川大版高数第三册答案(1)

1第一章行列式1.23154110103631254=520010=8(1)3(1)321(1)(2)(3)2441(1)3214243(1)321012)4nnnnnnnnmnmnnnmnmnnm（）该数列为奇排列（）该排列为偶排列（）当或时，为偶数，排列为偶排列当或时，为奇数，排列为奇排列（其中，，（）12(1)13521)246(2)0123(1)244113521)246(2)424313521)246(2)012)2.(1)(2)(nnnnnnnmnmnnnmnmnnmiiiknnn（当或时，（为偶数，排列为偶排列当或时，（为奇数，排列为奇排列（其中，，解：已知排列的逆序数为，这个数按从大到小排列时逆序数为111112(1)3)2(1)2xxxnxnxnnnnnnxiriiinxrinxnniiiiii个.设第数之后有个数比小，则倒排后的位置变为，其后个数比小，两者相加为故3证明：.因为：对换改变排列的奇偶性，即一次变换后，奇排列改变为偶排列，偶排列改变为奇排列当n2时，将所有偶排列变为奇排列，将所有奇排列变为偶排列因为两个数列依然相等，即所有的情况不变。偶排列与奇排列各占一半。4（1）13243341aaaa不是行列式的项14233142aaaa是行列式的项因为它的列排排列逆序列=(4321)=3+2+0+0=5为奇数，应带负号（2）5142332451aaaaa不是行列式的项1352413524aaaaa=1324354152aaaaa因为它的列排排列逆序列(34512)=2+2+2+0+0=6为偶数应带正号。5解：112332441223344114233142aaaaaaaaaaaa利用为正负数来做，一共六项，为正，则带正号，为负则带负号来做。6解：（1）因为它是左下三角形112122313233..........12300...00...0...0...nnnnnaaaaaaaaaa=112131411223242233433444...............0...00...000...0000...nnnnnnaaaaaaaaaaaaaaa=21231122331nnnaaaa=112233nnaaaa（2）1112314152122232425313241425152000000000aaaaaaaaaaaaaaaa=2223242511321142520001000000aaaaaaaa+21`23242521311241510001000000aaaaaaaa=1111112212211010aaaa=0（3）1200340021131751=1212121313451=32（4）000000000000000xyxyxyxyyx=01212023120000011000xyxyxyxyyxyxxyyx=55xy7.证明：11121212212............nnnnnnaaaaaaaaa将行列式转化为111221200...00...0...............0nnaaaaa若零元多于2nn个时，行列式可变为211200...00...0...0nnaaa故可知行列式为0.8.（1）20413611313121233152041361112302331=4310361112302331543105940123023315343143159452121063012313701122121212111212112122111112121212122112121122121.)().)1101=ymxbxyxyyymxxyyyxbxyxxyyxyyxyxyyxbbyxxxxxxyyxyxyyxxxxxxyxyxyyyxyx第一章高数3册9.(1).经过(，，斜率代入(，则又由左边2122112122112120xxyxyyyxyxyyxxxxx右边则问题特征：422222222sincoscos2sincoscos2sincoscos2cosc10.145oscos2.=+=221=bccaabbccaabbccababcacabbcacabbcacababcabcabc利用性质和分成六个行列式相加其余结合为零故原式性质222222222222222cos1coscos2coscoscos22cos1coscos2coscoscos22cos1cos1-2+(1)_cos2cos2coscos2cos2coscos1052cos2coscos2（）列列性质522222342222222222222000013.000040111011110101010101011.12324323yzxzxzxyzxyzxyzxyzxzyxxzxyyzxyyzxyyzxzxyzyxzyzxzxyzyzyxyzxyzzxzyzxzxyyxyxabcdaababcabcdaababcabcdaa列列列列1-122+323423+43-34463106300023243200203631063003630002000babcabcdabcdabcdaababcaababcaababcaabaababcaababcdaababcaaaba列加到行行列行行行行61-2+21-3+31-+1+111213112112232123311231231000-103-12622-1-20-1032-1-2-30-1002620321-1234!004200013nnnnnnnnnnnnnnnnnnxaaaaaxxaaxxxaxxxxx列列列列列列降阶3122322332312213311221331233223321-+21+131131-[phone]-111001000nnnnnnnnnnnnnxnnnnxnnnnaxaaxxaxxxaxaxaxaxaxaxaxaxaxaxaxxxaxax列列列列降阶习题一13（1）000000000000xyxyDxyyx根据“定义法”(2.3.4.5...)1(1)(1)nInnnnnDxyxy（2）1231110000220000011nnDnn7根据“降阶法”~n(1)n(n+1)23n-1n2n(n+1)34n12n(n+1)12n-2n-12D将第2列加到第列上得-1123n-1123n-1n011111341n(n+1)n(n+1)=01111221122101111nnnnnnn将前一行乘以加到后一行得(2)~(n)(1)1111-n-1111-n111-n1-111-n1n(n+1)(n-1)=211-n11-11111-n111将列加到列上得变为阶1111-n111-n1n(n+1)=-211-n111111-1(1)(2)~(n)110110(1)-2101000nnnnn列加到列2(1)(2)3222(1)2112222(1)11(1)(1)(1)(1)222nnnnnnnnnnnnnnnn（3）212122222111112111111a12111(1)(1)(1)(2)(1)12(2)(2)(1)(2)(1)11(1)(1)nnnnnnnnaaaaaanaaaaaaanaaaaaaanananan转置(1)2(-1)1!2!(1)!nnn范达蒙行列式注：根据范达蒙行列式原式=123(1)(1)(2)(1)(1)1!2!(1)!nnn(1)(2)(2)n-1=(1)2(1)1!2!(1)!nnn8（4）122111111111122122222222nn122-111111111annnnnnnnnnnnnnnnnnnnnnnaabababbaabababbnaabababb第行提出得12211111111112122222n-11212222211111211111111nnnnnnnnnnnnnnnnnnnnnnnbababababbbbaaaaaaabbbbaaaa=2111112111112122222n-11212222211111211111111nnnnnnnnnnnnnnnnnnnnnnnbbbbaaaabbbbaaaaaaabbbbaaaa=1231()()jnnnninjiijijbbaaaaababaa14（1）证明：cossincos222cossincos222+cossincos222sincossincos2222=coscos++22sincossincos2222sincos-22+cos++2sincos22++=cos(sincoscossin)cos(sincoscossin)2222222222+cos(sincoscossin)222229cossincossincossin222222111sin()sin()sin()2221sin()sin()sin()2(2)证明：123422221234444412341111xxxxxxxxxxxx12341xxxx（3）12(-1)(1)~()naxaaaaaaxaaanaaaaxaaaaaa最后一行乘以加到行得1212123000000000000nnnxxxxxaaxxxxxaaaaa（4）“递推法”01211000100001000nnaaxaxax01n+n112100100010100(-1)(1)00001nnnaaxxxaaxx降阶11nnxDa12221112011:nnnnnnDxDaDxDaDaxaxa由此类推1015.(1)=+=(ab+1)(cd+1)-[a(-d)]=(ab+1)(cd+1)+ad(2)==(4