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Step-by-Step Solution
Step 1: Understand the Problem
We are given a 3×3 matrix whose entries are the greatest integer (floor) values of certain expressions in terms of $x$. We need to determine for which values of $x$ the determinant of this matrix equals 192.
Step 2: Write Down the Matrix
Let $A$ be the matrix
$A =
\begin{pmatrix}
[x+1] & [x+2] & [x+3] \\
[x] & [x+3] & [x+3] \\
[x] & [x+2] & [x+4]
\end{pmatrix}
$
where $[t]$ denotes the greatest integer less than or equal to $t$.
Step 3: Apply Row Operations
To simplify the determinant, we use the following row transformations, which do not change the value of the determinant:
$R_1 \to R_1 - R_3$
$R_2 \to R_2 - R_3$
Before these operations, observe that for any integer $n$, $[n + k] = [n] + k$ if $x$ does not cross integer boundaries within that interval. Once we determine $[x]$, we can confirm this is consistent for $x$ in the correct interval.
Step 4: Form the New Matrix
After performing $R_1 - R_3$ and $R_2 - R_3$, the new matrix becomes:
$\begin{pmatrix}
[x+1] - [x] & [x+2] - [x+2] & [x+3] - [x+4] \\
[x] - [x] & [x+3] - [x+2] & [x+3] - [x+4] \\
[x] & [x+2] & [x+4]
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 & -1 \\
0 & 1 & -1 \\
[x] & [x+2] & [x+4]
\end{pmatrix}
$
This is based on the property that for non-integer $x$, $[x+k] = [x] + k$.
Step 5: Compute the Determinant
Let $D = \det(A)$. The determinant of the transformed matrix is the same as the determinant of the original matrix. So,
$D =
\begin{vmatrix}
1 & 0 & -1 \\
0 & 1 & -1 \\
[x] & [x+2] & [x+4]
\end{vmatrix}.
$
Expanding along the first row,
$
D
= 1 \times
\begin{vmatrix}
1 & -1 \\
[x+2] & [x+4]
\end{vmatrix}
- 0 \times
\begin{vmatrix}
0 & -1 \\
[x] & [x+4]
\end{vmatrix}
+ (-1) \times
\begin{vmatrix}
0 & 1 \\
[x] & [x+2]
\end{vmatrix}.
$
Calculate each minor:
$
\begin{vmatrix}
1 & -1 \\
[x+2] & [x+4]
\end{vmatrix}
= 1 \cdot [x+4] - (-1) \cdot [x+2]
= [x+4] + [x+2].
$
The second term is multiplied by zero, so it vanishes.
$
\begin{vmatrix}
0 & 1 \\
[x] & [x+2]
\end{vmatrix}
= 0 \cdot [x+2] - 1 \cdot [x]
= -[x].
$
Putting these together:
$
D = 1 \times \bigl([x+4] + [x+2]\bigr) + (-1)\times \bigl(-[x]\bigr)
= [x+4] + [x+2] + [x].
$
Step 6: Solve for $[x]$
We are given $D = 192$. Hence,
$[x+4] + [x+2] + [x] = 192.
$
For $x$ not an integer boundary within that interval, $[x+k] = [x] + k$. Therefore,
$([x] + 4) + ([x] + 2) + [x] = 192.
$
$3[x] + 6 = 192 \quad \Longrightarrow \quad 3[x] = 186 \quad \Longrightarrow \quad [x] = 62.
$
Since $[x] = 62$, $x$ must lie in the interval $[62, 63)$.
Step 7: Final Answer
The set of values of $x$ for which the determinant of the given matrix is 192 is:
$[62, 63)$.
