Discussion Forum

If the system of linear equation x+ky+3z=0;3x+ky-2z=0; 2x+4y-3z=0  has a non-zero solution ( x,y,z), then $\frac{xz}{y^{2}}$ is equal to

Answer:

Explanation:

We have, x+ky+3z=0; 3x+ky-2z=0;2x+4y-3z=0

System of the equation has non-zero solution, If

 $\begin{bmatrix}1 & k & 3 \\3 & k & -2 \\2 & 4 & -3 \end{bmatrix}=0$

$\Rightarrow$   (-3k+8)-k(-9+4)+3(12-2k)=0

$\Rightarrow$   -3k+8+9k-4k+36-6k=0

$\Rightarrow$    -4k+44=0 $\Rightarrow$ k=11

Let z=λ , then we get

        x+11y+3λ =0      ...... (i)

       3x+11y-2λ=0         .......(ii)

and   2x+4y-3λ =0      ........(iii)

Solving Eqs. (i) and (ii) , we get

   $x=\frac{5\lambda}{2},y=\frac{-\lambda}{2}, z=\lambda$

$\Rightarrow$   $\frac{xz}{y^{2}}=\frac{5\lambda^{2}}{2\times (-\frac{\lambda}{2})^{2}}=10$