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Question

Let A and B be n×nn \times n matrices, and suppose that B and AB are both invertible. Prove that A is also invertible.

Solution

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Since BB is invertible, so B1B^{-1} is also invertible by Theorem 3.23 part (a). Now (AB)B1=A(BB1)=AI=A(AB)B^{-1} = A(BB^{-1}) = AI = A, so we have A=(AB)B1A = (AB)B^{-1}. Since both ABAB and B1B^{-1} are invertible, so AA is product of two invertible matrix. Therefore by Theorem 3.23 part (b), AA is invertible.

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