线代书上习题答案第一章解答

写出四阶行列式中含有因子a11a23解四阶行列式中含有因子a11a23a11a23a32a44和a11a23a34a42 如果 21，计算下列各行列式之值

3

3x

3z2

y

z1x4y1z 111解

r

2

12 2112

1

2

3x 3z2x

y

z 3z 3y xy0z2111111设行列式D

解已知行列式的第4行各元素式之和的值相当于如下行列式的值3040222230402222000117 设行列式D 2，求 解

A31A32A33A33A35 A31A32A33= 00 112112313122；2310；123012110；000000．0012324124321；1202 a001a001abbb0a00；abab00a0baba100abbba解

123212323211232123232192140002130213

c2c4

4(1)43

c3c

1

1111304572r-r1012315+201=123011021104572r-

r3

4-42

(1)21 c1+2c3 a

1 0c4-ac1

(a21)a2.abbbabbbabababbbabbbababababbababababbba0b00

(ba) a 0(ba)(a

(a (6第一行取(1,1a1后,第四行只能取(4,4a4a2、a3000000b0a00 (a1a4axaxayazayazaxazaxay．aybz

aybzazbx

azbxaxbyxaz

aybzazbxax

azbxaxbyaybzxaybz yza2yazbxxb2zxzaxby xyxy yza3yzxb3zxzx xyxy xya3yzxb3yzzx zxxy(a3b3)yzxzx试计算n 11 11(n11(n解f(xx的n1f(x)a(x (x又在行列式中分别取x1

n1时，行列式的值为零，所以1

n1就是f(x)0n1个根为了确定a,再由原行列式按n列展开的情况知:a为xn1的代数式(1)2n 而Mnn正好是由1, n1构成的Vandermonde行列式,于是 ,n1)(x1)(x [x(n(n2)!(n 2!(x1)(x [x(na00a001(1)Dn00a00a00100aaaa0(2)Dnaaa00aaa0aaaanan2

(n1)an1a0001a00(3)D501a 1aa2a3a4a5001a0001aax xn1xnx(4)Dn

Dn

a00a0010a0000a010000a 0

0

(a21)an2anan2a 1

(n

Dn

(n

(n(n

11

100101001001001000

(n

00(n1)a(1)n1 00(n00(n1)a(1)n1 00

(n1)an1a0001a00`D5011a0001a00`D501a0001a00011a00001a0001a0001a001a00a000a00a000a0a000a1a00001a

DDa4a5

a3)a4 1aa2a3a4a52n2时D2

axx2命题成立 x 假设对于(n1)阶行列式命题成立 Dn1aax xn2xn1

Dn按第一列展开 Dn

00 00x(aax xn2xn1) aax xn1xn 因此n阶行列式命题成立

x12x23x32x12x2x313x4x3x解

D 12

1

D1 14，D22114，D3 12

31

所

D142，

D242,

D32

A(1,12B(320C(055解

axbyczd0A(1,12)B(320)C(0555b5cd设(x,yz)axbyczdab2cd 3a2bd 5b5cd因为abcdxyz111xyz11121301051整理 29x16y5z550所求平面方程为29x16y5z