Matrices and Determinants 4 Question 4
4. 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 $(x, y, z)$, then $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k$ is equal to
(2019 Main, 9 April II)
(a) -4
(b) $\frac{1}{2}$
(c) $-\frac{1}{4}$
(d) $\frac{3}{4}$
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Answer:
Correct Answer: 4. (b)
Solution:
- Given system of linear equations
$ \begin{gathered} 2 x+3 y-z=0, \\ x+k y-2 z=0 \end{gathered} $
and $2 x-y+z=0$ has a non-trivial solution $(x, y, z)$.
$ \therefore \Delta=0 \Rightarrow \begin{vmatrix} 2 & 3 & -1 \\ 1 & k & -2 \\ 2 & -1 & 1 \end{vmatrix}=0 $
$ \begin{aligned} & 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{aligned} $
So, system of linear equations is
and
$ \begin{aligned} 2 x+3 y-z & =0 \\ 2 x+9 y-4 z & =0 \end{aligned} $
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
$ \begin{aligned} & \quad 4 x+2 y=0 \Rightarrow \frac{x}{y}=-\frac{1}{2} \\ & \text { So, } \frac{x}{z}=\frac{x}{y} \times \frac{y}{z}=-\frac{1}{4} \Rightarrow \frac{z}{x}=-4 \quad [\because \frac{y}{z}=\frac{1}{2} \text { and } \frac{x}{y}=-\frac{1}{2} ]\\ & \therefore \frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k=-\frac{1}{2}+\frac{1}{2}-4+\frac{9}{2}=\frac{1}{2} . \end{aligned} $
