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x = [ x1 , x2 ] be a unit eigen vector when 

 √(x12 + x22 ) = 1      

So ....(x12 + x22 = 1)

Ax = ∆x  , where ∆ = eigen value

= x'Ax = x'∆x (here ' represent transpose)

=  ∆x'x ( since ∆ can be think as scalar value)

 = ∆ [x1,x2]' [x1,x2]

= ∆ [ x12 + x22 ]

= ∆(1) (since unit eigen vector)

= ∆

From this derivation we can say that max value of x'Ax is the maximum eigen value..

Just find the eigen values which one is max is the ans..

So Max . vakue of x'Ax= ∆ =( 5 + √5) /2.

http://math.stackexchange.com/questions/1553046/what-is-the-maximum-value-of-xt-ax

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