System of linear equations

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If the system of equations $2 x + 3 y - z = 0, x + k y - 2 z = 0$ and $2 x - y + z = 0$ has a non trivial solution. Then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k =$

Moderate

Solution

$|\begin{matrix} 2 & 3 & - 1 \\ 1 & k & - 2 \\ 2 & - 1 & 1 \end{matrix}| = 0 \Rightarrow K = \frac{9}{2}$
Given equations are $2 x + 3 y - z = 0 \to (1)$
$x + \frac{9}{2} y - 2 z = 0 \to (2), 2 x - y + z = 0 \to (3)$
$(1) - (3) \Rightarrow 4 y - 2 z = 0 \Rightarrow \frac{z}{y} = 2$
$(1) + (3) \Rightarrow \frac{x}{y} = \frac{- 1}{2} \to (5)$
$(4) \times (5) \Rightarrow \frac{y}{z} \times \frac{x}{y} = \frac{- 1}{4} \Rightarrow \frac{z}{x} = - 4$
$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k = - \frac{1}{2} + \frac{1}{2} - 4 + \frac{1}{2} = \frac{1}{2}$

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System of linear equations

Remember concepts with our Masterclasses.

If the system of equations 2x+3y−z=0, x+ky−2z=0and 2x−y+z=0has a non trivial solution. Then xy+yz+zx+k=

23-11k-22-11=0 ⇒K=92Given equations are 2x+3y−z=0→(1)x+92y−2z=0→(2), 2x−y+z=0→(3)(1)−(3)⇒ 4y−2z=0 ⇒zy=2(1)+(3)⇒ xy=−12→(5)(4) ×(5) ⇒yz×xy=−14 ⇒zx=−4xy+yz+zx+k= −12+12−4+12=12

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If the system of equations $2 x + 3 y - z = 0, x + k y - 2 z = 0$ and $2 x - y + z = 0$ has a non trivial solution. Then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k =$