Let S be the set of values of λ, for which the system of equations 6λx – 3y + 3z = 4λ, 2x + 6λy + 4z

Q: Wait, why does $\Delta = 0$ imply 'no solution'? Can't it also mean 'infinitely many solutions'?

Excellent question! You are absolutely right to be cautious. $\Delta = 0$ is a necessary condition for both 'no solution' and 'infinitely many solutions.' To distinguish between them, we check the auxiliary determinants $\Delta_x, \Delta_y, \Delta_z$. If $\Delta = 0$ AND at least one of these is non-zero, the system is inconsistent (no solution). If all are zero, it's usually dependent (infinitely many). In this problem, we are specifically looking for the 'no solution' case, so we focus on the roots of $\Delta = 0$.

Q: Do I really need to check $\Delta_x, \Delta_y, \Delta_z$ for every root? That seems like a lot of work.

In the heat of a JEE Advanced exam, time is gold. However, mathematically, it is the only way to be certain. In this specific problem, the question asks for the set of values for which there is 'no solution.' If you find the roots of $\Delta=0$, you have the candidates. For most JEE problems of this type, these roots will indeed lead to the 'no solution' case. But always keep the definition in mind—it's the bedrock of your understanding!

Q: How do I solve the cubic equation $9\lambda^3 - 7\lambda - 2 = 0$ without panicking?

Panic is the enemy of logic! When you see a cubic, always start with the Rational Root Theorem. Test simple integers like $1, -1, 2, -2$. Here, testing $\lambda = 1$ immediately gives $9(1)^3 - 7(1) - 2 = 0$. Once you find one root, use polynomial division or synthetic division to reduce it to a quadratic. You've got this!

Question

Let $S$ be the set of values of $\lambda$, for which the system of equations $6\lambda x – 3y + 3z = 4\lambda$, $2x + 6\lambda y + 4z = 1$ and $3x + 2y + 3\lambda z = \lambda$ has no solution. Then, $12 \sum_{\lambda \in S} |\lambda|$ is equal to ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have three planes, each defined by a linear equation. In a perfect world, these planes intersect at a single, beautiful point—a unique solution.

But today, we are looking for something different. We are looking for the moments where the system breaks down, where the planes refuse to meet, where the system is **inconsistent**.

This is the essence of the problem before us. We are given a system of three linear equations involving a parameter $\lambda$. Our goal is to find the values of $\lambda$ that render this system unsolvable.

### The Gatekeeper: The Determinant

To understand why a system fails, we must look at the coefficient matrix. The determinant of this matrix, denoted as $\Delta$, acts as the gatekeeper of existence.

If $\Delta 
eq 0$, the system is well-behaved and has a unique solution. But if $\Delta = 0$, the gatekeeper has closed the door. The planes are either parallel, or they form a prism-like structure where they never share a common point.

To begin our journey, we must construct this determinant:

$$\Delta = \begin{vmatrix} 6\lambda & -3 & 3 \ 2 & 6\lambda & 4 \ 3 & 2 & 3\lambda \end{vmatrix}$$

We set this equal to zero, because that is the condition for the system to be inconsistent.

### The Algebraic Grind: Expanding the Matrix

Now, we must expand this determinant. We expand along the first row.

The first term is $6\lambda$ multiplied by the minor:

$$\begin{vmatrix} 6\lambda & 4 \ 2 & 3\lambda \end{vmatrix}$$

This gives us $6\lambda(18\lambda^2 - 8)$.

Next, we move to the second term. Remember the alternating sign rule! The element is $-3$, so we subtract $(-3)$, which becomes $+3$. The minor is:

$$\begin{vmatrix} 2 & 4 \ 3 & 3\lambda \end{vmatrix}$$

This yields $3(6\lambda - 12)$. Finally, the third term is $+3$ multiplied by the minor:

$$\begin{vmatrix} 2 & 6\lambda \ 3 & 2 \end{vmatrix}$$

This gives $3(4 - 18\lambda)$. When we combine these, we get the expression:

$$\Delta = 6\lambda(18\lambda^2 - 8) + 3(6\lambda - 12) + 3(4 - 18\lambda) = 0$$

Expanding this, we get:

$$108\lambda^3 - 48\lambda + 18\lambda - 36 + 12 - 54\lambda = 0$$

Simplifying this, we arrive at $108\lambda^3 - 84\lambda - 24 = 0$. Dividing the entire equation by $12$, we find the elegant cubic:

$$9\lambda^3 - 7\lambda - 2 = 0$$

### The Cubic Challenge: Finding the Roots

We are now at the heart of the problem. We need to solve $9\lambda^3 - 7\lambda - 2 = 0$.

Look for the 'easy' roots first. Testing $\lambda = 1$:

$$9(1)^3 - 7(1) - 2 = 9 - 7 - 2 = 0$$

Success! Since $\lambda = 1$ is a root, $(\lambda - 1)$ is a factor. By performing polynomial division, we can factor the cubic into:

$$(\lambda - 1)(9\lambda^2 + 9\lambda + 2) = 0$$

Now, we solve the quadratic part: $9\lambda^2 + 9\lambda + 2 = 0$. Splitting the middle term, we get:

$$9\lambda^2 + 6\lambda + 3\lambda + 2 = 0$$

This factors beautifully into $(3\lambda + 2)(3\lambda + 1) = 0$. Thus, our roots are $\lambda = 1$, $\lambda = -\frac{2}{3}$, and $\lambda = -\frac{1}{3}$.

### The Final Victory

We have found our set $S = \{1, -\frac{1}{3}, -\frac{2}{3}\}$. The problem asks for $12 \sum_{\lambda \in S} |\lambda|$.

Let's calculate the sum of the absolute values:

$$|1| + |-\frac{1}{3}| + |-\frac{2}{3}| = 1 + \frac{1}{3} + \frac{2}{3} = 1 + 1 = 2$$

Finally, we multiply by $12$:

$$12 	imes 2 = 24$$

We have navigated the geometry, conquered the algebra, and emerged victorious. The final answer is **24**.

Animated Solution for Mathematics - Matrices and Determinants: Let $S$ be the set of values of $λ$ , for which the system of equations $6 λ x -3 y + 3 z = 4 λ$ , $2 x + 6 λ y + 4 z = 1$ and $3 x + 2 y + 3 λ z = λ$ has no solution. Then, $12 \sum_{λ \in S} ∣ λ ∣$ is equal to ________.

Enter Numerical Value:

Visualized Solution

Condition for No Solution

Setting up the Determinant $Δ$

Expanding $Δ$ : First Term

Expanding $Δ$ : Remaining Terms

Combining and Expanding

Simplifying the Cubic Equation

Finding the First Root

Factorizing the Cubic

The Set of Values $S$

Calculating the Final Sum

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

Analyzing the Setup

The Gatekeeper

The Algebraic Grind

The Cubic Challenge

The Final Victory

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Enter Numerical Value:

Visualized Solution

Condition for No Solution

Setting up the Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ: First Term

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ: Remaining Terms