If the system of linear equations x - 2y + kz = 1, 2x + y + z = 2, 3x - y - kz = 3 has a solution (x

Q: Wait, why do we reject the negative root here? Or is there no root to reject?

In this specific problem, we are not solving a quadratic equation, so there are no roots to reject! We are dealing with a system of linear equations. The condition $z \neq 0$ is given to ensure that the system is consistent and that the variable $z$ actually exists in the solution, preventing a trivial or degenerate case. We simply use linear algebra to find the locus.

Q: Why can't we just use Cramer's Rule or matrix inversion to solve for $x, y, z$?

You absolutely could, but that would be a long, tedious path! In JEE Advanced, time is your most precious resource. By observing the symmetry in the coefficients of $kz$ and $-kz$, we can bypass the heavy lifting of determinants and find the relationship between $x$ and $y$ in just one step. Always look for the 'shortcut' hidden in the symmetry of the equations.

Q: Does the value of $k$ matter for the final line equation?

That is the beauty of this problem! By adding the two equations, the term $kz$ is eliminated entirely. This means the relationship between $x$ and $y$ is independent of the value of $k$. Whether $k$ is $1, 5,$ or $100$, the point $(x, y)$ will always lie on the same line.

Question

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

Options:

(A) $3x - 4y - 1 = 0$

(B) $3x - 4y - 4 = 0$

(C) $4x - 3y - 4 = 0$

(D) $4x - 3y - 1 = 0$

Sigma Percentile · Accepted Answer

### The Geometry of Hidden Constraints

Welcome, future engineer. Today, we are not just solving a system of equations; we are uncovering a hidden geometric truth.

When you look at a system of linear equations like this, it is easy to feel overwhelmed by the variables $x$, $y$, $z$, and the parameter $k$. But remember, in the world of JEE Advanced, the most complex-looking problems often hide the most elegant solutions. Let us peel back the layers together.

### The Art of Observation

We are given the system:

$$(1) \quad x - 2y + kz = 1$$
$$(2) \quad 2x + y + z = 2$$
$$(3) \quad 3x - y - kz = 3$$

Our objective is to find the locus of the point $(x, y)$. This means we need to find a single equation that relates $x$ and $y$ directly, without the interference of $z$ or the parameter $k$.

Most students would immediately jump to Cramer's Rule or Gaussian elimination. While those methods are robust, they are also time-consuming.

Look closer at the equations. In equation $(1)$, we have a $+kz$ term. In equation $(3)$, we have a $-kz$ term. This is a mathematical gift!

### The Power of Linear Combination

When you see coefficients that are additive inverses, your instinct should immediately be to add the equations. By adding equation $(1)$ and equation $(3)$, we perform a strategic operation that simplifies the entire system.

Let us execute this step-by-step:

$$(x - 2y + kz) + (3x - y - kz) = 1 + 3$$

Watch the magic happen. The $+kz$ and $-kz$ terms cancel out perfectly, leaving us with:

$$(x + 3x) + (-2y - y) = 4$$

$$4x - 3y = 4$$

### The Final Revelation

We have successfully eliminated both $z$ and $k$ in a single stroke. We are left with the linear equation $4x - 3y = 4$.

To bring this into the standard form of a straight line, $Ax + By + C = 0$, we simply rearrange the terms:

$$4x - 3y - 4 = 0$$

This is the equation of the line on which any valid solution $(x, y)$ must lie. It is a beautiful result, isn't it?

We didn't need to find the specific values of $x$, $y$, or $z$. We didn't need to know the value of $k$. We simply recognized the symmetry and let the algebra do the work for us.

This is the essence of competitive mathematics: observation, strategy, and execution. Keep this mindset, and you will find that even the most intimidating problems have a path to the solution that is both simple and profound. The final locus is **$4x - 3y - 4 = 0$**.

Animated Solution for Mathematics - Matrices and Determinants: 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 :

Select Answer:

Visualized Solution

System of Equations

Identify the Goal

Strategic Observation

Setup the Operation

Compute $x$ Terms

Compute $y$ Terms

Eliminate $k z$ Terms

Compute Constants

Standard Form

Visualizing the Locus

Plotting the Line

Final Conclusion

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

The Geometry of Hidden Constraints

The Art of Observation

The Power of Linear Combination

The Final Revelation

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