JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 14)
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 ___________.
Answer
2020
Explanation
$$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right],\,B = \left[ {\matrix{
\alpha \cr
\beta \cr
} } \right]$$
$$AB = B$$
$$\left[ {\matrix{ a & b \cr c & d \cr } } \right]\left[ {\matrix{ \alpha \cr \beta \cr } } \right] = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$$
$$\left[ {\matrix{ {a\alpha + b\beta } \cr {c\alpha + d\beta } \cr } } \right] = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$$$$ \Rightarrow $$ $$\eqalign{ & a\alpha + b\beta = \alpha \,......(1) \cr & c\alpha + d\beta = \beta \,......(2) \cr} $$
$$\alpha (a - 1) = - b\beta $$ and $$c\alpha = \beta (1 - d)$$
$${\alpha \over \beta } = {{ - b} \over {a - 1}}$$ & $${\alpha \over \beta } = {{1 - d} \over c}$$
$$ \therefore $$ $${{ - b} \over {a - 1}} = {{1 - d} \over c}$$
$$ - bc = (a - 1)(1 - d)$$
$$ - bc = a - ad - 1 + d$$
$$ad - bc = a + d - 1$$
$$ = 2021 - 1 = 2020$$
$$AB = B$$
$$\left[ {\matrix{ a & b \cr c & d \cr } } \right]\left[ {\matrix{ \alpha \cr \beta \cr } } \right] = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$$
$$\left[ {\matrix{ {a\alpha + b\beta } \cr {c\alpha + d\beta } \cr } } \right] = \left[ {\matrix{ \alpha \cr \beta \cr } } \right]$$$$ \Rightarrow $$ $$\eqalign{ & a\alpha + b\beta = \alpha \,......(1) \cr & c\alpha + d\beta = \beta \,......(2) \cr} $$
$$\alpha (a - 1) = - b\beta $$ and $$c\alpha = \beta (1 - d)$$
$${\alpha \over \beta } = {{ - b} \over {a - 1}}$$ & $${\alpha \over \beta } = {{1 - d} \over c}$$
$$ \therefore $$ $${{ - b} \over {a - 1}} = {{1 - d} \over c}$$
$$ - bc = (a - 1)(1 - d)$$
$$ - bc = a - ad - 1 + d$$
$$ad - bc = a + d - 1$$
$$ = 2021 - 1 = 2020$$
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