矩阵B(入)相抵于对角入一矩阵dag{d1(),d2(入),…,dn(},且d1(入,Z2(入),……,dn(入) 是d1(入),d2(入,…,d(入)的一个置换,则A()相抵于B(入 可逆λ一矩阵 定义4若A(入),B()都是n阶入一矩阵,且 A(入)B(入)=In,B(4(A)=Ln, 则称A(入)为可逆入一矩阵,B(川是A(的逆入一矩阵,记为A-1(X) 初等入一矩阵是可逆阵事实上,P3P;=ln;P(c)P(C-1)=n;(3)T(f()T2(-f(A) 命题1A-矩阵的逆入矩阵若存在,必唯一 证明B(入),C(A均为A())的逆入-矩阵,则B())=B(川)n=B()A(C()= InC(入)=C() 命题2可逆入矩阵的乘积也是可逆入-矩阵,且(A()B()-1=B-1(入)A-1(入) 证明与数字矩阵一样证明.口 定理2一个n×n阶入-矩阵A())可逆的充分必要条件是|A(为一非零 常数 证明充分性设d=|A(刈≠0,则矩阵4()也是个A-矩阵,而 4()4(x)=a4(A)A(x)==4()4( 必要性.|4(米川B(川)=|=1,|A4(川,|B(川)都是A的多项式,所以 1A()|,|B(刈)为零次多项式,故只能是非零常数.口 四.入-矩阵的带余除法 引理2设M(入),N(入)为n阶λ-矩阵,且都不为零,又设B为n阶数字矩 阵,则必存在入一矩阵Q(A及S(λ)和数字矩阵R及T,使下式成立 M(从)=(MI-B)Q(入)+R N(A)=S(从)(A-B)+TV> B(λ) '3.O λ− V> diag{d ′ 1 (λ), d′ 2 (λ), · · · , d′ n (λ)}, o d ′ 1 (λ), d′ 2 (λ), · · · , d′ n (λ) ￾ d1(λ), d2(λ), · · · , dn(λ) $&9FC￾; A(λ) '3 B(λ). v
Ym λ− V> N[ 4 u A(λ),B(λ) -￾ n R λ− V>￾o A(λ)B(λ) = In, B(λ)A(λ) = In, ; A(λ) Ym λ− V>￾ B(λ) ￾ A(λ) $m λ− V>￾H A−1 (λ). % λ− V>￾Ym>
{w￾PijPij = In; Pi(c)Pi(c −1 ) = In; (3) Tij (f(λ))Tij (−f(λ)) = In. TX 1 λ− V>$m λ− V>u:￾ &
`S B(λ), C(λ) W A(λ) $m λ− V>￾; B(λ) = B(λ)In = B(λ)A(λ)C(λ) = InC(λ) = C(λ). 2 TX 2 Ym λ− V>$E$￾Ym λ− V>￾o (A(λ)B(λ))−1 = B−1 (λ)A−1 (λ). `S 5IV>&"Ah
2 NQ 2 &9 n × n R λ− V> A(λ) Ym$6 #M￾ |A(λ)| &5b 
`S 6
x d = |A(λ)| 6= 0, ;V> 1 dA⋆ (λ) $￾&9 λ− V>￾0 A(λ) 1 dA⋆ (λ) = 1 dA(λ)A⋆ (λ) = In = 1 dA⋆ (λ)A(λ). #
|A(λ)||B(λ)| = |In| = 1, |A(λ)|, |B(λ)| -￾ λ $/}￾ ( |A(λ)|, |B(λ)| b/}￾:El￾5b
2 
λ− V>$42 \Q 2 x M(λ), N(λ)  n R λ− V>￾o-b￾2x B  n RIV >￾; : λ− V> Q(λ) F S(λ) ?IV> R F T, |}℄ M(λ) = (λI − B)Q(λ) + R; (1) N(λ) = S(λ)(λI − B) + T. (2) 4