If $$B = \begin{pmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{pmatrix}$$ is the inverse of a 3$$\times$$3 matrix A, then the sum of all values of $$\alpha$$ for which det(A) + 1 = 0, is:

We are given that the matrix $$B=\begin{pmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{pmatrix}$$ is the inverse of a matrix $$A$$, that is, $$A^{-1}=B$$.

For any square matrices, we know the basic determinant relation $$\det(A)\,\det(A^{-1}) = 1$$. Since $$A^{-1}=B$$, this becomes

$$\det(A)\,\det(B)=1.$$

The condition given in the question is $$\det(A)+1=0$$. Re-writing, we have

$$\det(A) = -1.$$

Substituting this value of $$\det(A)$$ into the product relation, we get

$$(-1)\,\det(B) = 1 \quad\Longrightarrow\quad \det(B) = -1.$$

So our task reduces to finding all values of $$\alpha$$ for which the determinant of the matrix $$B$$ equals $$-1$$.

Now we compute $$\det(B)$$. Expanding along the first row, we obtain

$$ \det(B) = 5\, \begin{vmatrix} 2 & 1\\ 3 & -1 \end{vmatrix} \;-\; (2\alpha)\, \begin{vmatrix} 0 & 1\\ \alpha & -1 \end{vmatrix} \;+\; 1\, \begin{vmatrix} 0 & 2\\ \alpha & 3 \end{vmatrix}. $$

We evaluate each of the 2×2 determinants one by one.

First minor:

$$ \begin{vmatrix} 2 & 1\\ 3 & -1 \end{vmatrix} = 2(-1) - 1(3) = -2 - 3 = -5. $$

Second minor:

$$ \begin{vmatrix} 0 & 1\\ \alpha & -1 \end{vmatrix} = 0(-1) - 1(\alpha) = -\alpha. $$

Third minor:

$$ \begin{vmatrix} 0 & 2\\ \alpha & 3 \end{vmatrix} = 0\cdot 3 - 2\alpha = -2\alpha. $$

Substituting these into the expansion of $$\det(B)$$, we get

$$ \det(B) \;=\; 5(-5)\;-\;(2\alpha)(-\alpha)\;+\;1(-2\alpha). $$

Carrying out the multiplications:

$$ \det(B) = -25 + 2\alpha^{2} - 2\alpha. $$

We must have $$\det(B) = -1$$, so we set

$$ -25 + 2\alpha^{2} - 2\alpha = -1. $$

Adding 25 to both sides and simplifying,

$$ 2\alpha^{2} - 2\alpha - 24 = 0. $$

Dividing the entire equation by 2,

$$ \alpha^{2} - \alpha - 12 = 0. $$

This is a quadratic equation. Using the quadratic formula $$\alpha = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ with $$a=1,\; b=-1,\; c=-12$$, we find

$$ \alpha = \frac{1 \pm \sqrt{(-1)^{2}-4(1)(-12)}}{2} = \frac{1 \pm \sqrt{1 + 48}}{2} = \frac{1 \pm \sqrt{49}}{2} = \frac{1 \pm 7}{2}. $$

Thus we get two values:

$$ \alpha_{1} = \frac{1+7}{2} = \frac{8}{2} = 4,\quad \alpha_{2} = \frac{1-7}{2} = \frac{-6}{2} = -3. $$

The question asks for the sum of all such values, so

$$ \alpha_{1} + \alpha_{2} = 4 + (-3) = 1. $$

Hence, the correct answer is Option B.