1.21计算I110110001a100..01.a2Da+11000#-11000..an1一列开始，第n+1列乘以第n列乘以解从最后，第2列乘以江dar加到第1列，得211...11ai00.11aimr001...1a2D,+1 =-0000..1αn-10000anl1.22计算11y1+Tiy3Tiyn1y2Tiyn-1"2y#T1y22.y212.y312.yn-1X1y312y33.3T3ya-1T3ynD. =r29-1T3ya-1n-1yn-iT1yn-1Tn-iyn...r3y.an-1ynTtyn12ynXnyn第n列提出公因子y，解从D，的第1行提出公因子1，然后第1行乘以－，依次加到第i行（i=2，"，n），得1...y!y2y3Vn-11211y22.y212y3T2Yn-1...r3aiy312y3x3y323.yn1Dn=iyn1yu-1t2n-1T3yn-1Xn-1Y-1Xn-1X1n1.12yt3yn.X-1Yn21
1yiy3y2yu-100001y2y1-2000riyi12y3-y21300T1Y-1-V12-12Vn-1V22n-113-13-10.Tiyn-yixn12yn-Y2X3yn-Y3tnC-1y-yn-ixn=±1yh - (- 1)+1 , (z1y2 - y122) . (z2y3 - y273)(r.-1n - yn-1rn)TI(i+1y: -yi+1x,).Tiy-计算1.2311111+ai.11"111 + Q211111+a3..D,1111F.1 +an-1N111..11 +anVL将D，的最后一列写成两列的和解2111111o.1[1 + α1[1 + a1..11"1111101 + a21 + 42111011.111+ as..1+ asD.+..1111.1I....01 + an-111+a-111111..111an将第一个行列式的最后一列乘以（-1)加到其余各列，将第二个行列式按最后-列展开，得D,=a,Dn-1+a1a2"*an-1从而Dn-1=an-1Du-2+a1a2"*an-2,Dy=a3D2+a1a2=as(ar+a2+aa2)+aa2.所以D,=aia2"an-i+a1a2""an-2a,+...+anan-1""a4a2al+anan-1*agagai +anan-1asaga2+anau-1"agagazai=(a1a2an-1a,)(+++1+++++1)aan-la3a2-(a1a2*an--a,)(1+2),22
1.24计算a2a3a-1112a3a132an-1a,x3ala2+.+a-1aD, =a1a3a2Tn-1ana1a2a3Xnan-1解令，=（an一an）+an，将D,写成两个行列式的和011a2an-1ana2asr1a3an-1032aia1a3an-1a,2asan-10a23.an-1a1a2a3:.a1anan-1D,0a1a2ara1a23a3Eg1Xn-1alaia2a3anQ2a3an-1a,-1nan一行，并将第二个行列式按最后在第一个行列式中，从第一一行开始依次减去后一列展开，得到000..α1a22-32c10.00$2-42a3-±3.0000"13-a3D,=+(r,-a,)Dn-10000an-1Tn-1aiana2a3an=1an-1)+(rn-an)D.-1=an(ri-ai)(r2 a2)-(an-1从而Du-1 =(n-1-an-1)Dn-2 +an-1( - a1)(2 - a2)(n-2- an-2),D=(g"ag)D2+a3(t-a1)(2-a2)(-a)(i2-a)a(-a)(z-a2)a(1 -a,)(r2 -a2) +a2(/ - a)(g -ag)+i(2-a2)(-as).所以D=(r1-a1)(x2-a2)--(zn-1 -an-1)am+ (ri- a1)(x2 - az)-(zn-2- an-2)(rn - an)an-1 ++(-a1)(-as).(n-i-an-1)(,-an)az+(2-a2)(3-a)..(n-an)1(, - a)(_-a + 1),a-a.23
1.25计算[1-b-h-b-b...5126-36...(n-2)bna(n - 1)6b1a3h(n-1)a一(n-)2)6(n -1)bD, =1(n -2)aaa"..(n - 2)b(n - 1)6-n13aaa(n -1)b...2aaa..aa解从最后行开始，依次将后-行减去前一行，得[1-b-b-b...-b-b0na+b-6-26...(n - 3)b(n-2)b000...0a + 2bo0Dn =000Qα+3b0000a+(n-2)b010000..-aa+(n-1)bna+b-b- 26..(n-3)b- (n -2)b00a + 2b0-a...000a+3b..a0000++.a+(n2)b000.aa+(n-1)b将第2行(i=2,1)加到第一行，得,72a+ba+ba+b...atba+b000a+2ba0a + 36...00aD. =000+(n-2)ba0000..aα +(n-1)b将第ii=2,n-1)列乘以加到第一列，得a+ib+b)a(uta(a+b)a(a +b)Aa+ba+b...a+ba+b+26@ +36b(n-1)b000a + 2b0""00...00a+36D,=0000+(n - 2)b0006+(n-1)6aa(a+b)+a(a+b)α(a+b)2u+6+.(a+26)(a+3b)..[a+(n-1)b)a+2ha+364+(n-1)6)24
=(a + b)(a + 2b)(α + 3b)..[α +(n - 1)b] - [1+++*a+2+..++（n-1)b-(a+ 10)(1+2%),=1.26计算Q1*~1422"-2aor"anan-1r000b1..aoraoz200b2.airDa+1 *-20ba-1aor"air"a2"41~1Q2#~2aornb.an-1r解在D，+1中，将最后一行乘以（-1)加到第一行，其余从最后一行开始，每行减去前一行的文倍，得00000.an-b.000061...aot000.0b2air-brD.+100000bu-2..000bu-10an-2 — bn-20000...baan-1-bn-1r(= 1)n+1+1(4.b,)aorI(ar -br)--a:)E707I=1.27计算11[111...cl1c2c3ch-1.c2c3c3c+11D=C*-2c"-3c2-24c22371c"-1cn=lc23c2-2.解利用c=c-1+c-！.从最后一行开始依次减去前一行，得25