माना `A=[(0, -"tan" alpha/2),("tan" alpha/2,0)]` तथा I ,2 क्रम का तत्समक आव्यूह है, दर्शाइए की
`(I+A)=(I-A)[(cos alpha,-sin alpha),(sin alpha,cos alpha)]`

Explanation

`cos alpha =(1-tan^(2) (alpha//2))/(1+tan^(2) (alpha//2))=(1-t^(2))/(1+t^(2))`,
`sin alpha=(2 tan (alpha//2))/(1+tan^(2) (alpha//2))=(2t)/(1+t^(2))`
जहाँ `"tan" alpha/2=t`
पुनः `(I+A)=[(1,0),(0,1)]+[(0-t),(t,0)]=[(1,-t),(t,1)]`
`(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),(t,1)]=(I+A)`
इस प्रकार `(I+A)=(I-A)[(cos alpha,-sin alpha),(sin alpha,cos alpha)]`