《行列式的计算方法》ppt课件

1、第五节第五节 行列式的计算方法行列式的计算方法1. 利用行列式的性质化行列式为上下三角行列式利用行列式的性质化行列式为上下三角行列式2. 行列式按行或列展开，将行列式降阶行列式按行或列展开，将行列式降阶3. 利用利用Laplace定理将行列式降阶定理将行列式降阶4. 箭式行列式的求法箭式行列式的求法n210a0010a0100a1111aC1C21/a1C1C31/a2C1Cn1/ann21a1a1a10a0000a0000a0111an21naaanaaaaa111032121例15. 形如xaxaxaxaxaxaxaxaxannn2n12n22211n1211的行列式的计算原式nnn22n

2、221n12axaxxaxaxxaxaxx+nnn2n12n22211n1211axaxaaxaxaaxaxannn22n221n12aaxaaxaax+nnn12n211n11axxaaxxaaxxa+nnn2n12n22211n1211axaaaxaaaxaannn22n221n12aaxaaxaax+nnn12n211n11axaaxaaxaxaaxaaxaan2n122211211+nnn2n12n22211n1211aaaaaaaaa D + x A11+A21+An1 + x A12+A22+An2 + x A1n+A2n+Ann n1in1jijAxD+ 例2. 计算行列式2 2

3、2 22 2n n2 22 23 32 22 22 22 22 21 12 22 22 2解： 原式0 02 22 20 02 22)2)- -(n(n2 20 02 20 02 21 12 20 02 20 02 20 02 20 02 20 02 21 1- -2 20 02 20 02 20 02 20 00 00 02 2n n0 00 01 10 00 00 00 00 01 10 00 00 0+ 2n n1 1j ji,i,ijijA A一般的，对于0 00 00 0a a0 00 0a a0 00 0a a0 00 0a a0 00 00 0n1n12)2)3(n3(n1)1)

4、2(n2(n1n1nA11= A12 = A1n-1= 0, 除了i+ j = n+1，其它的Aij= 0对于行列式nnnn333322221111a a0 00 00 00 0a a0 00 00 00 0a a0 00 00 00 0a a除了i = j ,其它的Aij= 0回到例2. 原式0 00 00 02 2n n0 00 01 10 00 00 00 00 01 10 00 00 0n n1 1j ji,i,ijijA A+ 2 2A2n-10 00 02 2n n0 01 10 01 10 00 022)!1)(n(1)(1)(212)1)(n(n1n2)!2(n1)(123)(

5、n2)(nn2)!2(n1)(1)n(n21= 2-1n+1a2n-1例3. 求n阶行列式n21a1111a1111a1的值解： 原式n21a000a000a+ A11+ A22+ +Ann= a1a2an+ a2a3an+ a1a3a4an + + a1a2an-1)1 (1112121naaanaaa6. 晋级法n21a1111a1111a1n21a11101a11011a101111n21a0010a0100a11111n21a1a1a1a0000a0000a01111n21) )(1(1a aa aa an n2 21 1a a1 1a a1 1a a1 1n n2 21 1例4例5.

6、 求行列式n321an321na321n3a21n32a1的值解：原式n321an3210na3210n3a210n32a10n3211n21a0010a0100a1n211nanaaaaann000000000211212121)1 (212121nanaanaaaa1, a2,an 都不为0例6. 求行列式 nnn3n2n12nn2n32n22n12n232221n321nxxxxxxxxxxxxxxxx1111D的值解：D123n22222123nn 1n 2n 2n 2n 2n 2123nn 1n 1n 1n 1n 1123nnnnnn123n11111xxxxxxxxxxxxxxxx

7、xxxxxxxxx设 A1n+1 + xA2n+1 + x2A3n+1 +xn-1Ann+1 +xnAn+1n+1 Ann+1 = -1n+n+1Dn= - Dn2n 1nn 11(n 1)2(n 1)3(n 1)n(n 1)(n 1)(n 1)DAxAx AxAx A1jinji121nn)x(x)x)(xx(x)x)(xx(x等式两端xn-1的系数相等，可得1jinjin211)n(n)x(x)xx(xA1jinjin21)x(x)xx(xnD7. 递推公式法例7. 求行列式9400059000009400059400059n nD D的值解：原式n195004900900950049 1

8、n94004900009000541)(521nn 12950049009D2000950049 9Dn-1-20Dn-2Dn = 9Dn-1-20Dn-2 1 Dn- 4Dn-1 = 5Dn-1-20Dn-1 2 = 5Dn-1 4Dn-2= 52Dn-2 4Dn-3=5n-3D3- 4D29D61,9459D369,940594059D123所以有 Dn 4Dn-1 = 5n 2 2n1nn20D9DD2n1n1nn20D4D5DD3n233n2n1n4)5D(D4)5D4(Dn1nn45DD352 43: Dn5n+1 4n+1对于一般情况，假设n阶行列式Dn满足Dn= aDn-1- b

9、Dn-2设 a = p + q, b = pq那么有 Dn pDn-1 = qDn-1 bDn-2 = qDn-1 pDn-2 = q2Dn-2 pDn-3 = = qn-mDm- pDm-1Dn- qDn-1= pDn-1 bDn-2 = pDn-1 qDn-2= p2Dn-2- qDn-3= = pn-mDm - qDm-1其中Dm和Dm-1可以求出,122p 1q: p-qDn= pn-m+1Dm- pDm-1 qn-m+1Dm-qDm-1进而求得Dn例8. 求解n阶行列式yzxyyyyzxyyyzzxyy()zzzxyzzzzx 解：原式yy)(xzzzzy0 xzzzy0yxzzy0

10、yyxzy0yyyxyzzzzyxzzzyyxzzyyyxzyyyyxy)(xzzzz0 xzzz0yxzz0yyxz0yyyx原式左x yDn-10 0 x xz z0 00 00 00 0y yx x0 00 00 00 00 0y yx xx xz z0 00 00 00 0y yx xx xz zy yy yy yy yx x右右= yx-zn-1 Dnx-yDn-1+yx-zn-1 1zz)(xz0z0z0z0yxzzzyyxzzyyyxzyyyyx原式zzzzzyxzzzyyxzzyyyxzyyyyxz)(x0000yxzzzyyxzzyyyxzyyyyxDnx-z Dn-1+ zx-yn-1 21x-z-2x-y: y-zDn = yx-zn-zx-ynzzynnny(x)(x)Dyz小结1. 利用行列式的性质化行列式为上下三角行列式2. 行列式按行或列展开，将行列式降阶3. 利用Laplace定理将行列式降阶4. 箭式行列式的求法xaxaxaxaxaxaxaxaxannn2n12n22211n12115. 利用公式n1in1jijAxD6. 晋级法7. 递推公式法 Dn= mDn-1- nDn-2 假设p + q =