行列式化简计算技巧窍门实题

.\.\行列式化简计算技巧和实题操练Zachary.技巧:技巧1:行列式与它的转置行列式的值相等，即D=Dtaaaaaa11121n1121n1aaaaaa21222n=1222n2aaaaaan1n2nn1n2nnn技巧2:互换行列式的任意两行（列）,行列式的值将改变正负号aaaaaa11121n21222naaaaaa21222n=—11121naaaaaan1n2nnn1n2nn技巧3:行列式中某一行（列）的所有元素的公因子可以提到行列式记号的外面bababaaaa11111211n11121nbababaaaa22122222n=21222n11bni=1bababaaaann1nn2nnnn1n2nn技巧4:行列式具有分行（列）相加性a11a12a1na11a12a1na11a12a1nb+c11 11b+ct2 t2b+ctn tn=b11bt2btn+c11ct2ctnan1an2annan1an2annan1an2ann技巧5：将行列式的某一行（列）的各元素乘以同一数k后加到另一行（列）对应的元素上，行列式的值不变

a11a12a1na11a12a1nas1as2asna+kas1 11a+kas2 t2a+kasn tna11at2atna11at2atnan1an2annan1an2ann技巧6:分块行列式的值等于其主对角线上两个子块行列式的值的乘积a11a11a1m00aa00m1mmccbb111m111nccbbn1nmn1nnaabb111m111naabbm1mmn1nn技巧7:［拉普拉斯按一行（列）展开定理］行列式等于它的任一行（列）的各元素与其对应的代数余子式乘积之和D=ZaA(i=1,2,,n)=ZaA(j=1,2,,n)ikikk=1kjkjkikikk=1二.解题方法:方法1:对于2阶行列式和3阶行列式,可以直接使用对角线法则进行计算aa1112=aaa112122aaa111213aaa212223aaa313233a一aa22 1221=aaa+aaa+aaa112233122331132132一aaa一aaa一aaa112332 122133 132231方法2:上三角行列式,下三角行列式,主对角线行列式,副对角线行列式aii0a12a22a1na2nannaii0a12a22a1na2nann=乩a11a21a22i=1二九九12iian1an2ann=lnlaiii=1九（其余未写出元素均为零）,nijkl=0000ijkl=00000e

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hijkl=0n（n1）=（-1）nV九九九（其余未写出元素均为零）12n方法3:若行列式中有两行（列）对应元素相等,则此行列式的值等于零efgh方法4:若行列式中有一行（列）的元素全为零,则此行列式的值为零方法5:若行列式中有两行（列）元素成比例,则此行列式的值等于零kakbkckde

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hijkl=0实题操练：计算下列行列式的值:习题1:TOC\o"1-5"\h\z1 2 0-1 1 -43 -1 8解答:2 0-1 1 -4=1x1x8+2x(-4)x3+0x(-1)x(-1)-0x1x3-2x(-1)x8-1x(-4)x(-1)=-43 -1 8习题3:12 3 45678910111213141516解答:1 2 3567910111314151 2 3567910111314154 18c-c512；-「916^^13131171=011111151习题4:3 -3 3x-3TOC\o"1-5"\h\z3 -3 x+3 -33x-3 3 -3x+3 -3 3 -3解答:3 -33 -33x-3x+3 -3x—3-3-3-311=x113x-3r-r1-33x—3x+3-32100x一xr-rx3-3310x0一xr-r3-31000一x-3-3x-3-33x-3 x -3 3x+3 -3 £x -3 x+33 -3『1 i_2cx x-3 33 -3 | 「卜-3 31-33x1-3TOC\o"1-5"\h\z0 -x=x4x -x0 -x习题5:abababab11121314abababab12222324abababab13233334abababab14243444解答:ab按第一列展开,1

abi:ab13abab12ab22abab13ab2

ab22

ab3

abab14ab24ab34ab44=abab22abab23abab24ab一abababab+ababab12 13ababab14ab一abab12abab13abab14ab12 1aba3ba4b1123ab2433ab3434ab441223 3aba24 33 34bab4 4413a22 23bab24 3424ab441422ab2323ab3324ab34ab12ab22ab24ab13ab23ab34ab14ab24ab44ab12和ab22ab23ab13ab23ab33ab14b24b34由于行列式aa有两行元素成比例,因此值为0,D=aabb1214b2ab23b3ab33b4ab34一a2bb124b2ab23b3ab33b4ab34=(aabb1214一a2bb)124b2ab23b3ab33b4ab34aaaaaaaaa14423423423434=(aabb一a2bb)(aabb+aabb一aab2—a2bb)1214 124 3423 2334 24 3243=ab(ab一ab)[ab(ab-ab)+ab(ab一ab)]14 21 12 43 32 23 34 23 32=ab(ab一ab)(ab一ab)(ab一ab)14 21 12 32 23 43 34二一ab(ab一ab)(ab一ab)(ab一ab)14 12 21 23 3234 43一ab13114(abii+1一ab)i+1ii=1a4(a一1)4(a一2)4(a一3)4(a一4)4a3(a一1)3(a一2)3(a一3)3(a一4)3a2(a-1)2(a一2)2(a一3)2(a一4)2aa一1a—2a—3a一411111习题6:解答:将行列式上下翻转后再左右翻转,不难得11111aa-1a-2a-3a—4D=(-1)4+3+2+1a25a3(a-1)2(a-2)2(a-3)2(a-4)2(a-1)3(a-2)3(a-3)3(a-4)3a4(a-1)4(a-2)4(a-3)4(a-4)411111a—4a-3a-2a-1a=(-1)4+3+2+1(-1)4+3+2+1(a-4)2(a-3)2(a-2)2(a-1)2a2(a-4)3(a-3)3(a-2)3(a-1)3a3(a-4)4(a-3)4(a-2)4(a-1)4a4=4!3!2!1!=288习题7:x-10000x-10000x00000x-1aaaax+ann-1n-22解答:按第一列展开,得D的递推公式nTOC\o"1-5"\h\z-1 0 0 0x -1 0 0D=xD+a (-1)n+10 x 0 0n n-1n0 0 x -1nD=xD+an n-1 nD=xD+an-1 n-2 n-1D=xD+an-2 n-3 n-2D=xD+a2 12D=x+a11将上述各式的两边分别乘以1,x,x2,,xn-1后全部相加,并化简,得D=xn+axn-1+axn-2++ax+an 1 2 n-1 n

习题8:abcd习题8