If the system of equations 2x + 3y -- z = 0, x + ky
-- 2z = 0 and 2x -- y + z = 0 has a non-trival solution
(x, y, z), then \[{x \over y} + {y \over z} + {z \over x} + k\]
is equal to :-
Solutions · 1
Official solution
Given 2x + 3y -- z = 0,
x + ky -- 2z = 0
2x -- y + z = 0
For non trivial solution
\[\Delta = 0 \Rightarrow \left| {2123k−1−1−21 } \right| = 0\]
\[ \Rightarrow k = {9 \over 2}\]
\[ \therefore \] 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
\[ \Rightarrow \] \[{x \over y} = - {1 \over 2}\]
\[ \Rightarrow \] \[{x \over y} \times {y \over z} = - {1 \over 2} \times {1 \over 2} = - {1 \over 4}\]
\[ \Rightarrow \] \[{z \over x} = - 4\]
\[ \therefore \] \[{x \over y} + {y \over z} + {z \over x} + k = {{ - 1} \over 2} + {1 \over 2} - 4 + {9 \over 2}\] = \[{1 \over 2}\]