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I'd like to understand how to prove that the determinant of the matrix below is zero, without actually calculating it, just based on the determinant properties

$$ \begin{vmatrix} x & y & z \\ z+y & z+x & x+y \\ 1 & 1 & 1 \\ \end{vmatrix} $$

I searched everywhere for but couldn't find a question like mine, so if it's duplicate, I apologize in advance

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Adding row two to row one doesn't affect the determinant, so it equals $$\begin{vmatrix} x+y+z & x+y+z & x+y+z \\ z+y & z+x & x+y \\ 1 & 1 & 1 \\ \end{vmatrix}.$$ If one row is a multiple of another, the determinant is zero.

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    $\begingroup$ Equivalently, the row vector $\begin{bmatrix}1 & 1 & {-x-y-z} \end{bmatrix}$ is in the left nullspace of the matrix. $\endgroup$ Nov 23, 2019 at 20:22

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