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Using properties of determinants prove that:

contexto.140

Solution:

\begin{array}{l} \left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right| \\ =\left|\begin{array}{ccc} x+y+z & -x+y & -x+z \\ x+y+z & 3 y & z-y \\ x+y+z & y-z & 3 z \end{array}\right|\left[C_{1}^{\prime}=C_{1}+C_{2}+C_{3}\right] \\ =(x+y+z)\left|\begin{array}{ccc} 1 & -x+y & -x+z \\ 1 & 3 y & z-y \\ 1 & y-z & 3 z \end{array}\right|\left[C_{1}^{\prime}=C_{1} /(x+y+z)\right] \\ =(x+y+z)\left|\begin{array}{ccc} -x+y & 3 y & y-z \\ -x+z & z-y & 3 z \end{array}\right|[\text { transforming row and column }] \\ =(x+y+z)\left|\begin{array}{ccc} -x-2 y & 2 y+z & y-z \\ -x+y & -y-2 z & x \end{array}\right|\left[C_{1}^{\prime}=C_{1}-C_{2} \& C_{2}^{\prime}=C_{2}-C_{3}\right] \\ =(x+y+z)[0+0+(-x-2 y)(-y-2 z)-(-x+y)(2 y+z)][e x p a n s i o n \text { by first row }] \\ =(x+y+z)\left(x y+2 y^{2}+2 x z+4 y z+2 x y-2 y^{2}+x z-y z\right) \\ =(x+y+z)(3 x y+3 y z+3 x z) \\ =3(x+y+z)(x y+y z+z x) \end{array}