?6?a?7A?B,即??4a?4?6b0??11?10???A??15?2?，B??00?0?21??00????b，解得a?1,b?0，则0??0?因为A~B，所以矩阵A,B的特征值都为6???2????1?1，?2?0，?3?6.当??1时，由（E?A）X?0，得?1??0?；?1????-1???当??0时，由（0E?A）X?0，得?2??1?；?2????-1??2????1??当??6时，由（6E?A）X?0，得?3??-5?；令?1?1??0??15???2????1??2??5?-1??-1??3?1??1????2?2?1，????,3???0??3?-5?，再令P???1,?2，?2?36??30???2???2?1???5则有PTAP?B?1616261??30?5??，?30?2??30??
?1??1?????24、因为A的每行元素之和为5，所以有A?1??5?1?，即A有一个特征
?1??1??????1???值为?1?5，其对应的特征向量为?1??1?,A?1?5?1又AX=0的通解为
?1????2??1?????k1??1??k2?3?，则r(A)=1??2??3?0，其对应的特征向量为?3??4??????2??1??????2???1?,?3??3?,A?2?0,A?3?0?3??4?????。令
?1???x1?1?x2?2?x3?3??,解x1?8,x2??1.x3??2，则A??40?1?
?1???
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