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Using determinants, prove that if $A_1,A_2,...,A_m$ are invertible $nxn$ matrices, where $m$ is a positive integer, then $A_1A_2...A_m$ is an invertible matrix.

Need help starting the proof. Do I show that $A_1,A_2,..,A_m$'s determinant is not equal to zero and use the determinant property, $det(AB)=det(A)det(B)$. Not sure where to go from here.

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  • $\begingroup$ Well, you prove the determinant of $\;A_1\cdot\ldots\cdot A_n \;$ is not zero by using the determinant product theorem, of course. $\endgroup$ – Timbuc Jul 2 '15 at 22:29
  • $\begingroup$ Yes, you're on the right track. What does $m$ represent? Number of matrices may be? $\endgroup$ – Vectorizer Jul 2 '15 at 22:30
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A matrix is invertible if and only if its determinant is nonzero. Since $\det(A_i)\neq 0$, and $\det(A_1...A_n)=\det(A_1)...\det(A_n)$, we know that $\det(A_1...A_n)\neq 0$, so $A_1...A_n$ is invertible.

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