Let λ ∈ R. The system of linear equations 2x₁ - 4x₂ + x₃ = 1, x₁ - 6x₂ + x₃ = 2, λx₁ - 10x₂ + 4x₃ =

Q: Wait, if $\Delta = 0$, doesn't that automatically mean 'No Solution'?

That is the most common trap in JEE! $\Delta = 0$ only means the system is *not* independent. It could be inconsistent (no solution) or dependent (infinite solutions). You must always check the consistency of the constants to distinguish between these two cases.

Q: How do I check if it's infinite solutions or no solution without solving the whole thing?

Great question! Check if the third equation is a linear combination of the first two. If the LHS coefficients match the combination but the RHS constant doesn't, it's inconsistent. If both match, it's infinite solutions. It is a powerful way to verify your result.

Q: Why did we use $k_1$ and $k_2$ to check the equations?

We are testing for linear dependence. If the third row is a linear combination of the first two, the planes are 'locked' together in a specific way. If the constants don't follow the same rule, they are parallel and disjoint, creating the inconsistency we are looking for.

Question

Let $\lambda \in R$. The system of linear equations $2x_1 - 4x_2 + x_3 = 1 \ x_1 - 6x_2 + x_3 = 2 \ \lambda x_1 - 10x_2 + 4x_3 = 3$ is inconsistent for :

Options:

(A) exactly two values of $\lambda$.

(B) exactly one positive value of $\lambda$.

(C) every value of $\lambda$.

(D) exactly one negative value of $\lambda$.

Sigma Percentile · Accepted Answer

### The Geometry of Inconsistency: A Journey into 3D Space

Welcome, future engineer! Today, we are not just solving a system of equations; we are exploring the geometry of three-dimensional space. Imagine you are standing in a room, and each of the three equations provided represents a flat, infinite plane slicing through that space.

Our goal is to find the value of $\lambda$ that makes these planes refuse to meet at a single, common point. This is what we call an 'inconsistent' system.

### Phase 1: The Determinant as a Gatekeeper

To begin, we look at the coefficient matrix. The system is given by:

$$2x_1 - 4x_2 + x_3 = 1$$
$$x_1 - 6x_2 + x_3 = 2$$
$$\lambda x_1 - 10x_2 + 4x_3 = 3$$

For these planes to intersect at a unique point, the determinant of the coefficient matrix, denoted as $\Delta$, must be non-zero. If $\Delta 
eq 0$, the system has a unique solution.

Therefore, for the system to be inconsistent (or dependent), we must have $\Delta = 0$. Let us set up our determinant:

$$\Delta = \begin{vmatrix} 2 & -4 & 1 \ 1 & -6 & 1 \ \lambda & -10 & 4 \end{vmatrix}$$

### Phase 2: The Algebraic Dance

Now, we expand this determinant along the first row. Be meticulous here; a single sign error can derail the entire calculation. Expanding along $R_1$, we get:

$$\Delta = 2((-6)(4) - (1)(-10)) - (-4)((1)(4) - (1)(\lambda)) + 1((1)(-10) - (-6)(\lambda))$$

Simplifying this step-by-step:

$$\Delta = 2(-24 + 10) + 4(4 - \lambda) + 1(-10 + 6\lambda)$$
$$\Delta = 2(-14) + 16 - 4\lambda - 10 + 6\lambda$$
$$\Delta = -28 + 16 - 10 + 2\lambda$$
$$\Delta = 2\lambda - 22$$

Setting $\Delta = 0$, we find our candidate value: $2\lambda = 22$, which gives us $\lambda = 11$.

### Phase 3: The JEE Trap

Stop! Do not circle $\lambda = 11$ just yet. This is where many students lose marks.

As we discussed, $\Delta = 0$ only tells us that the system is *not* independent. It could be inconsistent (no solution), or it could be dependent (infinite solutions). We must verify if $\lambda = 11$ truly leads to an inconsistency.

We test for linear dependence by assuming the third equation is a linear combination of the first two: $k_1(	ext{Eq } 1) + k_2(	ext{Eq } 2) = 	ext{Eq } 3$. Comparing the coefficients of $x_1$ and $x_3$, we get:

$$2k_1 + k_2 = 11$$
$$k_1 + k_2 = 4$$

Subtracting these equations, we find $k_1 = 7$, which implies $k_2 = -3$. Now, let us check if these multipliers satisfy the $x_2$ coefficient:

$$7(-4) - 3(-6) = -28 + 18 = -10$$

It matches perfectly! The left-hand sides are indeed linearly dependent. Now, for the final test: do the constants on the right-hand side follow the same rule?

$$7(1) - 3(2) = 7 - 6 = 1$$

But the constant in our third equation is $3$. Since $1 
eq 3$, the planes are parallel and do not intersect. The system is truly inconsistent!

### Conclusion

We have found that the system is inconsistent only when **$\lambda = 11$**. Since $11$ is a positive number, we have exactly one positive value of $\lambda$ that satisfies the condition.

You have navigated the trap, verified the geometry, and arrived at the truth. Keep this rigor in your toolkit, and no JEE problem will ever stand in your way!

Animated Solution for Mathematics - Matrices and Determinants: Let $λ \in R$ . The system of linear equations $2 x_{1} - 4 x_{2} + x_{3} = 1 x_{1} - 6 x_{2} + x_{3} = 2 λ x_{1} - 10 x_{2} + 4 x_{3} = 3$ is inconsistent for :

Select Answer:

Visualized Solution

System of Linear Equations

Condition for Inconsistency

Setting up the Determinant

Expanding the Determinant

Simplifying the Determinant

Solving for $λ$

The JEE Trap: Verification

Linear Combination Setup

Finding the Multipliers

Verifying the $x_{2}$ Coefficient

Checking the Constants

Final Conclusion

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

The Geometry of Inconsistency

Phase 1

Phase 2

Phase 3

Conclusion

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