Matrices and Determinants 4 Question 13
14. If the system of linear equations
$ \begin{aligned} x+k y+3 z & =0,3 x+k y-2 z=0 \\ 2 x+4 y-3 z & =0 \end{aligned} $
has a non-zero solution $(x, y, z)$, then $\frac{x z}{y^{2}}$ is equal to
(a) -10
(b) 10
(c) -30
(d) 30
(2018 Main)
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Answer:
Correct Answer: 14. (b)
Solution:
- We have,
$x+k y+3 z=0 ; 3 x+k y-2 z=0 ; 2 x+4 y-3 z=0$
System of equation has non-zero solution, if
$ \begin{vmatrix} 1 & k & 3\\ 3 & k & -2 \\ 2 & 4 & -3 \end{vmatrix} $ =0
$ \begin{aligned} & \Rightarrow(-3 k+8)-k(-9+4)+3(12-2 k)=0 \\ & \Rightarrow \quad-3 k+8+9 k-4 k+36-6 k=0 \\ & \Rightarrow \quad-4 k+44=0 \Rightarrow k=11 \\ & \text { Let } \quad z=\lambda \text {, then we get } \\ & x+11 y+3 \lambda=0 \\ & 3 x+11 y-2 \lambda=0 \\ & \text { and } \quad 2 x+4 y-3 \lambda=0 \end{aligned} $
Solving Eqs. (i) and (ii), we get
$ x=\frac{5 \lambda}{2}, y=\frac{-\lambda}{2}, z=\lambda \Rightarrow \frac{x z}{y^{2}}=\frac{5 \lambda^{2}}{2 \times-\frac{\lambda}{2}}=10 $
