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Need a bit of help with this question.

We're given two invertible square $n\times n$ matrices $A$ and $B$ with entries in the reals.

We have to show that $AB$ is also invertible and then express $(AB)^{-1}$ in terms of $A$ and $B$.

I've managed to get the first part out.

Since $A$ is invertible, $Det(A)$ $\neq$ $0$. Similarily for $B, Det(B) \neq 0$

Also from the properties of Determinants: $Det(A)Det(B) = Det(AB)$ Hence $Det(AB) \neq 0$ and so $AB$ is invertible.

It's the second part that I need help. Thanks.

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  • $\begingroup$ Do you mean $(AB)^{-1}$ interms of $A$ and $B$? $\endgroup$ – minibuffer May 22 '14 at 13:05
  • $\begingroup$ Sigh, you don't need determinants... $\endgroup$ – Nishant May 22 '14 at 13:17
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Assuming you mean $(AB)^{-1}$, there's a well-known simile that can help you finding the answer.

Think of $A$ and $B$ representing actions, like $A$ means putting on socks, $B$ means putting on shoes.

In the morning you do $AB$, socks first, then shoes.

In the evening you need to undo this, that is, do $(AB)^{-1}$. You will need to take off the socks, which is $A^{-1}$, and take off the shoes, which is $B^{-1}$. But what do you do first, that is, do you do $A^{-1} B^{-1}$ or $B^{-1} A^{-1}$?

Once you've got the right idea, just compute the product to see it is correct.

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Consider $B^{-1}A^{-1}$. Multiply with this matrix on both sides of $AB$. You'll get the identity each time. Note that since $A$ and $B$ are invertible $A^{-1}$ and $B^{-1}$ are well defined.

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Here's another way: let C be the inverse of AB. then ABC = I and CAB = I. Now try to get C by multiplying on the left and/or multiplying or the right so that you are finally left with expression of the form C = f(A, B). This is the mathematical form of Andreas Caranti's answer.

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