Inverse Matrix....計極都極唔到....pls help!!thx

Let A =
(1 0 -2)
(2 3 -2)
(2 5 2)

det A = 6 – 20 + 12 + 10 = 8

cofactor of A =
(16 -8 4)
(-10 6 -5)
(6 -2 3)

adjoint of A =
transpose of (cofactor of A)
(16 -10 6)
(-8 6 -2)
(4 -5 3)

inverse of A
= (1/det A) x adjoint of A =

(1/8) [16 -10 6; -8 6 -2; 4 -5 3] =

(2 -5/4 3/4)
(-1 3/4 -1/4)
(1/2 -5/8 3/8)

I have checked the above answer with my calculator.
If there is still any mistake, please inform me.

Inverse of
1 0 -2
2 3 -2
2 5 2


=
( 16 -8 4 )T
(-10 6 -5 )
( 6 -2 3 )

=
( 16 -10 6 )
(- 8 6 -2 )
( 4 -5 3 )