Here's a detailed, educational solution to the problem, structured for clarity and understanding.

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Understanding the Problem

We are asked to find all real values of $\lambda$ for which a given system of three linear equations in three variables ($x, y, z$) has a solution. The notation $[\lambda]$ represents the greatest integer less than or equal to $\lambda$ (the floor function). A system of linear equations "has a solution" if it is consistent, meaning it has either a unique solution or infinitely many solutions.

Key Concept: Consistency of a System of Linear Equations (Cramer's Rule)

For a system of linear equations:

$$\begin{align*} a_1x + b_1y + c_1z &= d_1 \\ a_2x + b_2y + c_2z &= d_2 \\ a_3x + b_3y + c_3z &= d_3 \end{align*}$$

We define the following determinants:

1. Coefficient Determinant ($D$): $D = \left| {\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}} \right|$
2. Determinants for Variables ($D_x, D_y, D_z$): These are formed by replacing the respective coefficient column in $D$ with the constant terms column. $D_x = \left| {\begin{array}{ccc} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{array}} \right|, \quad D_y = \left| {\begin{array}{ccc} a_1 & d_1 & c_1 \\ a_2 & d_2 & c_2 \\ a_3 & d_3 & c_3 \end{array}} \right|, \quad D_z = \left| {\begin{array}{ccc} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\ a_3 & b_3 & d_3 \end{array}} \right|$

The consistency rules are as follows:

• Unique Solution (Consistent): If $D \neq 0$, the system has a unique solution given by $x = D_x/D$, $y = D_y/D$, $z = D_z/D$.
• No Solution (Inconsistent): If $D = 0$ AND at least one of $D_x, D_y, D_z$ is non-zero.
• Infinitely Many Solutions (Consistent): If $D = 0$ AND $D_x = D_y = D_z = 0$.

Our goal is to find $\lambda$ such that the system is consistent, meaning either $D \neq 0$ or ($D = 0$ and $D_x = D_y = D_z = 0$).

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Step 1: Identify Coefficients and Constants

First, let's write down the given system and identify the coefficients $a_i, b_i, c_i$ and the constants $d_i$:

1. $1x + 1y + 1z = 4$
2. $3x + 2y + 5z = 3$
3. $9x + 4y + (28 + [\lambda])z = [\lambda]$

Comparing these to the general form, we have:

• $a_1=1, b_1=1, c_1=1, d_1=4$
• $a_2=3, b_2=2, c_2=5, d_2=3$
• $a_3=9, b_3=4, c_3=(28+[\lambda]), d_3=[\lambda]$

Remember that $[\lambda]$ is an integer. For example, if $\lambda=3.7$, then $[\lambda]=3$. If $\lambda=-2.1$, then $[\lambda]=-3$.

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Step 2: Calculate the Determinant of the Coefficient Matrix ($D$)

The first crucial step is to calculate $D$. Its value will determine the path for further analysis. $D = \left| {\begin{array}{ccc} 1 & 1 & 1 \\ 3 & 2 & 5 \\ 9 & 4 & {28 + [\lambda]} \end{array}} \right|$ We can expand this determinant along the first row for simplicity: $D = 1 \cdot \left| {\begin{array}{cc} 2 & 5 \\ 4 & {28 + [\lambda]} \end{array}} \right| - 1 \cdot \left| {\begin{array}{cc} 3 & 5 \\ 9 & {28 + [\lambda]} \end{array}} \right| + 1 \cdot \left| {\begin{array}{cc} 3 & 2 \\ 9 & 4 \end{array}} \right|$ Now, let's compute the $2 \times 2$ determinants: $D = 1 \cdot [2(28 + [\lambda]) - 5(4)] - 1 \cdot [3(28 + [\lambda]) - 5(9)] + 1 \cdot [3(4) - 2(9)]$ $D = [56 + 2[\lambda] - 20] - [84 + 3[\lambda] - 45] + [12 - 18]$ $D = [36 + 2[\lambda]] - [39 + 3[\lambda]] + [-6]$ $D = 36 + 2[\lambda] - 39 - 3[\lambda] - 6$ Combine the constant terms and the terms involving $[\lambda]$: $D = (36 - 39 - 6) + (2[\lambda] - 3[\lambda])$ $D = -9 - [\lambda]$ So, the determinant $D$ is $D = -([\lambda] + 9)$.

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Step 3: Analyze Cases Based on the Value of $D$

We now consider the two main scenarios for $D$: $D \neq 0$ and $D = 0$.

Case 1: $D \neq 0$ (Unique Solution)

If $D \neq 0$, the system has a unique solution, and thus it is consistent. From our calculation: $D = -([\lambda] + 9) \neq 0$ This implies: $[\lambda] + 9 \neq 0$ $[\lambda] \neq -9$ So, for any value of $\lambda$ where its floor $[\lambda]$ is not equal to $-9$, the system has a unique solution and is consistent.

