First find bb(A^-1)
The easiest way to find the inverse of bb(A) is to find the determinant of bb(A):
This is just:
(3xx6)-(1xx2)=16
Next switch the elements on the leading diagonal of bb(A) and change the signs of the elements on the non-leading diagonal of bb(A)
So you should have:
[(6,-2),(-1,3)]
Divide each element by the determinant bb(16):
[(6/16,-2/16),(-1/16,3/16)]=[(3/8,-1/8),(-1/16,3/16)]
bb(A^-1)=[(3/8,-1/8),(-1/16,3/16)]
Now:
bb(YA)+bb(B)=bb(C)
bb(YA)=bb(C-B)
Using bb(A^-1)
bb(YA A^-1)=bb((C-B)A^-1)
bb(Y)=bb((C-B)A^-1)
Note we are post multiplying on both sides. This is important as Matrix multiplication is non-commutative.
i.e.
bb(AB)!=bb(BA) ( In general )
bb(C-B)=[(3,4),(2,6)]-[(4,-1),(2,2)]=[(-1,5),(0,4)]
:.
bb(Y)=[(-1,5),(0,4)][(3/8,-1/8),(-1/16,3/16)]=[(-11/16,17/16),(-1/4,3/4)]
bb(Y)=[(-11/16,17/16),(-1/4,3/4)]