Everything is 1 1

Algebra Level 1

If a 2 + b 2 = 1 a^2+b^2=1 , c 2 + d 2 = 1 c^2+d^2=1 and a c + b d = 1 , ac+bd=1, what is the value of a d − b c ? ad-bc?

-1 1 0 1 2 \frac{1}{2}

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Mahindra Jain by Mahindra Jain

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1 solution

a 2 + b 2 + c 2 + d 2 − 2 a c − 2 b d = 1 + 1 − 2 = 0 a^2 + b^2 + c^2 + d^2 - 2ac - 2bd = 1 + 1 - 2 = 0

( a − c ) 2 + ( b − d ) 2 = 0 (a-c)^2 + (b-d)^2 = 0

Since squares are always positive, ( a − c ) 2 = ( b − d ) 2 = 0 (a-c)^2 = (b-d)^2 = 0

Therefore, a = c , b = d a = c, b = d

⇒ a d − b c = a b − b a = 0 \Rightarrow ad - bc = ab - ba = \boxed{0}

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