यदि `A=[{:(0,-"tan"(alpha)/(2)),("tan"(alpha)/(2)," "0):}]` तथा I कोटि 2 का एक तत्समक आव्यूह है । तो सिद्ध कीजिए कि `I+A=(I-A)[{:(cosalpha,-sinalpha),(sinalpha,cosalpha):}]`

Explanation

हम जानते है कि `sin alpha=(2"tan"(alpha)/(2))/(1+"tan"^(2)(alpha)/(2))` और `cos alpha=(2"tan"(alpha)/(2))/(1+"tan"^(2)(alpha)/(2))` माना `"tan"(alpha)/(2)=x`
`therefore sinalpha=(2x)/(1+x^(2))` और `cosalpha=(1-x^(2))/(1+x^(2))`
`A=[{:(0,-x),(x,0):}]`
`thereforeI-A=[{:(1,0),(0,1):}]-[{:(0,-x),(x,0):}]=[{:(1,x),(-x,1):}]`
अब ,R.H.S. `=(I-A)[{:(cosalpha,-sinalpha),(sinalpha,cosalpha):}]`
`=[{:(1,x),(-x,1):}][{:((1-x^(2))/(1+x^(2)),(-2x)/(1+x^(2))),((2x)/(1+x^(2)),(1-x^(2))/(1+x^(2))):}]`
`=[{:((" "1-x^(2))/(1+x^(2))+,(2x^(2))/(1+x^(2)),(-2x)/(1+x^(2))+,(x(1-x^(2)))/(1+x^(2))),((-x(1-x^(2)))/(1+x^(2))+,(2x)/(1+x^(2)),(2x^(2))/(1+x^(2))+,(1-x^(2))/(1+x^(2))):}]`
`=[{:(1,-x),(x,1):}]=[{:(1,0),(0,1):}]+[{:(0,-x),(x," "0):}]`
=I+A=R.H.S`. यही सिद्ध करना था