If `S=[((sqrt(3)-1)/(2sqrt(2)),(sqrt(3)+1)/(2sqrt(2))),(-((sqrt(3)+1)/(2sqrt(2))),(sqrt(3)-1)/(2sqrt(2)))], A=[(1,0),(-1,1)]` and `P=S ("adj.A") S^(T)

Monjur Alahi

If `S=[((sqrt(3)-1)/(2sqrt(2)),(sqrt(3)+1)/(2sqrt(2))),(-((sqrt(3)+1)/(2sqrt(2))),(sqrt(3)-1)/(2sqrt(2)))], A=[(1,0),(-1,1)]` and `P=S ("adj.A") S^(T)`, then find matrix `S^(T) P^(10) S`.


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Salim Ahmed Kawsar

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`S=[((sqrt(3)-1)/(2sqrt(2)),(sqrt(3)+1)/(2 sqrt(2))),(-((sqrt(3)+1)/(2sqrt(2))),(sqrt(3)-1)/(2sqrt(2)))]`
`=[("sin "15^(@),cos 15^(@)),(-"cos "15^(@),sin 15^(@))]`
`:. SS^(T)=S^(T)S=I`
Now,
`S^(T) P^(10) S=S^(T)(S ("adj. A")S^(T))^(10)S`
`=S^(T)S("adj. A") S^(T) (S("adj. A")S^(T))^(9)S`
`=I ("adj. A")S^(T) (S("adj. A")S^(T))^(9)S`
`=("adj. A")S^(T)S("adj. A") S^(T) (S("adj. A")S^(T))^(8) S`
`=("adj. A")^(2)S^(T) (S("adj. A")S^(T))^(8)S`
...
...
`=("adj. A")^(10)`
`A=[(1,0),(-1,1)]`
`:.` adj. `A=[(1,0),(1,1)]`
`:. ("adj. A")^(2)=[(1,0),(1,1)][(1,0),(1,1)]=[(1,0),(2,1)]`
`:. ("adj. A")^(3)=[(1,0),(3,1)]`
And so on.
`:. ("adj. A")^(10)=[(1,0),(10,1)]=S^(T) P^(10) S`

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