Let A, B, C, D be n x n matrices, each with non-zero determinant. If ABCD = I, then B^(-1) is: A. D^(-1)C^(-1)A^(-1) B. CDA C. ADC D. Does not necessarily exist

GATE 2004 · Engineering Mathematics · Matrix · medium

Answer: B^(-1) = CDA. Answer: B.

1. Rearrange ABCD = I to find an expression for B: From ABCD = I: multiply A^(-1) on the left: BCD = A^(-1). This means B*(CD) = A^(-1).
2. Derive B^(-1): From BCD = A^(-1), we have B^(-1) = CD * (A^(-1))^(-1) = CD * A. Wait — re-derive directly: ABCD = I => (BCD)A * ... use another approach. Since ABCD = I and all invertible, note that CDAB is also related: start fresh. From ABCD = I, multiply both sides on right by A: ABCDA = A. Multiply on left by A^(-1): BCDA = I. So B * (CDA) = I, meaning B^(-1) = CDA.