First multiply by C inverse on the leftC−1CBATDBCTA=C−1CB?
Yielding:BATDBCTA=BT
Then Multiply by B inverse and A transpose inverse on the left yielding
DBCTA=B−1A−TBT
Now, multiply by B inverse, C transpose inverse and A inverse on the right
D=B−1A−TBTB−1C−TA−1
