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.