Let [λ] be the greatest integer less than or equal to λ. The set of all values of λ for which the sy

Q: Wait, why do we even care about $[\lambda]$? Can't we just treat it as a constant $k$?

You absolutely can! Treating $[\lambda]$ as a constant $k$ is a brilliant way to simplify the algebra. Just remember that at the very end, you must interpret what $k = [\lambda]$ implies for $\lambda$. Since $[\lambda]$ can take any integer value, if the system is consistent for all $k$, then it is consistent for all $\lambda$.

Q: If $D = 0$, doesn't that automatically mean 'no solution'?

Not necessarily! $D = 0$ only tells us that the system is not guaranteed a unique solution. It could be inconsistent (no solution) or dependent (infinitely many solutions). You must check the other determinants ($D_1, D_2, D_3$) to see if the planes are parallel or if they intersect along a common line.

Q: How do I know if I need to calculate $D_1, D_2, D_3$ or just use Gaussian elimination?

Both are valid! Cramer's rule is often faster for $3 \times 3$ systems when you have a parameter like $[\lambda]$. If $D=0$ and $D_1=D_2=D_3=0$, the system is consistent. If any of them were non-zero, it would be inconsistent. Choose the method that feels more intuitive to your algebraic flow.

Q: Is it possible for the system to have no solution at all?

In this specific problem, we found that even when $D=0$, the other determinants also vanish. This means the planes are not parallel and disjoint; rather, they intersect in a way that creates a line of solutions. Always check the math—don't assume inconsistency just because $D=0$!

Q: Why is the final answer $\mathbf{R}$ and not just a specific interval?

Because the system is consistent for all values of $[\lambda]$. Since $[\lambda]$ is an integer, and the system works for the critical value $[\lambda] = -9$ as well as all other values, there is no restriction on $\lambda$ itself. Thus, $\lambda$ can be any real number.

Question

Let $[\lambda]$ be the greatest integer less than or equal to $\lambda$. The set of all values of $\lambda$ for which the system of linear equations $x + y + z = 4, 3x + 2y + 5z = 3, 9x + 4y + (28 + [\lambda])z = [\lambda]$ has a solution is:

Options:

(A) $\mathbf{R}$

(B) $(-\infty, -9) \cup (-9, \infty)$

(C) $[-9, -8)$

(D) $(-\infty, -9) \cup [-8, \infty)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional room. You have three sheets of glass, each representing an equation: $x + y + z = 4$, $3x + 2y + 5z = 3$, and $9x + 4y + (28 + \lambda)z = \lambda$.

Your goal is to find where these sheets intersect. Do they meet at a single point, share a common line, or never touch at all? This is the heart of linear algebra.

### The Diagnostic Tool: The Determinant

Before we dive into the deep end, we need a compass. In linear algebra, that compass is the determinant of the coefficient matrix, $D$. If $D 
eq 0$, the system is "well-behaved" and possesses a unique solution.

We calculate $D$ as:

$$D = \begin{vmatrix} 1 & 1 & 1 \ 3 & 2 & 5 \ 9 & 4 & 28 + \lambda \end{vmatrix}$$

Expanding this along the first row, we get:

$$D = 1 \cdot (2(28 + \lambda) - 20) - 1 \cdot (3(28 + \lambda) - 45) + 1 \cdot (12 - 18)$$

As you simplify this, the terms collapse beautifully into:

$$D = -\lambda - 9$$

This is our "critical point." If $\lambda 
eq -9$, the system is guaranteed to have a unique solution.

### The Moment of Truth: When $D = 0$

When $\lambda = -9$, the determinant vanishes. The system loses its unique solution. To determine if it becomes impossible or infinitely flexible, we turn to the auxiliary determinants $D_1$, $D_2$, and $D_3$.

These are the gatekeepers of consistency. If $D = 0$ and $D_1 = D_2 = D_3 = 0$, the system is consistent, meaning the planes intersect at a line or coincide.

Substituting $\lambda = -9$ into our system, we calculate $D_1$:

$$D_1 = \begin{vmatrix} 4 & 1 & 1 \ 3 & 2 & 5 \ -9 & 4 & 19 \end{vmatrix} = 4(38 - 20) - 1(57 + 45) + 1(12 + 18) = 72 - 102 + 30 = 0$$

It is zero! The tension breaks. If you repeat this for $D_2$ and $D_3$, you will find they also vanish.

This tells us that even at the "danger zone" where $\lambda = -9$, the system remains consistent. It simply shifts from having a unique point of intersection to having an infinite number of solutions along a line.

### The Grand Conclusion

We checked the case where the system is unique ($\lambda 
eq -9$) and the case where it is dependent ($\lambda = -9$). In both scenarios, the system has a solution.

Since the system is consistent for every value of $\lambda$, there is no value that makes the system inconsistent. Therefore, $\lambda$ can be any real number.

The set of all values is **$\mathbb{R}$**. You have successfully navigated the geometry of these planes and proven that no matter how you tweak the parameter $\lambda$, the system holds together.

Animated Solution for Mathematics - Matrices and Determinants: Let $[λ]$ be the greatest integer less than or equal to $λ$ . The set of all values of $λ$ for which the system of linear equations $x + y + z = 4, 3 x + 2 y + 5 z = 3, 9 x + 4 y + (28 + [λ]) z = [λ]$ has a solution is:

Select Answer:

Visualized Solution

System of Linear Equations

Cramer's Rule: Coefficient Determinant $D$

Expanding the Determinant $D$

Simplifying $D$

Case 1: Unique Solution Condition

Case 2: When $D = 0$

Calculating $D_{1}$ for $[λ] = - 9$

Calculating $D_{2}$ and $D_{3}$

Conclusion for $[λ] = - 9$

Final Answer

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

Analyzing the Setup

The Diagnostic Tool

The Moment of Truth

The Grand Conclusion

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Visualized Solution

System of Linear Equations

Cramer's Rule: Coefficient Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

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

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Case 1: Unique Solution Condition

Case 2: When .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D=0

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D1​ for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [λ]=−9

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D2​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D3​

Conclusion for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [λ]=−9