Let A = begin{bmatrix} a & b c & d end{bmatr | vedclass

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(D) Given $AB = B$,which can be rewritten as $AB - B = O$,where $O$ is the zero matrix. This implies $(A - I)B = O$,where $I$ is the identity matrix $

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$2020$

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(D) Given $AB = B$,which can be rewritten as $AB - B = O$,where $O$ is the zero matrix. This implies $(A - I)B = O$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$. Since $B 
eq O$,the matrix $(A - I)$ must be singular,meaning its determinant must be zero: $|A - I| = 0$. Calculating the determinant: $|A - I| = \begin{vmatrix} a - 1 & b \ c & d - 1 \end{vmatrix} = (a - 1)(d - 1) - bc = 0$. Expanding this,we get $ad - a - d + 1 - bc = 0$. Rearranging the terms,we have $ad - bc = a + d - 1$. Given that $a + d = 2021$,we substitute this value: $ad - bc = 2021 - 1 = 2020$.

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 ...... .

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