※ 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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