Equation of the tangent to the circle, at the point $$(1, -1)$$, whose center is the point of intersection of the straight lines $$x - y = 1$$ and $$2x + y = 3$$ is:

First we locate the centre of the circle. It is given that the centre is the point of intersection of the two straight lines

$$x - y = 1 \qquad\text{and}\qquad 2x + y = 3.$$

We solve these simultaneous linear equations. From the first equation we have

$$y = x - 1.$$

Substituting this value of $$y$$ into the second equation, we obtain

$$2x + (x - 1) = 3.$$

Simplifying the left-hand side gives

$$3x - 1 = 3.$$

Adding $$1$$ to both sides,

$$3x = 4,$$

and dividing by $$3$$,

$$x = \dfrac{4}{3}.$$

Now we substitute this back into $$y = x - 1$$:

$$y = \dfrac{4}{3} - 1 = \dfrac{4}{3} - \dfrac{3}{3} = \dfrac{1}{3}.$$

So the centre of the circle is

$$C\left(\dfrac{4}{3},\,\dfrac{1}{3}\right).$$

The given point of tangency is

$$P(1,\,-1).$$

We next find the slope of the radius $$CP$$. The slope formula is

$$m_{r} = \dfrac{y_2 - y_1}{x_2 - x_1},$$

where $$C(x_1,y_1)$$ and $$P(x_2,y_2)$$. Substituting $$C\left(\dfrac{4}{3},\dfrac{1}{3}\right)$$ and $$P(1,-1)$$, we get

$$m_{r} = \dfrac{-1 - \dfrac{1}{3}}{\,1 - \dfrac{4}{3}\,} = \dfrac{-\dfrac{4}{3}}{-\dfrac{1}{3}} = 4.$$

For a tangent to a circle, the tangent line is perpendicular to the radius. Therefore, if the slope of the radius is $$m_{r}$$, the slope of the tangent $$m_{t}$$ satisfies

$$m_{t}\,m_{r} = -1 \quad\Longrightarrow\quad m_{t} = -\dfrac{1}{m_{r}}.$$

Using $$m_{r} = 4$$, we obtain

$$m_{t} = -\dfrac{1}{4}.$$

Now we write the equation of the straight line having slope $$m_{t} = -\dfrac{1}{4}$$ and passing through the point $$P(1,-1)$$. Using the point-slope form,

$$y - y_0 = m\,(x - x_0),$$

where $$P(x_0,y_0)=(1,-1)$$, we get

$$y - (-1) = -\dfrac{1}{4}\,\bigl(x - 1\bigr),$$

which simplifies to

$$y + 1 = -\dfrac{1}{4}\,(x - 1).$$

To remove the fraction, multiply every term by $$4$$:

$$4(y + 1) = -(x - 1).$$

Expanding both sides gives

$$4y + 4 = -x + 1.$$

Now we bring all terms to the left-hand side:

$$x + 4y + 4 - 1 = 0,$$

which combines to

$$x + 4y + 3 = 0.$$

This matches Option A.

Hence, the correct answer is Option A.