If the system of linear equations x + ky + 3z = 0, 3x + ky - 2z = 0, 2x + 4y - 3z = 0 has a non-zero

Q: Wait, why do we reject the trivial solution $(0, 0, 0)$?

The problem explicitly states that the system has a 'non-zero solution'. In a homogeneous system $AX = 0$, the trivial solution $(0, 0, 0)$ always exists. If the determinant of $A$ is non-zero, the trivial solution is the *only* solution. Therefore, for a non-zero solution to exist, the determinant must be zero, which makes the equations dependent and allows for infinitely many non-zero solutions.

Q: Why can't we just solve for $x, y, z$ as specific numbers?

Because the system is homogeneous and the determinant is zero, the equations are linearly dependent. This means there isn't a unique solution for $x, y, z$. Instead, there are infinitely many solutions that lie along a line in 3D space. We can only express two variables in terms of the third (e.g., $x$ and $y$ in terms of $z$), which is perfectly fine because the target expression $\frac{xz}{y^2}$ is a ratio where the specific value of $z$ will eventually cancel out.

Q: Does the value of $z$ matter for the final answer?

Not at all! Notice that the numerator $xz$ is of degree 2 (since $x$ is proportional to $z$) and the denominator $y^2$ is also of degree 2. When you substitute $x = c_1 z$ and $y = c_2 z$, the $z^2$ terms appear in both the numerator and denominator and cancel out completely. This confirms that the ratio is constant for any non-zero solution.

Question

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

Options:

(A) 30

(B) -10

(C) 10

(D) -30

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to unravel a problem that might look like a standard system of linear equations, but it hides a beautiful, symmetric secret.

When you see a system of equations where every constant term on the right-hand side is zero, like $x + ky + 3z = 0$, $3x + ky - 2z = 0$, and $2x + 4y - 3z = 0$, you are looking at a **homogeneous system**.

This is not just a collection of lines; it is a geometric structure where every plane passes through the origin $(0, 0, 0)$.

### Phase 1: The Determinant Key

In a homogeneous system, the origin is always a solution. But the problem whispers a secret: there is a *non-zero* solution. This is our golden ticket.

For a system of linear equations to have a non-trivial solution, the equations must be linearly dependent. Mathematically, this means the determinant of the coefficient matrix, which we denote as $\Delta$ or $D$, must be exactly zero.

If $D 
eq 0$, the only solution would be the trivial one, $(0, 0, 0)$, which contradicts our premise. So, we set our sights on the determinant:

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

### Phase 2: The Algebra of $k$

Now, let's expand this determinant along the first row. I know, I know—determinant expansion can feel tedious, but stay with me. It is the bridge to our answer.

Expanding along the first row, we get:

$$1(k(-3) - 4(-2)) - k(3(-3) - 2(-2)) + 3(3(4) - 2(k)) = 0$$

Let's simplify this step-by-step. The first term becomes $1(-3k + 8)$. The second term is $-k(-9 + 4)$, which simplifies to $-k(-5)$, or $+5k$. The third term is $3(12 - 2k)$, which is $36 - 6k$.

Putting it all together:

$$-3k + 8 + 5k + 36 - 6k = 0$$

Grouping the $k$ terms: $(-3 + 5 - 6)k = -4k$. Grouping the constants: $8 + 36 = 44$. Thus, $-4k + 44 = 0$, which leads us to the elegant result: **$k = 11$**.

### Phase 3: The Ratio Game

Now that we have $k = 11$, our system is fully defined. We have:

(1) $x + 11y + 3z = 0$

(2) $3x + 11y - 2z = 0$

We need to find the value of $\frac{xz}{y^2}$. A common trap here is to try and solve for $x, y,$ and $z$ as absolute values. But remember, we are dealing with a dependent system! We can only find the ratios.

Let's subtract equation (1) from equation (2):

$$(3x - x) + (11y - 11y) + (-2z - 3z) = 0 \ \Rightarrow 2x - 5z = 0 \ \Rightarrow x = \frac{5}{2}z$$

Now, substitute $x = \frac{5}{2}z$ back into equation (1):

$$\frac{5}{2}z + 11y + 3z = 0 \Rightarrow 11y = -\frac{11}{2}z \Rightarrow y = -\frac{1}{2}z$$

### The Final Victory

We have expressed $x$ and $y$ in terms of $z$. Now, let's look at our target expression: $\frac{xz}{y^2}$. Substitute our findings:

$$\frac{(\frac{5}{2}z) \cdot z}{(-\frac{1}{2}z)^2} = \frac{\frac{5}{2}z^2}{\frac{1}{4}z^2}$$

Look at that! The $z^2$ terms cancel out, leaving us with $\frac{5/2}{1/4} = \frac{5}{2} 	imes 4 = 10$.

The complexity vanishes, and we are left with the clean, crisp answer of **$10$**. This is the beauty of linear algebra—when you understand the underlying structure, the path clears itself.

Animated Solution for Mathematics - Matrices and Determinants: If the system of linear equations $x + k y + 3 z = 0, 3 x + k y - 2 z = 0, 2 x + 4 y - 3 z = 0$ has a non-zero solution $(x, y, z)$ , then $\frac{x z}{y ^{2}}$ is equal to :

Select Answer:

Visualized Solution

System of Homogeneous Equations

Condition for Non-Trivial Solution

Constructing the Determinant

Expanding the Determinant

Solving for $k$

Substituting $k$ Back

Expressing $x$ and $y$ in terms of $z$

Setting up the Target Expression

Final Calculation

Conclusion

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Victory

Similar Questions

If the system of equations $2 x + 3 y - z = 0, x + k y - 2 z = 0$ and $2 x - y + z = 0$ has a non-trivial solution $(x, y, z)$ , then $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$ is equal to :-

If the system of linear equations $x + y + 3 z = 0 x + 3 y + k^{2} z = 0 3 x + y + 3 z = 0$ has a non-zero solution $(x, y, z)$ for some $k \in R$ then $x + (\frac{y}{z})^{2}$ is equal to

If system of linear equations $x + y + z = 6$ , $x + 2 y + 3 z = 10$ , $3 x + 2 y + λ z = μ$ has more than two solutions, then $μ - λ^{2}$ is equal to ________.

If the system of linear equations, $x + y + z = 6 x + 2 y + 3 z = 10 3 x + 2 y + λ z = μ$ has more than two solutions, then $μ - λ^{2}$ is equal to . . . . .

If the system of equations $x - k y - z = 0$ , $k x - y - z = 0$ , $x + y - z = 0$ has a non-zero solution, then the possible values of $k$ are

If the system of linear equations $x - 2 y + k z = 1$ , $2 x + y + z = 2$ , $3 x - y - k z = 3$ has a solution $(x, y, z), z \neq = 0$ , then $(x, y)$ lies on the straight line whose equation is :

If the system of linear equations $2 x + y - z = 3$ , $x - y - z = α$ , $3 x + 3 y + β z = 3$ has infinitely many solution, then $α + β - α β$ is equal to .

The number of values of $k$ for which the linear equations $4 x + k y + 2 z = 0$ , $k x + 4 y + z = 0$ and $2 x + 2 y + z = 0$ possess a non-zero solution is

Let the system of linear equations $4 x + λ y + 2 z = 0, 2 x - y + z = 0, μx + 2 y + 3 z = 0, λ, μ \in R$ has a non-trivial solution. Then which of the following is true?

The following system of linear equations $7 x + 6 y - 2 z = 0$ , $3 x + 4 y + 2 z = 0$ , $x - 2 y - 6 z = 0$ , has