Invertibility of a product of invertible matrices

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SUMMARY

The product of three nxn invertible matrices A, B, and C, denoted as ABC, is guaranteed to be invertible. This conclusion is supported by the properties of matrix multiplication, specifically its associativity, which allows for the intermediate product AB to be established as invertible if both A and B are invertible. The non-zero determinant of each matrix further confirms that the product ABC retains invertibility. The argument extends to any number of invertible matrices through induction.

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  • Understanding of matrix multiplication and its associative property
  • Knowledge of invertible matrices and their properties
  • Familiarity with determinants and their significance in linear algebra
  • Basic concepts of linear maps and bijective functions
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  • Study the properties of determinants in relation to matrix invertibility
  • Learn about the implications of matrix multiplication in linear transformations
  • Explore induction proofs in linear algebra
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Students and professionals in mathematics, particularly those focusing on linear algebra, matrix theory, and related fields. This discussion is beneficial for anyone seeking to deepen their understanding of matrix properties and their applications in linear transformations.

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If A,B, and C are each nxn invertible matrices, will the product ABC be invertible?
 
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yea, if A, B are nxn invertible then AB is invertible. Since Matrix Multiplication is associative, (AB)C is invertible provided c is nxn invertible.
 
If your linear algebra is better than your basic set/function theory, remark that A,B,C have non-zero determinant, hence their product also.
 
If A and B are both invertible then (AB)(B^{-1}A^{-1})= A(B^{-1}B)A^{-1} because, as jakncoke says, matrix multiplication is associative. Of course, B^{-1}B= I so that becomes (AB)(B^{-1}A^{-1})= AA^{-1}= I. Simlarly, (B^{-1}A^{-1})(AB)= B^{-1}(A^{-1}A)B^{-1}= B^{1}B= I

You can extend that to any number of factors by induction and repeated use of associativity.
 
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As Landau referenced, if you like to think about functions, matrix multiplication is equivalent to applying a linear map, and invertible matrices define bijective linear maps (and vice versa), and a composition of bijective linear maps is a bijective linear map. so yes.
 

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