Value of Determinant

Algebra Level 3

∣ 1 / 1 1 / 3 1 / 15 1 / 3 1 / 15 1 / 35 1 / 5 1 / 35 1 / 63 ∣ \begin{vmatrix}{1/1} && {1/3} && {1/15} \\ {1/3} && {1/15} && {1/35} \\ {1/5} && {1/35} && {1/63}\end{vmatrix}

If the value of the determinant above equals A B \dfrac AB , where A A and B B are coprime positive integers, find A + B A+B .


The answer is 496189.

Rishabh Deep Singh by Rishabh Deep Singh , 23, India

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1 solution

D e t = [ 1 ( 1 15 ) ( 1 63 ) + ( 1 3 ) ( 1 35 ) ( 1 5 ) + 1 15 ( 1 3 ) ( 1 35 ) ] − [ 1 5 ( 1 15 ) ( 1 15 ) + 1 35 ( 1 35 ) ( 1 ) + 1 63 ( 1 3 ) ( 1 3 ) ] Det=\left[1\left(\dfrac{1}{15}\right)\left(\dfrac{1}{63}\right)+\left(\dfrac{1}{3}\right)\left(\dfrac{1}{35}\right)\left(\dfrac{1}{5}\right)+\dfrac{1}{15}\left(\dfrac{1}{3}\right)\left(\dfrac{1}{35}\right)\right]-\left[\dfrac{1}{5}\left(\dfrac{1}{15}\right)\left(\dfrac{1}{15}\right)+\dfrac{1}{35}\left(\dfrac{1}{35}\right)(1)+\dfrac{1}{63}\left(\dfrac{1}{3}\right)\left(\dfrac{1}{3}\right)\right]

= ( 1 945 + 1 525 + 1 1575 ) − ( 1 1125 + 1 1225 + 1 567 ) =\left(\dfrac{1}{945}+\dfrac{1}{525}+\dfrac{1}{1575}\right)-\left(\dfrac{1}{1125}+\dfrac{1}{1225}+\dfrac{1}{567}\right)

= 17 4725 − 1721 496125 =\dfrac{17}{4725}-\dfrac{1721}{496125}

= 64 496125 =\dfrac{64}{496125}

The desired answer is 64 + 496125 = 64+496125= 496189 \boxed{496189}

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