Proving ABC = I using matrix operations: Self-taught induction proof

[synkk]

I've never done a proof before so bare with me (I'll be self teaching myself proof by induction later this week):

Now in my textbook it says that (for a question) ABC = I and then it just goes on to say B^-1 = CA without any proof, so I'm trying to prove it; I've done two attempts below:

$$ABC = I$$
$$B^{-1}ABC = B^{-1}I$$
$$(B^{-1}B)AC =B^{-1}I$$
$$ACI=B^{-1}I$$
$$AC=B^{-1}$$


Now I know that matrices are non commutative so AC is not exactly CA so It seems wrong but I thought I would just get some criticism on it anyway. Heres my second attempt:

$$ABC = I$$
$$A^{-1}ABC = A^{-1}I$$
$$BC = A^{-1}$$
$$BCC^{-1} = A^{-1}C^{-1}$$
$$B = A^{-1}C^{-1} = (CA)^{-1}$$
$$B^{-1} = CA$$


Is this one any better? If anyone could correct me on any of these two it'd be great, general tips on how to improve also thanks.

 

[Mark44]

I've never done a proof before so bare with me (I'll be self teaching myself proof by induction later this week):

Now in my textbook it says that (for a question) ABC = I and then it just goes on to say B^-1 = CA without any proof, so I'm trying to prove it; I've done two attempts below:

$$ABC = I$$
$$B^{-1}ABC = B^{-1}I$$
$$(B^{-1}B)AC =B^{-1}I$$
$$ACI=B^{-1}I$$
$$AC=B^{-1}$$


Now I know that matrices are non commutative so AC is not exactly CA so It seems wrong but I thought I would just get some criticism on it anyway. Heres my second attempt:

$$ABC = I$$
$$A^{-1}ABC = A^{-1}I$$
$$BC = A^{-1}$$
$$BCC^{-1} = A^{-1}C^{-1}$$
$$B = A^{-1}C^{-1} = (CA)^{-1}$$
$$B^{-1} = CA$$


Is this one any better? If anyone could correct me on any of these two it'd be great, general tips on how to improve also thanks.

The second approach looks good to me. You don't mention it, but the problem should state some assumptions about A, B, and C being invertible.

Where you first try goes wrong is in the 3nd line. To get B^-1B, you had to commute AB, which as you already know, doesn't have to be equal to BA.

 

[micromass]

I'm not going to say anything on the contents of the proof. The contents is fine (of the second attempt at least). But let me say something on the style of the proof. Your proof is very readable, but it's not a proof you'll find in textbooks. Textbook-proofs have more inbetween texts and are not just a sequence of equalities.

If I were to rewrite your proof, I would do it like this:

We know by hypothesis that
$$ABC = I.$$
Multiplying both sides of the equation by $A^{-1}$, we get
$$A^{-1}ABC = A^{-1}I,$$
which yields immediately that
$$BC = A^{-1}.$$
If we multiply both sides of the equation by $C^{-1}$, we get that
$$BCC^{-1} = A^{-1}C^{-1}$$
and thus that
$$B = A^{-1}C^{-1} = (CA)^{-1}.$$
Taking inverses gives us the desired result that
$$B^{-1} = CA.$$

This has the great advantage that you also explain every step and tell us what it is you did.

 

[synkk]

thank you both for your feedback