In the following, assume all matrices involved (and their combinations) are square and invertible. Solve for X in terms of the other matrices and/or their inverses. AX = BX + C OA. X = (B – A) C В. X — В - А Oc. X = C(A – B)1 OD. X = C(B - A)-1 C Е. Х— A B OF. X = (A – B) C

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