3 and 4 .Determinants and Matrices
hard

यदि रैखिक समीकरण निकाय $x+k y+3 z=0$,$3 x+k y-2 z=0$,$2 x+4 y-3 z=0$ का एक शून्येतर हल $(x, y, z)$ है, तो $\frac{x z}{y^{2}}$ बराबर है

A

$10$

B

$-30$

C

$30$

D

$-10$

(JEE MAIN-2018)

Solution

For non zero solution of the system of linear equation;

$\left| {\begin{array}{*{20}{c}} 1&k&3\\ 3&k&{ – 2}\\ 2&4&{ – 3} \end{array}} \right| = 0$

$ \Rightarrow k = 11$ 

Now equations become 

$x+11y+3z=0$           ……$(1)$

$3x+11y-2z=0$           ……$(2)$

$2x+4y-3z=0$              …….$(3)$

Adding equations $(1)$ and $(3)$  we get

$3x + 15y = 0 \Rightarrow k = 5 – y$

Now put $x=-5y$ in equation $(1)$, we get 

$ – 5y + 11y + 3z = 0 \Rightarrow z =  – 2y$

$\therefore \frac{{xz}}{{{y^2}}} = \frac{{\left( { – 5y} \right)\left( { – 2y} \right)}}{{{y^2}}} = 10$

Standard 12
Mathematics

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