"Let $A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$ and $$B = \left[ {\matrix{ \alpha \cr \beta \cr } } \right] \ne \left[ {\matrix{ 0 \cr 0 \cr } } \right]$$ such that AB = B and a + d = 2021, then the value of ad $-$ bc is equal to ___________."

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"$$A = \left[ {\matrix{ a & b \cr c & d \cr } } \right],\,B = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$$

$AB = B$

$$\left[ {\matrix{ a & b \cr c & d \cr } } \right]\left[ {\matrix{ \alpha \cr \beta \cr } } \right] = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$$

$$\left[ {\matrix{ {a\alpha + b\beta } \cr {c\alpha + d\beta } \cr } } \right] = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$$$\Rightarrow$ $$\eqalign{ & a\alpha + b\beta = \alpha \,......(1) \cr & c\alpha + d\beta = \beta \,......(2) \cr} $$

$\alpha (a - 1) = - b\beta$ and $c\alpha = \beta (1 - d)$

${\alpha \over \beta } = {{ - b} \over {a - 1}}$ & ${\alpha \over \beta } = {{1 - d} \over c}$

$\therefore$ ${{ - b} \over {a - 1}} = {{1 - d} \over c}$

$- bc = (a - 1)(1 - d)$

$- bc = a - ad - 1 + d$

$ad - bc = a + d - 1$

$= 2021 - 1 = 2020$"

Let $A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$ and $$B = \left[ {\matrix{ \alpha \cr \beta \cr } } \right] \ne \left[ {\matrix{ 0 \cr 0 \cr } } \right]$$ such that AB = B and a + d = 2021, then the value of ad $-$ bc is equal to ___________.

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Let $A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$ and $$B = \left[ {\matrix{ \alpha \cr \beta \cr } } \right] \ne \left[ {\matrix{ 0 \cr 0 \cr } } \right]$$ such that AB = B and a + d = 2021, then the value of ad $-$ bc is equal to ___________.

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