माना `A=[(0,-“tan”(alpha)/2),(“tan”(alpha)/2,0)]` तथा `I` कोटि 2 का तत्समक आव्यूह है तो सिद्ध कीजिए कि- `(I+A)=(I-A).[(cos alpha, -sin alpha),(sin alpha, cos alpha)]`

Explanation

माना `”tan”(alpha)/2=t`
तब `cos alpha=(1-“tan”^(2)(alpha)/2)/(1+ “tan”^(2)(alpha)/2)=(1-t^(2))/(1+t^(2))`
और `sin alpha=(2 “tan”(alpha)/2)/(1+ “tan”(alpha)/2)=(2t)/(1+t^(2))`
`A=[(0,-“tan”(alpha)/2),(“tan”(alpha)/2,0)]=[(0,-t),(t,0)]`
`:.I+A[(1,0),(0,1)]+[(0,-t),(t,0)]=p[(1,-t),(t,1)]`
और `I-A=[(1,0),(0,1)]-[(0,-t),(t,0)]=[(1,t),(-t,1)]`
अब `(I-A).[(cos alpha, -sin alpha),(sin alpha, cos alpha)]`
`=[(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^(2))/(1+t^(2)),(-t(1+t^(2)))/(1+t^(2))),((t(1+t^(2)))/(1+t^(2)),(1+t^(2))/(1+t^(2)))]=[(1,-t),(t,1)]`
`=I+A`.