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I am working on a homework problem, but I am lacking some understanding. Here is the problem:

Let $A$ and $B$ be invertible $n \times n$ matrices with $\det(A) = 3$ and $\det(B) = 4$.

I know that the product matrix of two invertible matrices must be invertible as well, but I am not sure how to prove that. I am trying to show it through the product of determinants if possible.

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  • $\begingroup$ This really depends on what facts you have to work with. It turns out there are a lot of equivalent ways to say a matrix is invertible, but you may not have seen some of those in class yet and hence can't really use them. For instance, one of the answers here used the fact that $A$ is invertible iff $\det(A)\neq0$. Have you seen this before? You could also explicitly find an inverse, since you know $A$ and $B$ both have inverses. $\endgroup$ Commented Apr 28, 2016 at 6:16
  • $\begingroup$ Thanks Alex! I know how to prove it using the fact that A is invertible, but I am trying to understand how would the prove look like using the determinants since otherwise I do not see why would I need them in the problem. $\endgroup$ Commented Apr 28, 2016 at 6:20

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Yes, since $\det(AB) = \det(A)\cdot \det(B) = 3\cdot 4 = 12 \neq 0$.

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$C$ is invertible iff for all $y$ there is some $x$ such that $Cx=y$.

Suppose $A,B$ are invertible and choose some $y$. Then there is some $z$ such that $Az=y$, and there is some $x$ such that $Bx = z$. Hence $ABx=y$, and so we see that $AB$ is invertible.

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It is that $(AB)^{-1}=B^{-1}A^{-1}$, because $AB(AB)^{-1}=ABB^{-1}A^{-1}=1\!\!1$, but only for $n\times n$ matrices.

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lets assume that C is a product of two invertible matrices .i.e. $C = AB$, there exists $A^{-1}$ such that $A^{-1}A = I = AA^{-1}$ and there exists $B^-1$ such that $B^{-1}B = I = B^{-1}B$.

We need to prove that for C there exist a Right Inverse D such that $CD = I$ as well as a Left Inverse E such that $EC = I$.

Lets prove that there exists a Right Inverse :

we know, $C=AB$

Multiplying $B^{-1}$ from right on both sides,

$=> CB^{-1} = ABB^{-1} => CB^{-1} = AI => CB^{-1} = A $

Multiplying both sides by $A^{-1}$ from right,

$=> CB^{-1}A^{-1} = AA^{-1} => CB^{-1}A^{-1} = I $

This proves that the Right Inverse(D) for C is $B^{-1}A^{-1}$

Note: we could multiply $A^{-1}$ and $B^{-1}$ on both sides because they exist and along the way, we were just trying to reduce right hand side to Identity Matrix.

A similar proof can be given to prove that there exists a Left Inverse for C(this time multiplication has to be done from left side) and you will come to know that the Left Inverse is also the same.

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