Question 3715: Determinant
show that determinant
b+c,a,a,
b,c+a,b
c.c. a+b
=4abc
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! |b+c a a|
|b c+a b|
|c c a+b|
= (a+b)(b+c)(c+a) + 2abc - ca(c+a) - bc(b+c) - ab(a+b)
= (b^2+ab+ca+bc)(c+a) + 2abc - ca(c+a) - bc(b+c) - ab(a+b)
= b^2c+abc+c^2a+bc^2 + ab^2+a^2b+ca^2+abc + 2abc - ca(c+a) - bc(b+c) - ab(a+b)
= (b^2c+ bc^2)+(c^2a+ ca^2) +(a^2b+ ab^2)+42abc - ca(c+a) - bc(b+c) - ab(a+b)
= 4abc.
Or let this determinant be f(a,b,c).
We see that f(0,b,c) = (b+c) (cb - bc) = 0
Similarly, f(a,0,c) = f(a,b,0) = 0
Hence,abc is a divisor of f(a,b,c).
Since deg f(a,b,c) = 3, assume f(a,b,c) = k abc.
Set a=1,b=1,c =-1, we get
|0 1 1|
|1 0 1| = -1+(-1)-2 = -4 = k*1*1*(-1)= -k
|-1 -1 2|
So, k = 4 and f(a,b,c) = 4 abc.
I prefer the 2nd method and hope that you can understand it.
Kenny
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