Differential Equations Solutions 13

Differential Equations Solutions 13 - 23 CHALLENGE 5.6. for...

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23 CHALLENGE 5.6. for i=1:3, W = planerot(A(i:i+1,i)); % Note that the next instruction just operates on the part % of A that changes. It is wasteful to do multiplications % on the rest. A(i:i+1,i:n) = W * A(i:i+1,i:n); end CHALLENGE 5.7. For G to be unitary, we need I = G G = · c s ¯ sc ¸· cs ¯ ¸ = · | c | 2 + | s | 2 cs sc ¯ sc c ¯ s | c | 2 + | s | 2 ¸ . Therefore, it is sufficient that | c | 2 + | s | 2 =1. Now Gz = · cz 1 + sz 2 ¯ sz 1 + cz 2 ¸ = " | z 1 | ± z ± z 1 + sz 2 ¯ sz 1 + | z 1 | ± z ± z 2 ± . Therefore, if z 1 ± = 0, we make the second entry zero by setting
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