轾1犏犏5 故P2=犏犏2犏-犏臌525,P2TBP2=15骣00÷?? . ?÷÷?05桫TTTT 所以P1AP1=P2BP2，P2P1AP1P2=B
轾2犏犏5所以 Q=P1P2T=犏犏1犏犏臌5152-5轾1犏犏5犏犏2犏犏臌5-2轾43犏-犏555. =犏14犏3--犏犏55臌521.（1）由于P=(α,Aα)，α10，且lα1Aα 则α与Aα不成比例，且α10，故P可逆. （2）AP=A(α,Aα)=(Aα,Aα)=(Aα,-Aα+6α)
2轾06 即AP=P犏
犏1-1臌 故P-1AP=犏轾06
犏1-1臌轾06 所以A:犏=B
犏1-1臌 B-lE=-l16-1-l=(l+3)(l-2)
故l1=2，l2=-3，故B可以有两个不同的特征值，可以相似对角化，因此A
可以相似对角化.
22.
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2222?当x?y时，F(x,y)?综上所述：
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x?y
x?yF(y)?p{Y?y}?1111p?X2?y??p{X1?y)??(y)??(y)??(y) 2222即Y服从标准正态分布
23.
t??mtm?1????????，t…0f(t)???me
?0，其他?mP?T?s?t|T?s??
P{T?s?t,T?s?P{T?s?t}1?p{T?s?t}??
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似然函数为
L(?)??i?1n1m?1?nn?m?timm???i?1?te?i??f?ti????mn?i?1??0,?n,t1,t2,...tn?0其他
当t1,t2,...tn?0时
lnL(?)?nlnm?mnln??(m?1)?lnti?i?1n1?m?ti?1nmi
dlnL(?)mnmnm???m?1?ti?0
d???i?1$??
m?ti?1nmi n
2024年全国硕士研究生招生考试(数学一)--答案解析
