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$$\begin{vmatrix} y+z & x & x\\ y & z+x & y\\ z & z & x+y \end{vmatrix}=k(xyz)$$ 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?

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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 $4xyz$.

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