线性代数课后练习答案.pdf

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1、线性代数练习答案练习一一、1.AB2.ABD3.D4.C5.BD二、1234123412340521105211052111.D===n0−10−10−1000−61200−6120−5−14−1700−12−6000−30=⋅⋅−15(6)(30)⋅−=900111⋯1010⋯02.D=002⋯0=⋅⋅⋅112⋯⋅(n−1)=(n−1)!n⋯⋯⋯⋯⋯000⋯n−13.n(n+1)23⋯n123⋯n123⋯n2n(n+1)34⋯1134⋯1134⋯12n(n+1)n(n+1)n(n+1)Dn=245⋯2=2145⋯2=2145⋯2⋯⋯

2、⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯n(n+1)12⋯n−1112⋯n−1112⋯n−12123⋯n−1n11⋯11−n011⋯11−n11⋯1−n1n(n+1)n(n+1)=011⋯1−n1=⋯⋯⋯⋯⋯22⋯⋯⋯⋯⋯⋯11−n⋯1101−n1⋯111−n1⋯11(n−1)11⋯11−n11⋯1−100⋯−nn00⋯−n0n(n+1)n(n+1)=⋯⋯⋯⋯⋯=⋯⋯⋯⋯⋯220−n⋯0n0−n⋯00−n0⋯0n−n0⋯00(n−1)(n−1)(n−1)(n−2)n(n+1)n−12n−2=(−1)(−1)n20xyz0xyzxyzx−x00x00

3、021+4.D=(∵r−r)=(∵c+c)=−(1)xy+z0x42121yz0xyy+z0xy+zx0zyx0zy+zx0xyzxyy+zxy+z32+=−xy+z0x(∵r−r)=−xy+z0x(∵c+c)=−−x(1)x3232y+zx0x−x0x0222422=xx[−(y+z)]=x−xy(+z)0−2−37−16023−71620−54−9−205−4955.D=350−3−6=−(1)−3−5036=−D⇒D=0555−7−430−874−308169680−16−9−6−80三、22a2a+12a+32a+5a2a+1

4、22c−c2212b2b+12b+32b+3b2b+122c3−c2证：左=(∵c−c)=(∵)=02322c2c+12c+32c+3c2c+122c4−c3c−c2432d2d+12d+32d+3d2d+122练习二一、1.ACD2.CD3.ABD4.D5.C二、−100⋯00x00⋯00x−10⋯00x−10⋯00010−⋯000x0⋯001.D=−(1)n+1a0x−1⋯00+−(1)n+2a0x−1⋯00+−(1)n+3a00−1⋯00nn⋯⋯⋯⋯⋯⋯n−1⋯⋯⋯⋯⋯⋯n−2⋯⋯⋯⋯⋯⋯000⋯−10000⋯−10000⋯−1

5、0000⋯x−1000⋯x−1000⋯x−1x−1⋯000x−1⋯000x−1⋯0000x⋯0000x⋯0000x⋯000nn+−200⋯000nn+−100⋯000nn+00⋯000+⋯+−(1)a3⋯⋯⋯⋯⋯⋯+−(1)a2⋯⋯⋯⋯⋯⋯+−(1)a1⋯⋯⋯⋯⋯⋯00⋯0−1000⋯0x000⋯0x−100⋯0x−100⋯00−100⋯00xn+1n−1n+2n−22n−1n−22nn−1=−(1)a(1)−+−(1)a(1)−x+⋯+−(1)a(1)−x+−(1)axnn−1212n2n2nn−22nn−1=−(1)a+−(1)a

6、x+⋯+−(1)ax+−(1)axnn−121nn−2n−1ni−=an+an−1x+⋯+ax2+ax1=∑axii=12.11⋯1nn(+1)aa−1⋯an−nn(+1)nn(+1)222222Dn+1=−(1)a(a−1)⋯(an−)=−(1)∏[(ai−+1)(−a−+j1)](1)=−∏(ji−)⋯⋯⋯⋯n+≥>≥1ij1n+≥>≥1ij1nnna(a−1)⋯(an−)3.ab⋯00b0⋯000a⋯00ab⋯0011+n+1n−1n+1n−1nn+1nD=−(1)a⋯⋯⋯⋯⋯+−(1)b⋯⋯⋯⋯⋯=aa+−(1)bb=a+−(

7、1)bn00⋯ab00⋯b000⋯0a00⋯ab三、1111511112−14−22−141.D==−142≠⇒0方程组有唯一解；D==−1422315−−−1−−−−23153121101211DDDD1234D=−284,D=−426,D=142⇒x==1,x==2,x==3,x==−12341234DDDD⎧−a+b−c+d=0⎪32⎪a+b+c+d=42.设f(X)=ax+bx+cx+d由已知可得⎨⎪8a+4b+2c+d=3⎪⎩27a+9b+3c+d=16求得D=48，D1=96，D2=--240，D3=0，D4=336，32

8、则a=2,b=−5,c=0,d=7,于是f(x)=2x−5x+7四、33ab+ab22a−b22证：n=2⇒D==a+abb+,右==a+abb+=左21ab+ab−假设对于n-1阶结论成立abab+0⋯001ab0⋯0