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Here's what I did

I know it equals 0 beause I solved it using minors, but I should solve it using determinants' properties. I have just detarted the determinant into 2, but I can't do nothing else.

prishila
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    Note that the sum of the second & third columns equals $a+b+c$ times the first column. – copper.hat Oct 15 '15 at 15:06
  • I don't know what you're using to make those images, but it might be easier to type it out in MathJax. For example, the determinant $$\begin{vmatrix} 1 & a & b+c \ 1 & b & c+a \ 1 & c & a+b\end{vmatrix}$$ is obtained by typing $$\begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{vmatrix}$$. –  Oct 15 '15 at 15:24

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Following the comment by copper.hat, let's do the following

$$\begin{align}\begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{vmatrix} &= \begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{vmatrix} + 0 \\ &= \begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{vmatrix} + \begin{vmatrix} 1 & a & a \\ 1 & b & b \\ 1 & c & c\end{vmatrix} \\ &= \begin{vmatrix}1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c\end{vmatrix} \\ &= (a+b+c)\begin{vmatrix}1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1\end{vmatrix} \\ &= 0\end{align}$$

We took advantage of $2$ properties of determinants here:

  1. The determinant of any matrix with a repeated row or column is $0$.
  2. Determinants are multilinear in their columns and rows.
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