Confused.. why we cannot use determinant addition property for this question

Question

$$ \begin{vmatrix} x+y & z & z \\ x & y+z & x \\ y & y & x+z \\ \end{vmatrix} $$ The answer turns out to be 4xyz but after some transfomations. But here, why can we not use the addition property of matrices in 1st step and get the answer, which is giving 0.

Answer 1

The additive property of the determinant (or at least what I guess you're saying) applies when you have two matrices that share all but one row/column. Let's say $$A = \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}, \\ B = \begin{bmatrix} b_{11} & b_{12} & b_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix},\\ C = \begin{bmatrix} a_{11} + b_{11} & a_{12}+b_{12} & a_{13}+b_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix}, $$ Then $\det C = \det A + \det B$

Can you use this to solve the above problem? Yes, but you have to be clever $$ \begin{aligned} &\begin{vmatrix} x+y & z & z\\ x & y+z & x\\ y & y & x+z \end{vmatrix}\\ &= \begin{vmatrix} x+y & z & z\\ x & z-y & -x\\ y & y & x+z \end{vmatrix} + \begin{vmatrix} x+y & z & z\\ 0 & 2y & 2x\\ y & y & x+z \end{vmatrix}\\ &= 0 + 2\begin{vmatrix} x+y & z & z\\ 0 & y & x\\ y & y & x+z \end{vmatrix}\\ &= 2\begin{vmatrix} y & 0 & z\\ 0 & y & x\\ y & y & x+z \end{vmatrix} + 2\begin{vmatrix} x & z & 0\\ 0 & y & x\\ y & y & x+z \end{vmatrix}\\ &= 0 + 2\left(xy(z+x) - x^2y + xyz\right)\\ &= 4xyz \end{aligned} $$