Assuming that all matrices are n x n and invertible, solve for D. ABCT DBA¹C = ABT. D = (B¹)¯'BA¹C(B¹)¯¹A¯¹ D = (CT) 'B-¹BTC-¹ (A¹)¯¹B-¹ D = B-¹CTAC-¹ (A¹)-¹C-¹ -1 D = (A¹) ¹C-¹BTC-¹ (A¹)¯¹B-¹ D = CTB-¹A-¹C-¹A¹B-¹ -1 -1

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