The determinant \[\left| {\begin{array}{*{20}{c}} {1 + {\text{b}}}&{\text{b}}&1 \\ {\text{b}}&{1 + {\text{b}}}&1 \\ 2&{2{\text{b}}}&1 \end{array}} \right|\] equals to A. 0 B. 2b(b – 1) C. 2(1 – b) (1 + 2b) D. 3b(1 + b)

The correct answer is $\boxed{\text{C}}$.

To find the determinant of a 3×3 matrix, we can use the formula:

$$\det \begin{bmatrix}
a & b & c \
d & e & f \
g & h & i
\end{bmatrix} = (aei – bg) + (bf – ch) + (cd – ae)$$

In this case, we have:

$$\begin{align}
\det \begin{bmatrix}
1 + b & b & 1 \
b & 1 + b & 1 \
2 & 2b & 1
\end{bmatrix} &= (1 + b)(1 + b) – b(1) + (1)(2b) – (b)(2) \
&= (1 + b)^2 – b^2 + 2b – 2b \
&= 2(1 – b)(1 + 2b) \
&= \boxed{2(1 – b)(1 + 2b)}
\end{align}$$

Option A is incorrect because $b$ does not appear in the determinant. Option B is incorrect because $b^2$ does not appear in the determinant. Option D is incorrect because $3b$ does not appear in the determinant.