发布时间 : 星期四 文章《线性代数》第1章习题详解更新完毕开始阅读
a11(2)
a12a22c12c2200b11b2100a?11b12a21b22az?bxxyzxzx
ya21c11c21a12b11a22b21b12 b22ax?by(3) ay?bzay?bzaz?bxaz?bxax?by=(a3?b3)yax?byay?bzz1a(4) 2aa41bb2b41cc2c41d?(a?b)(a?c)(a?d)(b?c)(b?d)?(c?d)(a?b?c?d) 2dd4 证明 （1）左式?a1b2c3?b1c2a3?c1a2b3?a3b2c1?a2b1c3?a1b3c2 ?a1(b3c3?b3c2)?b1(a2c3?a3c2)?c1(a2b3?a3b2)
?a1b2b3c2c3?b1a2a3c2c3?c1a2b2a3b3
?右式
a11a21（2）
c11c21a12a22c12c2200b11b2100b12b22按第一行展开a22a11c12c220b11b210a210b11b210b12 b22b12?a12c11b22c21a11a21 ?a11a22b11b12b21b22az?bx?a12a21b11b12b21b22?a12b11b12
a22b21b22ax?by(3) ay?bzay?bzaz?bx按第一列分开az?bxax?by ax?byay?bzxay?bzaz?bxyay?bzaz?bxayaz?bxax?by ?bzzax?byay?bzxx(a2az?bxax?by
ax?byay?bzzaz?bx分别再分ay?bzzyyaz?bxzax?byx?0)?(0?bzyxxax?by)
yay?bz 5
分别再分xa3yzyzxzyzxyxxyzxzxyzxzx(?1)2?右边 yx?b3zyxy?a3yzzx?b3yyz1a(4) 2aa41bb2b41cc2c41dd2d4c2?c1c3?c1c4?c11a 2aa4c?a0b?ab2?a2b4?a40c?ac2?a2c4?a40d?a 22d?ad4?a4按第一列展开b?ad?ab2?a2c2?a2d2?a2 b2(b2?a2)c2(c2?a2)d2(d2?a2)111每列都提取公因式(b?a)(c?a)(d?a)b?ac?ad?a b2(b?a)c2(c?a)d2(d?a)100?c1?c2?c1?c3(b?a)(c?a)(d?a) b?ac?bd?bb2(b?a)c2(c?a)?b2(b?a)d2(d?a)?b2(b?a)按第一列展开(b?a)(c?a)(d?a)(c?b)(d?b)11
(c2?bc?b2)?a(c?b)(d2?bd?b2)?a(d?b)?(a?b)(a?c)(a?d)(b?c)(b?d)(c?d)(a?b?c?d)
?310.设行列式500423,求含有元素2的代数余子式的和. ?21解 含有元素2的代数余子式是A12?A22?A23?A13
???1?3534004?35?345?1?11?6?10??26 ???1????1????1?21212?22?230422211. 设行列式D?0?7053?202,求第四行各元素余子式之和的值是多少? 02解 解法一：第四行各元素余子式之和的值为
M41?M42?M43?M44
6
040340300304
?222?222?222?222?7000000?700?70??7?8?0?3?14?(?7)?(?1)?(?2)??28
解法二：第四行各元素余子式之和的值为
M41?M42?M43?M44??A41?A42?A43?A44
304222?0?70?11?1按第2行展开0201按第3行展开340(?7)(?1)3?2222?1?11r2?2r33704004
?1?11?2834??28
?1?11?112.已知 D?1?10112101523 ,试求: 04(1) A12?A22?A32?A42 (2) A41?A42?A43?A44 解 (1)方法一：
虽然可以先计算处每个代数余子式,然后再求和,但是这很烦琐.利用引理知道，第一列每个元素乘以第二列的代数余子式的和等于零。
a12A12?a21A22?a31A32?a41A42?A12?A22?A32?A42?0
11?1?1方法二：构造一个新的行列式,即D1?1111由性质可知道D1?0；
101123 01D,D1的代数余子式A41,A4,2A4,3A是完全一样的，按照第二列展开得A12?A22?A32?A42 由性质和展开式可知A12?A22?A32?A42?0
7
1?1(2)由于Aij,aij无关,可构造一个新的行列式,即D1?011023 ,则有D,D1的11101111代数余子式A41,A42,A43,A44是完全一样的. 而
D1?1?A41?1?A42?1?A43?1?A44?A41?A42?A43?A44
10121012按第4行展开101D?1103?r3?r4?11031?11101110?110??1
1111000111113. 计算下列行列式.
32?141?a111(1)
2?35111?a10?23 (2) 11111?b1 54131111?b2aaaa122?2a2aaa222?2(3) aa2aa (4) 223?2
aaa2a?????aaaa2222?nxa1a2?an?11a1?aa001xa3?an?111?aa(5) Daa2x?a?101n?11n+1??????? (6) D5?0?11?aaaa0?11?a2a013?x1aa00?12a3?a01n1解
32?14025?5按第一列展开25?5(1)
2?351?2r3?r210?23?3r0?39?53?r15413?5r10?23?39?5 3?r40411?12411?12 8
000
a?a1