This problem is a classic example of applying concepts from matrices and determinants, particularly involving eigenvalues and eigenvectors, or solving systems of linear equations.

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1. Understanding the Problem Statement

We are given a $2 \times 2$ matrix $A$ and a non-zero column vector $B$: $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \ne \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ We are provided with two crucial conditions:

1. $AB = B$
2. $a + d = 2021$ (This is the trace of matrix $A$)

Our goal is to find the value of $ad - bc$, which is the determinant of matrix $A$, i.e., $\det(A)$.

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2. Key Concept: Eigenvalues and Eigenvectors

The condition $AB=B$ can be rewritten as $AB = 1 \cdot B$. This equation is of the form $A\mathbf{x} = \lambda\mathbf{x}$, where $\mathbf{x} = B$ and $\lambda = 1$. This means that $B$ is an eigenvector of matrix $A$ corresponding to the eigenvalue $\lambda = 1$.

For a non-zero vector $B$ to be an eigenvector corresponding to an eigenvalue $\lambda$, the following must hold: $A B = \lambda B$ $A B - \lambda B = \mathbf{0}$ $(A - \lambda I) B = \mathbf{0}$ where $I$ is the identity matrix of the same dimension as $A$.

For a homogeneous system of linear equations $(A - \lambda I) B = \mathbf{0}$ to have a non-trivial solution (i.e., $B \ne \mathbf{0}$), the determinant of the coefficient matrix $(A - \lambda I)$ must be zero. $\det(A - \lambda I) = 0$ This is the characteristic equation of the matrix $A$.

In our problem, $\lambda = 1$. Therefore, for a non-zero vector $B$ to exist such that $AB=B$, we must have: $\det(A - I) = 0$

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3. Step-by-Step Solution - Method 1: Using the Characteristic Equation (Recommended Approach)

This method is generally more robust and less prone to case-specific errors (like division by zero) compared to directly manipulating individual equations.

Step 1: Form the matrix $(A - I)$. Given $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. $A - I = \begin{bmatrix} a & b \\ c & d \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a-1 & b \\ c & d-1 \end{bmatrix}$

Step 2: Set the determinant of $(A - I)$ to zero. Since $B \ne \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ is a solution to $(A-I)B = \mathbf{0}$, the matrix $(A-I)$ must be singular. $\det(A - I) = 0$ $\begin{vmatrix} a-1 & b \\ c & d-1 \end{vmatrix} = 0$

Step 3: Calculate the determinant and simplify the expression. For a $2 \times 2$ matrix $\begin{bmatrix} p & q \\ r & s \end{bmatrix}$, the determinant is $ps - qr$. So, $(a-1)(d-1) - bc = 0$

Step 4: Expand and rearrange the equation to find $ad - bc$. $ad - a - d + 1 - bc = 0$ Group the terms $ad - bc$: $(ad - bc) - (a + d) + 1 = 0$ Now, isolate $ad - bc$: $ad - bc = a + d - 1$

Step 5: Substitute the given value of $a+d$. We are given $a + d = 2021$. $ad - bc = 2021 - 1$ $ad - bc = 2020$

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4. Step-by-Step Solution - Method 2: Solving the System of Linear Equations Directly (As in the original solution)

This method involves explicitly writing out the system of equations and solving for the required relationship.

Step 1: Perform the matrix multiplication $AB$. $A B = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \end{bmatrix} = \begin{bmatrix} a\alpha + b\beta \\ c\alpha + d\beta \end{bmatrix}$

Step 2: Equate $AB$ to $B$. $\begin{bmatrix} a\alpha + b\beta \\ c\alpha + d\beta \end{bmatrix} = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}$ This gives us a system of two linear equations:

1. $a\alpha + b\beta = \alpha$
2. $c\alpha + d\beta = \beta$

Step 3: Rearrange the equations into a standard homogeneous form. Subtract $\alpha$ from the first equation and $\beta$ from the second:

1. $(a-1)\alpha + b\beta = 0 \quad \ldots (1)$
2. $c\alpha + (d-1)\beta = 0 \quad \ldots (2)$

Step 4: Analyze the system for non-trivial solutions. Since $B = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \ne \begin{bmatrix} 0 \\ 0 \end{bmatrix}$, at least one of $\alpha$ or $\beta$ must be non-zero. This means the system has a non-trivial solution. For a homogeneous system of two linear equations in two variables to have a non-trivial solution, the determinant of its coefficient matrix must be zero. This leads back to Method 1.

Alternatively, we can proceed by expressing the ratio $\alpha/\beta$ from both equations (as in the original solution). From equation (1): $(a-1)\alpha = -b\beta$ If $\beta \ne 0$ and $a-1 \ne 0$, we can write: $\frac{\alpha}{\beta} = \frac{-b}{a-1} \quad \ldots (3)$ From equation (2): $c\alpha = -(d-1)\beta$ If $\beta \ne 0$ and $c \ne 0$, we can write: $\frac{\alpha}{\beta} = \frac{-(d-1)}{c} = \frac{1-d}{c} \quad \ldots (4)$

Step 5: Equate the expressions for $\alpha/\beta$. $\frac{-b}{a-1} = \frac{1-d}{c}$

Step 6: Cross-multiply and simplify. $-bc = (a-1)(1-d)$ $-bc = a - ad - 1 + d$ Rearrange to find $ad - bc$: $ad - bc = a + d - 1$

Step 7: Substitute the given value of $a+d$. Given $a + d = 2021$. $ad - bc = 2021 - 1$ $ad - bc = 2020$

Important Note on Edge Cases for Method 2 (Ratio Method): The ratio method requires careful consideration of cases where denominators might be zero (e