If the system of equations 2x + 3y - z = 0,x | vedclass

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(C) For the system of equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.$D = \begin{vmatrix} 2 & 3 & -1

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$\frac{1}{2}$

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(C) For the system of equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.$D = \begin{vmatrix} 2 & 3 & -1 \ 1 & k & -2 \ 2 & -1 & 1 \end{vmatrix} = 0$Expanding along the first row:$2(k - 2) - 3(1 - (-4)) - 1(-1 - 2k) = 0$$2k - 4 - 3(5) + 1 + 2k = 0$$4k - 18 = 0 \Rightarrow 4k = 18 \Rightarrow k = \frac{9}{2}$Substituting $k = \frac{9}{2}$ into the equations:$(1) 2x + 3y - z = 0$$(2) x + \frac{9}{2}y - 2z = 0$$(3) 2x - y + z = 0$Subtracting $(3)$ from $(1)$:$(2x + 3y - z) - (2x - y + z) = 0 \Rightarrow 4y - 2z = 0 \Rightarrow z = 2y \Rightarrow \frac{y}{z} = \frac{1}{2}$Substituting $z = 2y$ into $(1)$:$2x + 3y - 2y = 0 \Rightarrow 2x + y = 0 \Rightarrow \frac{x}{y} = -\frac{1}{2}$Since $z = 2y$ and $x = -\frac{1}{2}y$,then $\frac{z}{x} = \frac{2y}{-0.5y} = -4$Now,calculate the expression:$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k = -\frac{1}{2} + \frac{1}{2} - 4 + \frac{9}{2} = \frac{1}{2}$

If the system of equations $2x + 3y - z = 0$,$x + ky - 2z = 0$ and $2x - y + z = 0$ has a non-trivial solution $(x, y, z)$,then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ is equal to

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