Question
If the system of linear equations
$$\eqalign{
& x + ky + 3z = 0 \cr
& \,3x + ky - 2z = 0 \cr
& 2x + 4y - 3z = 0 \cr} $$
has a non-zero solution $$(x, y, z),$$ then $$\frac{{xz}}{{{y^2}}}$$ is equal to:
A.
10
B.
$$- 30$$
C.
30
D.
$$- 10$$
Answer :
10
Solution :
For non - zero solution of the system of linear equations;
\[\left| \begin{array}{l}
1\,\,\,\,\,\,\,\,\,\,\,\,k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\\
3\,\,\,\,\,\,\,\,\,\,\,k\,\,\,\,\,\,\,\,\, - 2\\
2\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\, - 3
\end{array} \right| = 0\]
$$ \Rightarrow \,\,k = 11$$
Now equations become
$$\eqalign{
& x + 11y + 3z = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr
& \,3x + 11y - 2z = 0\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr
& 2x + 4y - 3z = 0\,\,\,\,\,\,\,\,\,\,.....\left( 3 \right) \cr} $$
Adding equations (1) & (3) we get
$$\eqalign{
& 3x + 15y = 0 \cr
& \Rightarrow \,\,x = - 5y \cr} $$
Now put $$x = - 5y$$ in equation (1), we get
$$\eqalign{
& - 5y + 11y + 3z = 0 \cr
& \Rightarrow \,\,z = - 2y \cr
& \therefore \frac{{xz}}{{{y^2}}} = \frac{{\left( { - 5y} \right)\left( { - 2y} \right)}}{{{y^2}}} = 10 \cr} $$