If `f (x)={:((cos x,-sin x,0),(sin x,cos x,0),(0,0,1)):},"show that "(f(x))^(-1) = f(-x)`.

Explanation

`|f(x)|=|(cos x,-sinx,0),(sinx,cos x,0),(0,0,1)|`
`= cos x(cos x - 0)+sin x(sin x - 0)+0`
` = cos^(2) x+sin^(2) x = 1 ne 0`
` :. (f(x))^(-1)`exists.
Consider `f (x)*(f(x))^(-1) = I`
`:. {:((cos x,-sin x,0),(sin x,cos x,0),(0,0,1)):} (f(x))^(-1) = {:((1,0,0),(0,1,0),(0,0,1)):}`
By `(cos x)R_(1)`, we get ,
` {:((cos ^(2)x,-sin xcos x,0),(sin x,cos x,0),(0,0,1)):} (f(x))^(-1) = {:((cos x,0,0),(0,1,0),(0,0,1)):}`
By `R_(1) +(sin x)R_(2)`, we get,
` {:((1,0,0),(sin x,cos x,0),(0,0,1)):} (f(x))^(-1) = {:((cos x,sin x,0),(0,1,0),(0,0,1)):}`
By `R_(2) - (sin x)R_(1)`, we get,
` {:((1,0,0),(0,cos x,0),(0,0,1)):} (f(x))^(-1) = {:((cos x,sin x,0),(-sin x cos x,cos^(2)x,0),(0,0,1)):}`
By `(1/(cos x)) R_(2)`, we get ,
` {:((1,0,0),(0,1,0),(0,0,1)):} (f(x))^(-1) = {:((cos x,sin x,0),(-sin x ,cosx,0),(0,0,1)):}`
`:. (f(x))^(-1)= {:((cos x,sinx,0),(-sin x,cos x,0),(0,0,1)):} `...(1)
Also, `f(-x)={:((cos(-x),-sin(-x),0),(sin(-x),cos(-x),0),(" "0," "0,1)):}`
`:.f(-x)={:((cosx,-sinx,0),(-sinx,cosx,0),(" "0," "0,1)):}`....(2)
Form (1) and (2) , we get,
`(f(x))^(-1) = f(-x)`.