Question

If the system of linear equations :

$$\begin{gathered} x+ky+3z=0 \\ 3x+ky-2z=0 \\ 2x+4y-3z=0 \end{gathered}$$

has a non-zero solution $\left(x,y,z\right)$, then $\frac{xz}{y^2}$ is equal to :

• A. $-30$
• B. $30$
• C. $-10$
• D. $10$

Answer 1

The correct option is D

$10$

Explanation for the correct option.

Step 1. Find the value of $k$.

The system of equation $$\begin{gathered} x+ky+3z=0 \\ 3x+ky-2z=0 \\ 2x+4y-3z=0 \end{gathered}$$

As the system of equations has non-zero solutions.

So,$\Delta \Rightarrow \left|\begin{array}{ccc}1& k& 3\\ 3& k& -2\\ 2& 4& -3\end{array}\right|=0$

On solving

$$\begin{gathered} 1(-3k+8)-k(-9+4)+3(12-2k)=0 \\ \Rightarrow -3k+8+5k+36-6k=0 \\ \Rightarrow 4k=44 \\ \Rightarrow k=11 \end{gathered}$$

Step 2. Find the solution of the system.

Now the given equations become

$$\begin{gathered} x+11y+3z=0 \\ 3x+11y-2z=0 \\ 2x+4y-3z=0 \end{gathered}$$

The solution of the system is given as:

$\begin{array}{rcl}\frac{x}{(-22-33)}& =& \frac{y}{-(-2-9)}=\frac{z}{(11-33)}=L\\ & \Rightarrow & \frac{x}{-55}=\frac{y}{11}=\frac{z}{-22}=L\end{array}$

So, $x=-55L,y=11L,z=-22L$

Step 3. Find the value of $\frac{xz}{y^2}$.

In the expression $\frac{xz}{y^2}$, substitute $x=-55L,y=11L,z=-22L$

$\begin{array}{rcl}\frac{xz}{y^2}& =& \frac{\left(-55L\right)\left(-22L\right)}{{\left(11L\right)}^2}\\ & =& 5\times 2\\ & =& 10\end{array}$

Hence, the correct option is D.