QJEEM19Ma56c0: JEE Main 2019 Mathematics - Class 12 JEEM19Ma56c0 Solution: If the system of equations 2x + 3y – z = 0, x + ky – 2z = 0 and …

Given 2x + 3y – z = 0,

x + ky – 2z = 0

2x – y + z = 0

For non trivial solution

$$\mathrm{\Delta }=0\Rightarrow \left|\begin{array}{ccc}2& 3& -1\\ 1& k& -2\\ 2& -1& 1\end{array}\right|=0$$

$\Rightarrow k = {9 \over 2}$

$\therefore$ Equations are 2x + 3y – z = 0 ...(i)

2x – y + z = 0 ...(ii)

2x + 9y – 4z = 0 ...(iii)

By (i) – (ii) we get,

4y - 2z = 0

$\Rightarrow$ 2y = z .......(iv)

$\Rightarrow$ ${y \over z} = {1 \over 2}$

From equation (i) and (iv)

2x + 3y - 2y = 0

$\Rightarrow$ 2x + y = 0

$\Rightarrow$ ${x \over y} = - {1 \over 2}$

$\Rightarrow$ ${x \over y} \times {y \over z} = - {1 \over 2} \times {1 \over 2} = - {1 \over 4}$

$\Rightarrow$ ${z \over x} = - 4$

$\therefore$ ${x \over y} + {y \over z} + {z \over x} + k = {{ - 1} \over 2} + {1 \over 2} - 4 + {9 \over 2}$ = ${1 \over 2}$