Question 549363
(-1  1 3)      (x11 x12 x13)=(1 0 0)
(0  0 -1)      (x21 x22 x23)=(0 1 0)
(2  0  1)      (x31 x32 x33)=(0 0 1)


From the definition above we want that middle matrix----(that's what an inverse is).   


Let's first find the determinant of A which is 
(-1  1 3)      
(0  0 -1)      
(2  0  1)


So DetA=-1*detB+0*detC-0*detD

Where B is 
(0 -1)
(2 1)
and DetB=0*1-(-1*2)=2

C is 
(-1 3)
(2 1)
and detC=-1*1-3*2=-1-6=-7

and D is 
(-1 3) 
(0 -1)
and detD=-1*-1+3*0=1


So DetA=-1*2=-2



Now we need to find AdjA
CoeffMatrix
=(+detE  -detB  +detF)
(-detG  +detC  -detH)
(+detI  -detD  +detJ)
=AdjA(transpose)


where B,C and D are the same as above, 
E=(0 -1)
  (0 1)
detE=0*1-(-1*0)=0


From here on, I will just write the determinants, they are calculated the same way we have been doing above

F=(0 0)
  (2 0)
detF=0

G=(1 3)
  (0 1)
detG=1

H=(-1 1)
  (2 0)
detH=-2

I=(1 3)
  (0 -1)
detI=-1

J=(-1 1)
  (0 0)
detJ=0


Leaving 
CoeffMatrix
=(0  -2  0)
 (-1  -7  2)
 (-1  -1  0)

And AdjA is the transpose of CoeffMatrix
So AdjA=
(0 -1 -1)
(-2 -7 -1)
(0 2 0)





So finally, to find our inverse we have:
InverseA=(1/detA)*AdjA
=(1/-2)*AdjA
=(0 1/2 1/2)
 (1 7/2 1/2)
 (0 -1 0)




Hopefully this helps a bit:)