F ind aba -1 w here (i) a = (5 , 7 , 9) , b = (1 , 2, 3) (ii) a = (1 , 2,5)(3 , 4) , b = (1 , 4 , 5) .
"aba^{-1}=(ab)a^{-1}\n\\\\=(ba)a^{-1}\n\\\\=baa^{-1}\n\\\\=b(aa^{-1})\n\\\\=b(1)\n\\\\=b"
(i):
a = (5 , 7 , 9) , b = (1 , 2, 3)
"aba^{-1}=b=(1,2,3)"
(ii):
a = (1 , 2,5)(3 , 4) , b = (1 , 4 , 5)
"aba^{-1}=b=(1,4,5)"
Comments
Leave a comment