Find the value of the determinant.

Question

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.

Answer 1

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.