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Show that $$ \begin{vmatrix} b_1+c_1 & c_1+a_1& a_1+b_1 \\ b_2+c_2 & c_2+a_2 & a_2+b_2 \\ b_3+c_3 & c_3+a_3 & a_3+b_3 \\ \end{vmatrix} = 2\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} $$

I can only see the brute-force approach. However, it really seems to have a more attractive and smarter way, but cannot find how to do that. Any help is appreciated.

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Just to illustrate what @RobertIsrael pointed out: since$$\left(\begin{array}{ccc} b_{1}+c_{1} & c_{1}+a_{1} & a_{1}+b_{1}\\ b_{2}+c_{2} & c_{2}+a_{2} & a_{2}+b_{2}\\ b_{3}+c_{3} & c_{3}+a_{3} & a_{3}+b_{3} \end{array}\right)=\left(\begin{array}{ccc} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3} \end{array}\right)\left(\begin{array}{ccc} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{array}\right),$$we need only calculate$$\left|\begin{array}{ccc} 0 & 1 & 1\\ 1 & 0 & 1\\ 1 & 1 & 0 \end{array}\right|=-\left|\begin{array}{cc} 1 & 1\\ 1 & 0 \end{array}\right|+\left|\begin{array}{cc} 1 & 0\\ 1 & 1 \end{array}\right|=1+1=2.$$

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Hint: if $A$ is the matrix on the right, matrix on the left is $AB$ for a certain $B$.

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  • $\begingroup$ Wouldn't that take longer to find? $\endgroup$ – UmbQbify Jun 21 at 16:51
  • $\begingroup$ @AshWhole Longer than what? It's much quicker than VIVID's approach, as my answer shows. But if you know a quicker way, e.g. with the property you linked to, it'd be lovely to see an answer going through it. $\endgroup$ – J.G. Jun 21 at 16:53
  • $\begingroup$ @J.G., I thought of separating the Determinant (many would lead to 0 that way) but it's slightly hefty. Martund, has an even better solution. $\endgroup$ – UmbQbify Jun 21 at 17:03
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Apply \begin{align*} \begin{vmatrix} b_1+c_1 & c_1+a_1& a_1+b_1 \\ b_2+c_2 & c_2+a_2 & a_2+b_2 \\ b_3+c_3 & c_3+a_3 & a_3+b_3 \\ \end{vmatrix} =&\ \ \begin{vmatrix} 2(a_1+b_1+c_1) & c_1+a_1 & a_1+b_1\\ 2(a_2+b_2+c_2) & c_2+a_2 & a_2+b_2 \\ 2(a_3+b_3+c_3) & c_3+a_3 & a_3+b_3 \end{vmatrix}\\ =&2\begin{vmatrix} a_1+b_1+c_1 & b_1 & c_1 \\ a_2+b_2+c_2 & b_2 & c_2 \\ a_3+b_3+c_3 & b_3 & c_3 \end{vmatrix}\\ =&2\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \end{align*} EXPLANATION

  1. $C_1=C_1+C_2+C_3$
  2. $C_2=C_1-C_2$ , $C_3=C_1-C_3$
  3. $C_1=C_1-C_2-C_3$
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  • $\begingroup$ Thank you! Very helpful. $\endgroup$ – VIVID Jun 21 at 17:12
  • $\begingroup$ @VIVID, You're welcome :) $\endgroup$ – Martund Jun 21 at 17:14

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