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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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Answer» `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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