The set of all values of λ for which the system of linear equations: 2x₁ - 2x₂ + x₃ = λx₁, 2x₁ - 3x₂

Q: Why do we set the determinant to zero? Can't we just use substitution or elimination like in school?

That is a fantastic question! In a standard system of equations, substitution works perfectly. However, this is a homogeneous system ($AX = 0$). If you try substitution, you will often find yourself chasing your tail, ending up with $0=0$ because the equations are linearly dependent. Setting the determinant to zero is the powerful, elegant shortcut that tells us exactly when those equations 'collapse' into a state where non-zero solutions exist.

Q: I found $\lambda = 1$ and $\lambda = -3$. But the root $\lambda = 1$ appears twice. Does the set contain three elements or two?

This is a classic trap! Remember the definition of a set in mathematics: it contains only distinct elements. Even though $\lambda = 1$ is a repeated root (algebraic multiplicity of 2), it is still just one unique value in the set of solutions. The set is $\{1, -3\}$, which contains exactly two elements.

Q: Why does the problem ask for a 'non-trivial' solution? What is a 'trivial' one?

The 'trivial' solution is the boring one: $x_1 = 0, x_2 = 0, x_3 = 0$. It works for any value of $\lambda$, but it doesn't tell us anything interesting about the system. A 'non-trivial' solution means at least one of the variables is non-zero. We are hunting for those specific values of $\lambda$ that allow the system to 'come alive' with non-zero values.

Question

The set of all values of $\lambda$ for which the system of linear equations: $2x_1 - 2x_2 + x_3 = \lambda x_1$, $2x_1 - 3x_2 + 2x_3 = \lambda x_2$, $-x_1 + 2x_2 = \lambda x_3$ has a non-trivial solution

Options:

(A) contains two elements

(B) contains more than two elements

(C) is an empty set

(D) is a singleton

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are not just solving a system of equations; we are peeling back the curtain on the behavior of linear systems.

When you see a problem like this, where a parameter $\lambda$ is dancing around the variables, it is easy to feel overwhelmed. But take a breath. Let us break this down, not as a chore, but as a discovery.

### The Homogeneous Transformation

We are given the system:

$$2x_1 - 2x_2 + x_3 = \lambda x_1$$
$$2x_1 - 3x_2 + 2x_3 = \lambda x_2$$
$$-x_1 + 2x_2 = \lambda x_3$$

At first glance, the $\lambda$ on the right side feels like an intruder. It is not a constant, and it is not a variable; it is a parameter that dictates the very nature of the system. Our first instinct must be to bring order to this chaos by moving everything to the left side.

This transforms our system into the standard homogeneous form: $(A - \lambda I)X = 0$. By rearranging, we obtain:

$$(2-\lambda)x_1 - 2x_2 + x_3 = 0$$
$$2x_1 - (3+\lambda)x_2 + 2x_3 = 0$$
$$-x_1 + 2x_2 - \lambda x_3 = 0$$

This is the moment where physics and math converge. We are looking for a **non-trivial solution**. If the matrix $(A - \lambda I)$ were invertible, we could simply multiply by its inverse to find $X = (A - \lambda I)^{-1}0$, which forces $X$ to be the zero vector.

That is the 'trivial' solution. We want to avoid that; we want the system to be 'singular'—we want it to have depth. For that to happen, the **determinant must be zero**.

### The Determinant's Power

Now, we construct our determinant. This is where precision is your best friend. We take the coefficients and place them into our grid:

$$\begin{vmatrix} 2-\lambda & -2 & 1 \ 2 & -(3+\lambda) & 2 \ -1 & 2 & -\lambda \end{vmatrix} = 0$$

Do not rush this. Expanding a $3 	imes 3$ determinant is a test of patience. Let us expand along the first row:

$$(2-\lambda)[\lambda(3+\lambda) - 4] - (-2)[-2\lambda - (-2)] + 1[4 - (3+\lambda)] = 0$$

### The Algebraic Grind

This is the part where many students stumble, not because they do not understand the math, but because they lose a negative sign. Let us simplify carefully.

Inside the first bracket, we have $\lambda(3+\lambda) - 4$, which simplifies to $\lambda^2 + 3\lambda - 4$. The second term becomes $2(-2\lambda + 2)$, and the last term is $(1-\lambda)$.

Expanding these, we get:

$$(2-\lambda)(\lambda^2 + 3\lambda - 4) + 2(-2\lambda + 2) + (1-\lambda) = 0$$

Multiplying these out yields:

$$(2\lambda^2 + 6\lambda - 8 - \lambda^3 - 3\lambda^2 + 4\lambda) + (-4\lambda + 4) + (1-\lambda) = 0$$

Grouping the terms by powers of $\lambda$ gives us the beautiful, clean cubic equation:

$$\lambda^3 + \lambda^2 - 5\lambda + 3 = 0$$

### The Final Revelation

Now, we solve for $\lambda$. By inspection, we test $\lambda = 1$. Plugging it in, we get $1 + 1 - 5 + 3 = 0$. It works!

Since $\lambda = 1$ is a root, $(\lambda - 1)$ is a factor. Dividing our cubic by $(\lambda - 1)$ leaves us with a quadratic: $\lambda^2 + 2\lambda - 3$. Factoring this further, we get $(\lambda - 1)(\lambda + 3)$.

So, our equation is:

$$(\lambda - 1)^2(\lambda + 3) = 0$$

The roots are $\lambda = 1$ (with multiplicity 2) and $\lambda = -3$. Finally, we address the question: the set of all values of $\lambda$. A set contains only distinct elements.

Therefore, our set is **$\{1, -3\}$**. It contains exactly two elements. You have navigated the matrix, conquered the determinant, and mastered the set theory. Well done.

Animated Solution for Mathematics - Matrices and Determinants: The set of all values of $λ$ for which the system of linear equations: $2 x_{1} - 2 x_{2} + x_{3} = λ x_{1}$ , $2 x_{1} - 3 x_{2} + 2 x_{3} = λ x_{2}$ , $- x_{1} + 2 x_{2} = λ x_{3}$ has a non-trivial solution

Select Answer:

Visualized Solution

System of Linear Equations

Standard Homogeneous Form

Condition for Non-Trivial Solution

Determinant Setup

Expanding the Determinant

Simplifying the Expansion

Forming the Cubic Equation

Finding the First Root

Factorizing the Polynomial

Final Set of Values

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

Analyzing the Setup

The Homogeneous Transformation

The Determinant's Power

The Algebraic Grind

The Final Revelation

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