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Step-by-Step Solution
Step 1: Write down the system of linear equations
We have the following system:
1) $x + y + z = 4$
2) $3x + 2y + 5z = 3$
3) $9x + 4y + (28 + [\lambda])z = [\lambda]$
Here, $[\lambda]$ is the greatest integer less than or equal to $\lambda$.
Step 2: Form the determinant of the coefficient matrix
The coefficient matrix is
$$
\begin{pmatrix}
1 & 1 & 1\\
3 & 2 & 5\\
9 & 4 & 28 + [\lambda]
\end{pmatrix}.
$$
Let $D$ be its determinant. Then,
$$
D
= \begin{vmatrix}
1 & 1 & 1\\
3 & 2 & 5\\
9 & 4 & 28 + [\lambda]
\end{vmatrix}.
$$
We expand this determinant:
$$
D = 1 \times
\begin{vmatrix}
2 & 5\\
4 & 28 + [\lambda]
\end{vmatrix}
- 1 \times
\begin{vmatrix}
3 & 5\\
9 & 28 + [\lambda]
\end{vmatrix}
+ 1 \times
\begin{vmatrix}
3 & 2\\
9 & 4
\end{vmatrix}.
$$
Calculate each minor:
• First minor: $2 \cdot (28 + [\lambda]) - 5 \cdot 4 = 56 + 2[\lambda] - 20 = 36 + 2[\lambda].$
• Second minor: $3 \cdot (28 + [\lambda]) - 5 \cdot 9 = 84 + 3[\lambda] - 45 = 39 + 3[\lambda].$
• Third minor: $3 \cdot 4 - 2 \cdot 9 = 12 - 18 = -6.$
Therefore,
$$
D = (36 + 2[\lambda]) - (39 + 3[\lambda]) - 6.
$$
Simplify:
$$
D = 36 + 2[\lambda] - 39 - 3[\lambda] - 6 = -[\lambda] - 9.
$$
Step 3: Determine conditions for the system to have a solution
We have $D = -[\lambda] - 9.$
• If $D \neq 0$, i.e., if $[\lambda] + 9 \neq 0$, then the system has a unique solution.
• If $D = 0$, i.e., $[\lambda] + 9 = 0$, then $[\lambda] = -9.$ In that case, we must check the system for consistency (for infinite solutions or no solutions). Substituting $[\lambda] = -9$ into the system, one finds that the resulting equations are consistent and lead to infinitely many solutions (because all the relevant minors also vanish appropriately).
Step 4: Conclude the range of $\lambda$
• When $[\lambda] \neq -9,$ the system has a unique solution.
• When $[\lambda] = -9,$ the system has infinitely many solutions.
Therefore, for every real value of $\lambda,$ the system will always have a solution (either unique or infinitely many), because for any $\lambda$ we pick, the integer part $[\lambda]$ will be some integer. Even if that integer is $-9,$ we still get infinitely many solutions.
Hence, the set of all values of $\lambda$ for which the system has a solution is $\mathbb{R}$.
