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Let `A=[[0,-tan(alpha//2)],[tan(alpha//2),0]]`and `I`be the identity matrixof order 2. Show that `I+A=(I-A)[[cosalpha,-sinalpha],[sinalpha,cosalpha]]`. |
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Answer» Let `"tan"(alpha)/(2)=t.` Then, `cosalpha=(1-tan^(2)(alpha//2))/(1+tan^(2)(alpha//2))=(1-t^(2))/(1+t^(2))` and, `sinalpha=(2tan(alpha//2))/(1+tan^(2)(alpha//2))=(2t)/(1+t^(2)).` `:." "(I+A)=[{:(1,0),(0,1):}]+[{:(0,-t),(t," "0):}]=[{:(1,-t),(t," "1):}].` And, `:." "(I-A)=[{:(1,0),(0,1):}]-[{:(0,-t),(t," "0):}]=[{:(1," "t),(-t," "1):}].` `:." "(I-A).[{:(cosalpha,-sinalpha),(sinalpha,cosalpha):}]` `=[{:(1,t),(-t,1):}][{:((1-t^(2))/(1+t^(2)),(-2t)/(1+t^(2))),((2t)/(1+t^(2)),(1-t^(2))/(1+t^(2))):}]` `=[{:((1-t^(2))/(1+t^(2))+(2t^(2))/(1+t^(2)),(-2t)/(1+t^(2))+(t(1-t^(2)))/(1+t^(2))),((-t(1-t^(2)))/(1+t^(2))+(2t)/(1+t^(2))," "(2t^(2))/(1+t^(2))+(1-t^(2))/(1+t^(2))):}]` `=[{:(1,-t),(t," "1):}]=(I+A).` Hence, `(I+A)=(I-A)[{:(cosalpha,-sinalpha),(sinalpha,cosalpha):}].` |
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