Assuming that all matrices are n x n and invertible, solve for D .

1 answer:

3 0

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

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