JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 17)
Let $$A = \left( {\matrix{
1 & { - 1} & 0 \cr
0 & 1 & { - 1} \cr
0 & 0 & 1 \cr
} } \right)$$ and B = 7A20 $$-$$ 20A7 + 2I, where I is an identity matrix of order 3 $$\times$$ 3. If B = [bij], then b13is equal to _____________.
Answer
910
Explanation
Let $$A = \left( {\matrix{
1 & { - 1} & 0 \cr
0 & 1 & { - 1} \cr
0 & 0 & 1 \cr
} } \right) = I + C$$
where, $$I = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right),C = \left( {\matrix{ 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr 0 & 0 & 0 \cr } } \right)$$
$${C^2} = \left( {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right),$$
$${C^3} = \left( {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right) = {C^4} = {C^5} = ........$$
$$B = 7{A^{20}} - 20{A^7} + 2I$$
$$ = 7{(I + C)^{20}} + 20{(I + C)^7} + 2I$$
$$ = 7(I + 20C + {}^{20}{C_2}{C^2}) - 20(I + 7C + {}^7{C_2}{C^2}) + 2I$$
So
$${b_{13}} = 7 \times {}^{20}{C_2}{C^2} - 20 \times {}^7{C_2} = 910$$
where, $$I = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right),C = \left( {\matrix{ 0 & { - 1} & 0 \cr 0 & 0 & { - 1} \cr 0 & 0 & 0 \cr } } \right)$$
$${C^2} = \left( {\matrix{ 0 & 0 & 1 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right),$$
$${C^3} = \left( {\matrix{ 0 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0 \cr } } \right) = {C^4} = {C^5} = ........$$
$$B = 7{A^{20}} - 20{A^7} + 2I$$
$$ = 7{(I + C)^{20}} + 20{(I + C)^7} + 2I$$
$$ = 7(I + 20C + {}^{20}{C_2}{C^2}) - 20(I + 7C + {}^7{C_2}{C^2}) + 2I$$
So
$${b_{13}} = 7 \times {}^{20}{C_2}{C^2} - 20 \times {}^7{C_2} = 910$$
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