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)]#
