`A=[(a,b),(c,d)] , B=[(alpha),(beta)]` such that `AB=B` and `a+d=2021` , B is a non-zero matrix. find `ad-bc`

Answer

To solve the problem, we have the matrices \( A \) and \( B \) defined as follows:

\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \quad B = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} \]

We know that \( AB = B \) and \( a + d = 2021 \). We need to find \( ad - bc \).

### Step 1: Set up the equation from \( AB = B \)

From the equation \( AB = B \), we can rewrite it as:

\[ AB - B = 0 \]

This can be factored as:

\[ (A - I)B = 0 \]

where \( I \) is the identity matrix.

### Step 2: Analyze the implications of \( (A - I)B = 0 \)

Since \( B \) is a non-zero matrix, the only way for the product \( (A - I)B \) to be zero is if the determinant of \( (A - I) \) is zero. Thus, we have:

\[ \text{det}(A - I) = 0 \]

### Step 3: Calculate \( A - I \)

The identity matrix \( I \) for a 2x2 matrix is:

\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

So, we compute \( A - I \):

\[ A - I = \begin{pmatrix} a - 1 & b \\ c & d - 1 \end{pmatrix} \]

### Step 4: Find the determinant of \( A - I \)

The determinant of \( A - I \) is given by:

\[ \text{det}(A - I) = (a - 1)(d - 1) - bc \]

Setting this equal to zero gives:

\[ (a - 1)(d - 1) - bc = 0 \]

### Step 5: Expand the determinant equation

Expanding the determinant, we have:

\[ ad - a - d + 1 - bc = 0 \]

Rearranging this, we find:

\[ ad - bc = a + d - 1 \]

### Step 6: Substitute the value of \( a + d \)

We know from the problem that \( a + d = 2021 \). Substituting this into our equation gives:

\[ ad - bc = 2021 - 1 = 2020 \]

### Conclusion

Thus, the value of \( ad - bc \) is:

\[ \boxed{2020} \]