Let $$A = \{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R} : |\alpha - 1| \le 4 \text{ and } |\beta - 5| \le 6\}$$ and $$B = \{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \le 144\}$$. Then

To find the relationship between the sets $$A$$ and $$B$$, we can analyze the geometrical regions they represent in the $$\alpha\beta$$-plane.

The set $$A$$ is defined by the system of inequalities:

$$|\alpha - 1| \le 4 \quad \text{and} \quad |\beta - 5| \le 6$$

Expanding these inequalities, we get:

$$-4 \le \alpha - 1 \le 4 \implies -3 \le \alpha \le 5$$

$$-6 \le \beta - 5 \le 6 \implies -1 \le \beta \le 11$$

Geometrically, set $$A$$ represents a rectangular region bounded by the intervals $$\alpha \in [-3, 5]$$ and $$\beta \in [-1, 11]$$.

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The set $$B$$ is defined by the inequality:

$$16(\alpha - 2)^2 + 9(\beta - 6)^2 \le 144$$

Dividing both sides of the equation by 144 to put it in standard form:

$$\frac{16(\alpha - 2)^2}{144} + \frac{9(\beta - 6)^2}{144} \le 1$$

$$\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \le 1$$

$$\frac{(\alpha - 2)^2}{3^2} + \frac{(\beta - 6)^2}{4^2} \le 1$$

Geometrically, set $$B$$ represents the interior and boundary of a vertical ellipse centered at $$(2, 6)$$.

The boundaries (extremes) of the elliptical region $$B$$ can be found using its semi-minor axis $$a = 3$$ and semi-major axis $$b = 4$$, from here we get :

$$\alpha_{\text{range}} = [2 - 3, 2 + 3] \implies \alpha \in [-1, 5]$$

$$\beta_{\text{range}} = [6 - 4, 6 + 4] \implies \beta \in [2, 10]$$

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Comparing the coordinate bounds of both regions:

For the horizontal coordinate: the ellipse spans $$[-1, 5]$$, which is entirely inside the rectangle's range of $$[-3, 5]$$.

For the vertical coordinate: the ellipse spans $$[2, 10]$$, which is entirely inside the rectangle's range of $$[-1, 11]$$.

Since every coordinate point $$(\alpha, \beta)$$ that satisfies the ellipse inequality also satisfies the rectangular boundary inequalities, the entire elliptical region $$B$$ fits completely within the rectangular region $$A$$.

Therefore, the correct relationship is $$B \subset A$$.