# Solving a determinant using properties of a determinant

$$\begin{vmatrix} y+z & x & x\\ y & z+x & y\\ z & z & x+y \end{vmatrix}=k(xyz)$$ Find the value of $k$.

I solved this question by substituting $x=y=z=1$ and then expanding the determinant to get $k=4$ which is the correct answer.

Is there any method other than substitution and expanding the determinant right away?

Use the property that adding multiples of rows does not affect the determinant. Subtracting row 2 and 3 from row 1, we get $$\,\,\,\,\,\,\begin{vmatrix} y+z & x & x\\ y & z+x & y\\ z & z & x+y \end{vmatrix}\\= \begin{vmatrix} 0 & -2z & -2y\\ y & z+x &y\\ z & z & x+y \end{vmatrix}$$ Add half row 1 to row 2 and 3, $$=\begin{vmatrix} 0 & -2z & -2y\\ y & x &0\\ z & 0 & x \end{vmatrix}$$ And now the rule of Sarrus will quickly get you $4xyz$.