First slide

System of linear equations

Question

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

Moderate

A

$- 4$
B

$\frac{1}{2}$
C

$- \frac{1}{4}$
D

$\frac{3}{4}$

Solution

Given system of linear equations
$\begin{aligned} 2 x + 3 y - z = 0 \\ x + k y - 2 z = 0 \end{aligned}$
$And 2 x - y + z = 0 has a non-trivial solution (x, y, z)$
$\begin{array}{c} ∴ Δ = 0 \Rightarrow |\begin{array}{ccc} 2 & 3 & - 1 \\ 1 & k & - 2 \\ 2 & - 1 & 1 \end{array}| = 0 \\ 2 (k - 2) - 3 (1 + 4) - 1 (1 - 2 k) = 0 \\ \Rightarrow 2 K - 4 - 15 + 1 + 2 K = 0 \\ \Rightarrow 4 K = 18 \Rightarrow K = \frac{9}{2} \end{array}$
So, system of linear equations is
$\begin{aligned} 2 x + 3 y - z = 0 - - - - - (1) \\ 2 x + 9 y - 4 z = 0 - - - - (2) \end{aligned}$
$And 2 x - y + z = 0 - - - - - (3)$
From Eqs. (i) and (ii) , we get
$6 y - 3 z = 0, \frac{y}{z} = \frac{1}{2}$
From Eqs. (i) and (iii), we get
$4 x + 2 y = 0 \Rightarrow \frac{x}{y} = - \frac{1}{2}$
so, $\frac{x}{z} = \frac{x}{y} \times \frac{y}{z} = - \frac{1}{4} \Rightarrow \frac{z}{x} = - 4$ $[∵ \frac{y}{z} = \frac{1}{2} a n d \frac{x}{y} = - \frac{1}{2}]$
$∴ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k = - \frac{1}{2} + \frac{1}{2} - 4 + \frac{9}{2} = \frac{1}{2}$

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