线性代数第五版答案(全)(6)

xay+bzaz+bxyay+bzaz+bx=ayaz+bxax+by+bzaz+bxax+byzax+byay+bzxax+byay+bzxay+bzzyzaz+bx=a2yaz+bxx+b2zxax+by

zax+byyxyay+bzxyzyzx=a3yzx+b3zxy

zxyxyzxyzxyz=a3yzx+b3yzx

zxyzxyxyz=(a3+b3)yzx.

zxya22b(3)2cd2

证明

(a+1)2(b+1)2(c+1)2(d+1)2

(a+2)2(b+2)2(c+2)2(d+2)2

(a+3)2(b+3)2

=0;(c+3)2(d+3)2

a2b2c2d2(a+1)2(b+1)2(c+1)2(d+1)2(a+2)2(b+2)2(c+2)2(d+2)2(a+3)2(b+3)2

(c4 c3,c3 c2,c2 c1得)(c+3)2(d+3)2

a22b=c2d22a+12b+12c+12d+12a+32b+32c+32d+32a+5

2b+5(c4 c3,c3 c2得)2c+52d+

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