LA2.1.11

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Prove that {\begin{vmatrix}-2a&a+b&c+a\\b+a&-2b&b+c\\c+a&c+b&-2c\end{vmatrix}}=4(a+b)(b+c)(c+a)\,

put b+a=0,b=-a\, in d. Then A={\begin{vmatrix}-2a&0&c+a\\0&2a&c-a\\c+a&c-a&-2c\end{vmatrix}}\,

Applying R_{1}\rightarrow R_{1}+R_{2},R_{3}\rightarrow R_{3}+R_{2}\,

A={\begin{vmatrix}c-a&c-a&-c+a\\0&2a&c-a\\c+a&c+a&-c-a\end{vmatrix}}=(c-a)(c+a){\begin{vmatrix}1&1&-1\\0&2a&c-a\\1&1&-1\end{vmatrix}}=0\,

Therefore (a+b) is a factor of A.Similarly, (c+a) and (b+c) are also.

Observe that A is of third degree.Also the above mentioned factors multiplied also third degree.

Thus A=k(a+b)(b+c)(c+a)\, where k is a number.

To find k,put a=0,b=1,c=2 then {\begin{vmatrix}0&1&2\\1&-2&3\\2&3&-4\end{vmatrix}}=k(1)(2)(3)\,

0(8-9)-1(-4-6)+2(3+4)=6k,24=6k,k=4\,

Thus A=4(a+b)(b+c)(c+a)\,


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