α=(a1,a2,a3)T,β=(b1,b2,b3)T,则|αβT+βαT|=0?结论正确吗?为什么?
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α=(a1,a2,a3)T,β=(b1,b2,b3)T,则|αβT+βαT|=0?
结论正确吗?为什么?
αβT= a1b1 a1b2 a1b3
a2b1 a2b2 a2b3
a3b1 a3b2 a3b3
βαT= b1a1 b1a2 b1a3
b2a1 b2a2 b2a3
b3a1 b3a2 b3a3
没有任何迹象表明|αβT+βαT|=0
|A|=|a1b1+b1a1 a1b2+b1a2 a1b3+b1a3|=a1*a2*a3*|b1+b1 b2+b1*a2/a1 b3+b1*a3/a1|
|a2b1+b2a1 a2b2+b2a2 a2b3+b2a3| |b1+b2*a1/a2 b2+b2 b3+b2*a3/a2|
|a3b...
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|A|=|a1b1+b1a1 a1b2+b1a2 a1b3+b1a3|=a1*a2*a3*|b1+b1 b2+b1*a2/a1 b3+b1*a3/a1|
|a2b1+b2a1 a2b2+b2a2 a2b3+b2a3| |b1+b2*a1/a2 b2+b2 b3+b2*a3/a2|
|a3b1+b3a1 a3b2+b3a2 a3b3+b3a3| |b1+b3*a1/a3 b2+b3*a2/a3 b3+b3|
=a1*a2*a3*|b1+b1 b2+b1*a2/a1 b3+b1*a3/a1 |
|b2*a1/a2-b1 b2-b1*a2/a1 b2*a3/a2-b1*a3/a1|
|b3*a1/a3-b1 b3*a2/a3-b1*a2/a1 b3-b1*a3/a1|
=a1*a2*a3*|b1+b1 b2+b1*a2/a1 b3+b1*a3/a1 |
|(b2/a2-b1/a1)*a1 (b2/a2-b1/a1)*a2 (b2/a2-b1/a1)*a3|
|(b3/a3-b1/a1)*a1 (b3/a3-b1/a1)*a2 (b3/a3-b1/a1)*a3|
=a1*a2*a3*(b2/a2-b1/a1)*(b3/a3-b1/a1)*|b1+b1 b2+b1*a2/a1 b3+b1*a3/a1|=0
|a1 a2 a3 |
|a1 a2 a3 |
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