??i?1n解:原式=
010xi11?10 =(b?a)(c?a)1b2?ab?a2221c2?ac?a2
x10?00x2?0????000?xn =n?xx11x2?n(?).
i?1xi四、证明题
11.设a,b,c是互异的实数,证明aa3件是a+b+c=0.
11110证明:abc?ab?aa3b3c3a3b3?a3 =
b?ac?ab3?a3c3?a3
11bc?0的充分必要条b3c30c?a
c3?a3=(b?a)(c?a)(c?b?ac?ab) =(b?a)(c?a)(c?b)(a?b?c)=0,
由于a,b,c是互异的实数,故要上式成立,当且仅当a+b+c=0.
abcd2.证明aa+ba?b?ca?b?c?da2a?b3a?2b?c4a?3b?2c?d?a4
a3a?b6a?3b?c10a?6b?3c?drabcd4?r3证明:左边r0aa?ba?b?c3?r2r0a2a?b3a?2b?c
2?r10a3a?b6a?3b?cabcdabcdr4?r30aa?ba?b?c0aa?ba?b?cr3?r2100a2a?br4?r300a2a?b?a400a3a?b000a=右边
克莱姆法则
一、选择题
?1.方程组??x1?x2?x3?1, ?x1??x2?x3?1,, 有唯一解,则( ).
??x1?x2??x3?1 (A)???1且???2; (B) ??1且???2;(C) ??1且??2; (D) ???1且??2.
?11解析:由克莱姆法则,当1?1?(2??)(??1)2?0,即
11???1且???2,选B.
?ax?z?0,2.当a?( )时,方程组??2x?ax?z?0,只有零解.
??ax?2y?z?0(A) -1 ;(B) 0 ;(C) -2 ;(D) 2. 解析:由克莱姆法则,
a01001当2a1?2?aa1?2(a?2)?0
a?210?21即a?2,选D.
三、解答题
1.用克莱姆法则下列解方程组.
?x?2y?z?2,(1)??x?2y?2z?3,
??2x?y?z?3;12?1解: D?1?22?3?0, 2?11由克莱姆法则知,此方程组有唯一解,
22?1D1?3?22?3, 3?11
12?1122D2?132?6,D3?133?9,
231233因此方程组的解为
x?D1DDD?1,y?2D?2,z?3D?3.
??x1?2x2?x3?x4?1,(2)??2x1?3x2?x3?2x4?3,?3x..
?x12?2x3?x4?2,??2x1?4x2?3x3?3x4?2121?1解:D?23?121321?4?0
243?3由克莱姆法则知,此方程组有唯一解,
121?1111?1D33?1223?121?2321?8, D2?1221??2,
243?3223?3121?11211D33223?133?21321?2,D4?1322?2.
242?32432因此方程组的解为
xD1?2,xD1D1?D2,x1D12?2??3?3D?2,x4D4?D?2.?2x1?2x22.判断线性方程组??x3?0,?x1?2x2?4x3?0,是否有非零解
??5x1?8x2?2x3?022?11?24解:因为系数行列式D?1?24??22?1 58?258?2 1?241?24=?06?9?06?9??30?0, 018?22005所以,方程组只有零解.
?x1?kx2?x3?0,3.已知齐次线性方程组? ?kx1?x2?x3?0,有非零解,求k的值.
??2x1?x2?x3?0解:因为齐次线性方程组有非零解,所以该方程组的系数行列式
必为零,即
1k?11k?1k11?01?k21?k
2?110?1?2k3=3(1?k2)?(1?k)(1?2k) =(1?k)(4?k)?0 解得,k=-1或k=4.
?2x1?4x2?(??1)x3?04.当?取何值时,齐次线性方程组??(??3)x1?x2?2x3?0有非
???x1?(1??)x2?x3?0零解
解:由齐次线性方程组有非零解的条件可知,
24??1??31?2?0,解得??0,2,3. ?11???1
第一章综合练习
一、判断题
1. n阶行列式Dn中的n最小为2.( ╳ )
2. 在n阶行列式D?aij中元素aij(i,j?1,2,L)均为整数,则D必为整数.( √ )