2. Let the set $M=\{x \mid 0 \leqslant x \leqslant 11, x \in \mathbf{Z}\}$, $F=\{(a, b, c, d) \mid a, b, c, d \in M\}$, and the mapping $f: F \rightarrow \mathbf{Z}$, such that $(a, b, c, d) \xrightarrow{f} a b - c d$. Given that $(u, v, x, y) \xrightarrow{f} 39$, $(u, y, x, v) \xrightarrow{f} 66$. Then the values of $x, y, u, v$ are $\qquad$
2. 1,9,8,6.

From the problem, we know
$$
\begin{array}{l}
u v - x y = 39, \\
u y - x v = 66,
\end{array}
$$
where \( u, v, x, y \) are non-negative integers and do not exceed 11. Therefore,
$$
\begin{array}{l}
(y + v)(u - x) = 105. \\
(y - v)(u + x) = 27.
\end{array}
$$

Given \( 0 \leqslant y \leqslant 11, 0 \leqslant v \leqslant 11 \), we have
$$
\left[\frac{10 \leqslant}{11}\right] + 1 \leqslant y + v \leqslant 22,
$$

which means \( 10 \leqslant y + v \leqslant 22 \).
Thus, \( 5 \leqslant u - x \leqslant 10 \). Therefore, \( u + x \geqslant 5 \).
Also, \((y - v)(u + x) = 1 \times 27 = 3 \times 9 = 9 \times 3\), so \( u + x = 9 \) or 27.
Since \( 0 \leqslant u + x \leqslant 22 \), we have \( u + x = 9 \).
From \( u + x = 9 \), we get \( y - v = 3 \). From \((y + v)(u - x) = 105 = 3 \times 35 = 5 \times 21 = 15 \times 7 = 21 \times 5\), we get \( y + v = 15 \) or 21.

When \( y + v = 15 \), \( u - x = 7 \), then \( y = 9, v = 6, u = 8, x = 1 \);

When \( y + v = 21 \), \( u - x = 5 \), then \( u = 7, x = 2, y = 12, v = 9 \). But \( 0 \leqslant y \leqslant 11 \), so this result is not valid.

Therefore, the values of \( x, y, u, v \) are \( x = 1, y = 9, u = 8, v = 6 \).