If D = left| begin{array}{ccc} frac{1}{z} & f | vedclass

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(D) Let the determinant be $D$. Multiply $C_1$ by $x$,$C_2$ by $y$,and $C_3$ by $z$,and divide the determinant by $xyz$:$D = \frac{1}{xyz} \left| \beg

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$D$ is dependent on $x, y, z$

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(D) Let the determinant be $D$. Multiply $C_1$ by $x$,$C_2$ by $y$,and $C_3$ by $z$,and divide the determinant by $xyz$:$D = \frac{1}{xyz} \left| \begin{array}{ccc} \frac{x}{z} & \frac{y}{z} & -\frac{x+y}{z} \ -\frac{y+z}{x} & \frac{y}{x} & \frac{z}{x} \ -\frac{y(y+z)}{x^2} & \frac{y(x+2y+z)}{xz} & -\frac{yz(x+y)}{xz^2} \end{array} ight|$.By performing row and column operations,or by simplifying the determinant,we find that the rows or columns are linearly dependent.Specifically,if we evaluate the determinant,we find $D = 0$.Since $D = 0$,the value of the determinant is constant (zero) and does not depend on $x, y,$ or $z$.Therefore,the statement '$D$ is dependent on $x, y, z$' is incorrect.

If $D = \left| \begin{array}{ccc} \frac{1}{z} & \frac{1}{z} & -\frac{(x+y)}{z^2} \\ -\frac{(y+z)}{x^2} & \frac{1}{x} & \frac{1}{x} \\ -\frac{y(y+z)}{x^2z} & \frac{x+2y+z}{xz} & -\frac{y(x+y)}{xz^2} \end{array} \right|$,then the incorrect statement is:

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