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3 and 4 .Determinants and Matrices
hard
If the system of linear equations $x + ky + 3z = 0;3x + ky - 2z = 0$ ; $2x + 4y - 3z = 0$ has a non-zero solution $\left( {x,y,z} \right)$ then $\frac{{xz}}{{{y^2}}} = $. . . . .
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
