Let $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ and $$B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ such that $$AB = B$$ and $$a + d = 2021$$, then the value of $$ad - bc$$ is equal to ________.

We have $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ and $$B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ such that $$AB = B$$.

Writing this out: $$\begin{bmatrix} a\alpha + b\beta \\ c\alpha + d\beta \end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$$. This gives us $$a\alpha + b\beta = \alpha$$ and $$c\alpha + d\beta = \beta$$, i.e., $$(a-1)\alpha + b\beta = 0$$ and $$c\alpha + (d-1)\beta = 0$$.

Since $$B \neq 0$$, this homogeneous system has a non-trivial solution, which means the determinant of the coefficient matrix must be zero: $$\det\begin{bmatrix} a-1 & b \\ c & d-1 \end{bmatrix} = 0$$.

Expanding: $$(a-1)(d-1) - bc = 0$$, which gives $$ad - a - d + 1 - bc = 0$$, so $$ad - bc = a + d - 1$$.

Since $$a + d = 2021$$, we get $$ad - bc = 2021 - 1 = 2020$$.