Case 2: $D = 0$ (Possibility of Infinite Solutions or No Solution)

If $D = 0$, we need to calculate $D_x, D_y, D_z$ to determine consistency. Setting $D=0$: $-([\lambda] + 9) = 0$ $[\lambda] + 9 = 0$ $[\lambda] = -9$ Now, we substitute $[\lambda] = -9$ into the original system and into the definitions of $D_x, D_y, D_z$. The system becomes:

1. $x + y + z = 4$
2. $3x + 2y + 5z = 3$
3. $9x + 4y + (28 - 9)z = -9 \implies 9x + 4y + 19z = -9$

Let's calculate $D_x, D_y, D_z$ for $[\lambda] = -9$:

Calculate $D_x$: $D_x = \left| {\begin{array}{ccc} 4 & 1 & 1 \\ 3 & 2 & 5 \\ -9 & 4 & 19 \end{array}} \right|$ Expand along the first row: $D_x = 4(2 \cdot 19 - 5 \cdot 4) - 1(3 \cdot 19 - 5 \cdot (-9)) + 1(3 \cdot 4 - 2 \cdot (-9))$ $D_x = 4(38 - 20) - 1(57 + 45) + 1(12 + 18)$ $D_x = 4(18) - 1(102) + 1(30)$ $D_x = 72 - 102 + 30 = 0$

Calculate $D_y$: $D_y = \left| {\begin{array}{ccc} 1 & 4 & 1 \\ 3 & 3 & 5 \\ 9 & -9 & 19 \end{array}} \right|$ Expand along the first row: $D_y = 1(3 \cdot 19 - 5 \cdot (-9)) - 4(3 \cdot 19 - 5 \cdot 9) + 1(3 \cdot (-9) - 3 \cdot 9)$ $D_y = 1(57 + 45) - 4(57 - 45) + 1(-27 - 27)$ $D_y = 1(102) - 4(12) + 1(-54)$ $D_y = 102 - 48 - 54 = 0$

Calculate $D_z$: $D_z = \left| {\begin{array}{ccc} 1 & 1 & 4 \\ 3 & 2 & 3 \\ 9 & 4 & -9 \end{array}} \right|$ Expand along the first row: $D_z = 1(2 \cdot (-9) - 3 \cdot 4) - 1(3 \cdot (-9) - 3 \cdot 9) + 4(3 \cdot 4 - 2 \cdot 9)$ $D_z = 1(-18 - 12) - 1(-27 - 27) + 4(12 - 18)$ $D_z = 1(-30) - 1(-54) + 4(-6)$ $D_z = -30 + 54 - 24 = 0$

Since we found that $D=0$ AND $D_x=0, D_y=0, D_z=0$ when $[\lambda] = -9$, the system has infinitely many solutions for $[\lambda] = -9$. This means the system is also consistent when $[\lambda] = -9$.

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Step 4: Combine the Results and Determine the Set of $\lambda$

Let's summarize our findings regarding the consistency of the system:

• If $[\lambda] \neq -9$ (i.e., for all integers $[\lambda]$ except $-9$), the system has a unique solution (consistent).
• If $[\lambda] = -9$, the system has infinitely many solutions (consistent).

This implies that for any integer value that $[\lambda]$ can take, the system of equations will have a solution. The greatest integer function, $[\lambda]$, can take any integer value. For any real number $\lambda$, $[\lambda]$ will be some integer. Since the system is consistent for every possible integer value of $[\lambda]$, it means the system is consistent for every real value of $\lambda$.

Therefore, the set of all values of $\lambda$ for which the system has a solution is the set of all real numbers, $\mathbb{R}$.

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Step 5: Match with Options

The derived set of all values of $\lambda$ is $\mathbb{R}$. This corresponds to option (A).

The final answer is $\boxed{\text{R}}$.

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Important Tips and Common Mistakes:

• Don't Stop at $D=0$: A frequent error is to assume that if $D=0$, there are no solutions. Always remember to check $D_x, D_y, D_z$ when $D=0$ to distinguish between no solution and infinitely many solutions. Both of these cases are important for problems asking "has a solution" (i.e., consistent).
• Careful with Determinant Calculations: Determinants, especially $3 \times 3$ ones, are prone to arithmetic errors. Double-check your calculations, paying close attention to signs. Using row/column operations to introduce zeros can sometimes simplify the expansion process and reduce errors.
• Understanding the Floor Function $[\lambda]$: Remember that $[\lambda]$ always yields an integer. If the problem involved properties of $\lambda$ directly, the range would be different (e.g., if $[\lambda]=-9$, then $-9 \le \lambda < -8$). However, since the consistency depends only on the integer value of $[\lambda]$, and $[\lambda]$ can be any integer, the domain of $\lambda$ is $\mathbb{R}$.
• Systematic Approach: Break down the problem into clear steps: calculate $D$, analyze $D \neq 0$, analyze $D=0$ (if applicable, calculate $D_x, D_y, D_z$), and finally combine the results.