

?010??100??1?43? (4)?100?X?001???20?1??001??010??1?20????????1

?1?010??1?43??100? 解 X??100??20?1??001?

?001??1?20??010????????010??1?43??100??2?10? ??100??20?1??001???13?4??001??1?20??010??10?2????????? 13

利用逆矩阵解下列线性方程组

x?2x2?3x3?1??1 (1)?2x1?2x2?5x3?2??3x1?5x2?x3?3

解 方程组可表示为

??1?22235????xx1??1?2???2?

?351????x??3??3??故 ??x?11?x??123??12?????10?2???225???

?x??3??351????3????0??x从而有 ??1?1?x2?0

??x3?0 (2)??x1?x2?x3?2?2x1?x2?3x3?

?1?3x1?2x2?5x3?0 解 方程组可表示为 ??1?2??11??13????x x1??2?2???

?32?5???1??x??3??0???1故 ??x?x1??1??11??13????21?????50?2???2?

?x??3??32?5????0????3??故有 ??x1?5?x2?0

??x3?3 14 设

AkO (k为正整数)(EA)1

EAA2

Ak1

证明 因为AkO 所以EAkE 又因为 EAk(EA)(EAA2

Ak1)所以 (EA)(EAA2

Ak1)E

由定理2推论知(EA)可逆 且

证明

(E

A)

1

EAA2

有Ek2

Ak1

证明 一方面 另一方面 E(EA)

(EA)1(EA)

2

由AO 有

(AA)A2

Ak1

(Ak1

Ak)

(EAA

A k1)(EA)

故 (EA)1(EA)(EAA2

Ak1)(EA)

两端同时右乘(EA) (E 15逆

设方阵A满足A1

2

1

就有

A)1(EA)EAA2Ak1

A2EO 证明A及A2E都可

并求A及(A2

2

2E)

1

证明 由A AA2EO得

且A?1?1(A?E)2

A2E 即A(AE)2E

或 A?1(A?E)?E2由定理2推论知A可逆 由A22

A2EO得 A6E4E 即(A2E)(A3E)

4E

A或 (A?2E)?1(3E?A)?E

4由定理2推论知(A

2E)可逆

且(A?2E)?1?1(3E?A)4