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matrix theory_basic results and techniques_exercise_1.2.2,1.2.3

Evaluate the determinant
|1+x1111+y1111+z|


Answer:将矩阵的第一行乘以-1加到第二行,得到
|1+x11xy0111+z|
将矩阵的第一行乘以-1加到第三行,得到
|1+x11xy0x0z|=xyz+xy+yz+zx

 

 

 

Show the 3×3 Vandermonde determinant identity

|111a1a2a3a21a22a23|=(a1a2)(a2a3)(a3a1)

 

我们先看该矩阵的转置:
|1a1a211a2a221a3a23|
然后第一行乘以-1加到第二行上,将第一行乘以-1加到第三行上:

|1a1a210a2a1a22a210a3a1a23a21|=(a2a1)(a23a21)(a3a1)(a22a21)=(a1a2)(a3a1)(a2a3)
And evaluate the determinant
|1aa2bc1bb2ca1cc2ab|


Answer:将矩阵的第一行乘以-1加到第二行上,将矩阵的第二行乘以-1加到第三行上,将矩阵的第三行乘以-1加到第一行上:

|0ac(ac)(a+b+c)0ba(ba)(a+b+c)0cb(cb)(a+b+c)|
可见,最后的结果是0.

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