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[" 18."11A=[[0,-con(alpha)/(2)],[cos(alp...

[" 18."11A=[[0,-con(alpha)/(2)],[cos(alpha)/(2),0]]],[1+A=0-A/[[cos a],[sin alpha]]]

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If A=,[[0,-tan((alpha)/(2))]tan((alpha)/(2)),0]], then I+A=(I-A)[[cos alpha,-sin alphasin alpha,cos alpha]]

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