推 ruby791104:smartlwj大大：真的太謝謝你了！感激不盡！（鞠躬 11/26 23:36
※ 引述《ruby791104 (阿年：）)》之銘言：
: 嗯……
: 我承認這是作業，
: 可是我真的不會寫，
: 拜託好心的大大幫忙！
: 以下三題，討論並證明”only if”的情形。
: 1.Suppose A, B are in R^n＊n and A is nonsingular.
: Prove that AB is singular if B is singular.
det(AB) = det(A)det(B)
since A is nonsingular and B is singular
=> det(A)=/=0 and det(B)=0
so, det(AB) = 0
thus AB is singular.
: 2.Suppose A, B are in R^n＊n and A-B is nonsingular.
: Prove that┌ ┐ is nonsingular.
: ｜I I ｜
: ｜ ｜
: ｜A B ｜
: └ ┘
consider [0 I] = [I I][ I 0]
[A-B B] [A B][-I I]
then det([0 I]) = det([I I])det([I 0])
[A-B B] [A B] [-I I]
=> -det(A-B)=det([I I])det(I)
[A B]
=> det([I I]) = -det(A-B)
[A B]
since A-B is nonsingular
=> det(A-B) =/= 0
=> det([I I]) =/= 0
[A B]
hence [I I] is nonsingular
[A B]
: 3.Suppose A, E, F are in R^n＊n and that E and F are elementary matrices.
: Prove that if A is nonsingular then EAF is nonsingular.
: 先在此謝過好心的大大囉！（鞠躬
Since E and F are elementary matrices, so E and F are invertable
and A is nonsingular, then det(EAF)=det(E)det(A)det(F) =/= 0
so, EAF is nonsingular.
我猜應該是這樣做吧?? 好像有點抖 囧...
有錯請指正
(only if 是指逆命題嗎?)
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