माना कि `A = [(0,-"tan"(alpha)/(2)),("tan"(alpha)/(2),0)]` तथा I, 2 कोटि वाला त्तस्मक आव्यूह हैं । दिखाएँ कि `(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`
पुन: `(I+A)=[(1,0),(0,1)]+[(0,-t),(t,0)]=[(1,-t),(t,1)]`
तथा, `(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)`
अत:, `(I+A)=(I-A)[(cos alpha,-sin alpha),(sin alpha,cos alpha)]`