Let $$A = \begin{bmatrix} [x+1] & [x+2] & [x+3] \\ [x] & [x+3] & [x+3] \\ [x] & [x+2] & [x+4] \end{bmatrix}$$, where $$[x]$$ denotes the greatest integer less than or equal to $$x$$. If det$$(A) = 192$$, then the set of values of $$x$$ is in the interval:

First of all, recall the definition of the greatest-integer (floor) function: for any real number $$x$$ we write $$[x]=n$$ when $$n$$ is the unique integer satisfying $$n\le x<n+1$$. We denote the fractional part of $$x$$ by $$f=x-n$$, so $$0\le f<1$$ and $$x=n+f$$.

In the matrix $$ A=\begin{bmatrix} [x+1] & [x+2] & [x+3]\\ [x] & [x+3] & [x+3]\\ [x] & [x+2] & [x+4] \end{bmatrix}, $$ let us put $$n=[x]$$. Because $$f\ge 0$$, each time we add a positive integer to $$x$$ we move at least that far to the right on the number line but never cross an extra integer boundary beyond the one we expect. Concretely,

$$ \begin{aligned} [x] &= n,\\[4pt] [x+1] &= [n+f+1]=n+1,\\[4pt] [x+2] &= [n+f+2]=n+2,\\[4pt] [x+3] &= [n+f+3]=n+3,\\[4pt] [x+4] &= [n+f+4]=n+4. \end{aligned} $$

Substituting these values, the matrix becomes

$$ A=\begin{bmatrix} n+1 & n+2 & n+3\\ n & n+3 & n+3\\ n & n+2 & n+4 \end{bmatrix}. $$

We now compute the determinant of this matrix. To simplify the arithmetic we perform elementary row operations that do not change the determinant: replace the second row by (second row) - (first row) and the third row by (third row) - (first row).

$$ \begin{aligned} R_2 &\leftarrow R_2-R_1: &(-1,\;1,\;0),\\ R_3 &\leftarrow R_3-R_1: &(-1,\;0,\;1). \end{aligned} $$

After these operations the matrix is

$$ \begin{bmatrix} n+1 & n+2 & n+3\\ -1 & 1 & 0\\ -1 & 0 & 1 \end{bmatrix}. $$

Using the cofactor (Laplace) expansion along the first row, and remembering the alternating signs $$+ - +$$, we have

$$ \begin{aligned} \det(A)&=(n+1)\begin{vmatrix}1&0\\0&1\end{vmatrix} \;-\;(n+2)\begin{vmatrix}-1&0\\-1&1\end{vmatrix} \;+\;(n+3)\begin{vmatrix}-1&1\\-1&0\end{vmatrix}. \end{aligned} $$

The 2 × 2 determinants are straightforward:

$$ \begin{aligned} \begin{vmatrix}1&0\\0&1\end{vmatrix}&=1\cdot1-0\cdot0=1,\\[6pt] \begin{vmatrix}-1&0\\-1&1\end{vmatrix}&=(-1)\cdot1-0\cdot(-1)=-1,\\[6pt] \begin{vmatrix}-1&1\\-1&0\end{vmatrix}&=(-1)\cdot0-1\cdot(-1)=1. \end{aligned} $$

So

$$ \det(A)=(n+1)(1)-(n+2)(-1)+(n+3)(1) =(n+1)+(n+2)+(n+3) =3n+6. $$

The problem states that $$\det(A)=192$$, therefore

$$ 3n+6=192\;\Longrightarrow\;3n=186\;\Longrightarrow\;n=62. $$

Since $$n=[x]$$, the definition of the floor function immediately gives

$$ 62\le x<63. $$

Thus the required set of values of $$x$$ is the interval $$[62,63)$$.

Hence, the correct answer is Option A.