if[{:(2,1),(3,2):}]A[{:(-3,2),(5,-3):}]=...

`if[{:(2,1),(3,2):}]A[{:(-3,2),(5,-3):}]=[{:(1,0),(0,1):}],"then" A=?`

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We have, `[{:(2,1),(3,2):}]_(2xx2)A.[{:(-3,2),(5,-3):}]_(2xx2)=[{:(1,0),(0,1):}]_(2xx2)`
Let `A=[{:(a,b),(c,d):}]_(2xx2)`
`therefore [{:(2,1),(3,2):}] [{:(a,b),(c,d):}][{:(-3,2),(5,-3):}]=[{:(1,0),(0,1):}]`
`rArr [{:( 2a+c,2b+d),(3a+2c,3b+2d):}][{:(-3,2),(5,-3):}]=[{:( 1,0),(0,1):}]`
`rArr [{:(-6a-3c_10b+5d,a+2c-6b-3d),(-9a-6c+15b+10d,6a+4c-9b-6d):}]=[{:(1,0),(0,1):}]`
`rArr -6a-3c+10b+5d=1`
`rArr 4a+2c-6b-3d=0`
`rArr -9a-6c+15b+10d=0`
`rArr 6a+4c-9b-6d=1`
On edding Eqs. i and iv, we get
`c+b-d=2rArrd=c+b-2`
On adding Eqs. v and vii
`v+b-2=1-arArra+b+c=3`
`rArr a=3-b-c`
Now, using values of a and d in eq. ii, we get
`4(3-b-c)+2c-6b-3(b+c-2)=0`
`rArr 12-4b-4c+2c-6b-3b-3c+6=0`
`rArr -13b-5c=-18`
On multiplying Eq. ix by 5 and Eq. x by 13, then adding we get
`=169b-65c=-234`
`170b+65c=235`
b=1
`rArr -13xx1-5c=-18`
`rArr -5c =-18+13=-5rArrc=1`
`therefore a=3-1-1=1` and d=1-1=0
`therefore A=[{:(1,1),(1,0):}]`

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