方阵的逆矩阵满足下列运算规律: (i)若A可逆,则A-1可逆,且(A-1)-1=A 由逆矩阵的唯一性知AA-=E→A-1的逆矩阵必为A.A与A-1互为 31矩阵 逆矩阵。 §2矩阵的运算 53逆矩阵 (i)若A可逆,数入≠0,则λA可逆,且(A4)-1=A1 §4矩阵分块法 本章总结 证因为A≠0,A≠0,所以AA=入A≠0=→入A可逆 (λA)-1(4)=E=(A)-1A=E=(A)1A=1E 主讲:张少 (AA)-AA-=JEA=(A)-=JA-1 标题页 (ⅲi)若A和B为同阶方阵且均可逆,则AB也可逆,且(AB)-1=B-1A-1 44 证(AB)(B-1A-1)=A(BB1)A-1=AEA-1=A-1=E (AB)-1=B-1A1 第11页共36页 注:依次类推,A1,A2,An均可逆,则有(A142…An)-1 A-A AoA 全屏显示 (iy)若A可逆,则A亦可逆,且(A1)-1=(A-1) A(A-1)=(A-14)T=E=E=→(A-1)T=(A-1)
天津师范大学 §1 Ý ✡ §2 Ý ✡ ✛ ✩ ➂ §3 ❴ Ý ✡ §4 Ý ✡ ➞ ➡ ④ ✢Ù♦✭ ❒ù: Ü✟r ■ ❑ ➄ JJ II J I ✶ 11 ➄ ✁ 36 ➄ ❼ ↔ ✜ ➯ ✇ ➠ ✬ ✹ ò Ñ ➄✡✛❴Ý✡÷✈❡✎✩➂✺➷➭ (i) ❡A➀❴➜❑A−1➀❴➜❹(A−1 ) −1 = A. ❞❴Ý✡✛➁➌✺⑧AA−1 = E =⇒ A−1 ✛❴Ý✡✼➃A. A❺A−1♣➃ ❴Ý✡✧ (ii) ❡A➀❴➜êλ 6= 0, ❑λA➀❴➜❹(λA) −1 = 1 λA−1 . ② Ï➃|A| 6= 0, λ 6= 0, ↕➧|λA| = λ n |A| 6= 0 =⇒ λA ➀❴✧ (λA) −1 (λA) = E =⇒ λ(λA) −1A = E =⇒ (λA) −1A = 1 λE =⇒ (λA) −1AA−1 = 1 λEA−1 =⇒ (λA) −1 = 1 λA−1 . (iii) ❡AÚB➃Ó✣➄✡❹þ➀❴➜❑AB➃➀❴➜❹(AB) −1 = B−1A−1 . ② (AB)(B−1A−1 ) = A(BB−1 )A−1 = AEA−1 = AA−1 = E =⇒ (AB) −1 = B−1A−1 . ✺ ➭ ➑ ❣ ❛ í ➜A1, A2, . . . An þ ➀ ❴ ➜ ❑ ❦(A1A2 · · · An) −1 = A−1 n A −1 n−1 · · · A −1 2 A −1 1 . (iv) ❡A➀❴➜❑AT➼➀❴➜❹(AT) −1 = (A−1 ) T. ② AT(A−1 ) T = (A−1A) T = E T = E =⇒ (A−1 ) T = (A−1 ) T.
当|A4|≠0时,对于方阵A,还可以定义A0=E,A-k=(4-1),k∈Z+.因 此,当A|≠0,A,μ∈Z时(上节定义为正整数,这一节用逆矩阵推广所有 的整数)有 A=A[(A)"=A §1矩阵 §2矩阵的运算 例:A4A-2=A2,其实A4A aaaaa-JA AAA(AA-)A-1 3逆矩阵 AA(AA)=AA=A §4矩阵分块法 本章总结 123 例10求方阵A=221的逆矩阵 主讲:张少 343 标题页 22-3r1 123 23 44 解A=221 0-2-5 2-5=2≠0,所 343 0-2-6 00-1 以A-1存在.再计算伴随矩阵 第12页共36页 23 2;,A21=(-1) 6;A3 43 43 全屏显示 23 12 3;A 6;A2=5;A13=2;A23=2; 33
天津师范大学 §1 Ý ✡ §2 Ý ✡ ✛ ✩ ➂ §3 ❴ Ý ✡ §4 Ý ✡ ➞ ➡ ④ ✢Ù♦✭ ❒ù: Ü✟r ■ ❑ ➄ JJ II J I ✶ 12 ➄ ✁ 36 ➄ ❼ ↔ ✜ ➯ ✇ ➠ ✬ ✹ ò Ñ ✟|A| 6= 0➒➜é✉➄✡A, ❸➀➧➼➶A0 = E, A−k = (A−1 ) k , k ∈ Z +. Ï ❞➜✟|A| 6= 0, λ, µ ∈ Z➒(þ✦➼➶➃✔✒ê➜ù➌✦❫❴Ý✡í✷↕❦ ✛✒ê), ❦ AλAµ = Aλ+µ (Aλ ) µ = Aλµ ⑦➭A4A−2 = A2 , Ù➣A4A−2 = AAAAA−1A−1 = AAA(AA−1 )A−1 = AA(AA−1 ) = AA = A2 . ⑦10 ➛➄✡A = 1 2 3 2 2 1 3 4 3 ✛❴Ý✡. ✮ |A| = 1 2 3 2 2 1 3 4 3 r2−2r1 r3−3r1 === 1 2 3 0 −2 −5 0 −2 −6 r3−r2 === 1 2 3 0 −2 −5 0 0 −1 = 2 6= 0, ↕ ➧A−1⑧✸. ✷❖➂❾➅Ý✡ A11 = (−1)1+1 2 1 4 3 = 2; A21 = (−1)2+1 2 3 4 3 = 6; A31 = (−1)3+1 2 3 2 1 = −4; A12 = −3; A22 = −6; A32 = 5; A13 = 2; A23 = 2; A33 = −2;