# Math Help - Invertible matrices

1. ## Invertible matrices

Let A, B, and C denote n x n matrices. Show that if A, C, and ABC are invertible, then B is invertible.

Any help would be appreciated!

2. A^-1 exists and B^-1 exists. ABC and so (ABC)^-1 exists.
Now (ABC) x (C^-1 x B^-1 x A^-1) = I (by associativity of matrix mutiplication)
So (ABC)^-1 = C^-1 x B^-1 x A^-1.
Therefore, since A^-1 exists and B^-1 exists and (ABC)^-1 exists, the so does B^1.
Therefore B is invertible.

3. Originally Posted by Debsta
A^-1 exists and B^-1 exists.
Did you mean "and C^-1 exists"?

ABC and so (ABC)^-1 exists.
Now (ABC) x (C^-1 x B^-1 x A^-1) = I (by associativity of matrix mutiplication)
That is only true if B^-1 exists, which is what you are trying to prove.

So (ABC)^-1 = C^-1 x B^-1 x A^-1.
Therefore, since A^-1 exists and B^-1 exists and (ABC)^-1 exists, the so does B^1.
Therefore B is invertible.
B is invertible if and only if det(B) is not 0. Since ABC is invertible, det(ABC)= det(A)det(B)det(C) is not 0 so det(B) cannot be 0.
Actually, given that ABC is invertible, it follows that each of A, B, and C is invertible. You don't need to be given that A and C are also invertible.

4. Originally Posted by Debsta
A^-1 exists and B^-1 exists. ABC and so (ABC)^-1 exists.
Now (ABC) x (C^-1 x B^-1 x A^-1) = I (by associativity of matrix mutiplication)
So (ABC)^-1 = C^-1 x B^-1 x A^-1.
Therefore, since A^-1 exists and B^-1 exists and (ABC)^-1 exists, the so does B^1.
Therefore B is invertible.

This cannot be correct since you're using in the "proof" what you must prove, namely: that $B^{-1}$ exists! If you begin by saying " $A^{-1}\,\,and\,\,B^{-1}$ exist" then all the following is already wrong, unless that was a typo and you actually meant $C^{-1}$ instead of $B^{-1}$, but then you use $B^{-1}$ in the next line so it is wrong, not to mention that your conclusion is that $B^1=B$ exists, which of course is trivially true.

I think we could argue as follows: since $ABC$ is invertible there exists an invertible matrix $D$ s.t. $ABC\cdot D=I$, but using associativity of matrix multiplication and the fact that $A^{-1}\,,\,\,C^{-1}$ exist we can get:

$A(BC)D=I\Longrightarrow BCD=A^{-1}\Longrightarrow BC=A^{-1}D^{-1}\Longrightarrow B=A^{-1}D^{-1}C^{-1}$, and as B is the product of invertible matrices then it itself is invertible. Q.E.D.

Tonio

5. Yes sorry ...I DID mean "and C^-1 exists" in the first line. And the second last line should be "Therefore since A^-1 and C^-1 exists and (ABC)^-1 exists, then so does B^-1."
Note to self: Proofread before hitting send button.

With those changes I think the proof is valid.

6. Originally Posted by Debsta
Yes sorry ...I DID mean "and C^-1 exists" in the first line. And the second last line should be "Therefore since A^-1 and C^-1 exists and (ABC)^-1 exists, then so does B^-1."
Note to self: Proofread before hitting send button.

With those changes I think the proof is valid.

I don't think so since in "the proof" still appears $B^{-1}$ twice more: here "(ABC) x (C^-1 x B^-1 x A^-1) = I" and here "(ABC)^-1 = C^-1 x B^-1 x A^-1} , so in fact you use $B^{-1}$ ALL along your "proof" and this, as already noted, is incorrect.
Anyway, you already have two different approaches to prove what you want.

Tonio

7. We are restricted in what we can use to write the proof for this question. Basically, all we can use is the Theorem that states:

If A is an invertible matrix, then A^-1 is invertible and (A^-1)^-1 = A

If A and B are n x n invertible matrices, then so is AB and the inverse of AB is the product of the inverses of A and B in the reverse order. (AB)^-1 = B^-1 A^-1

If A is an invertible matrix, then so is A^T , and the inverse of A^T is the transpose of A^-1 . That is (A^T)^-1 = (A^1)^T

The product of n x n invertible matrices is invertible, and the inverse is the product of their inverses in the reverse order.

We cannot use (ABC)D = I implies D(ABC) = I.

Also, another part to this question that I find confusing:

Let A, B, and C denote n x n matrices. Show that if A and AB are invertible, B is invertible. Same rules apply - we can only use the Theorem I posted.

8. Originally Posted by BrownianMan
We are restricted in what we can use to write the proof for this question. Basically, all we can use is the Theorem that states:

If A is an invertible matrix, then A^-1 is invertible and (A^-1)^-1 = A

If A and B are n x n invertible matrices, then so is AB and the inverse of AB is the product of the inverses of A and B in the reverse order. (AB)^-1 = B^-1 A^-1

If A is an invertible matrix, then so is A^T , and the inverse of A^T is the transpose of A^-1 . That is (A^T)^-1 = (A^1)^T

The product of n x n invertible matrices is invertible, and the inverse is the product of their inverses in the reverse order.

We cannot use (ABC)D = I implies D(ABC) = I.

Also, another part to this question that I find confusing:

Let A, B, and C denote n x n matrices. Show that if A and AB are invertible, B is invertible. Same rules apply - we can only use the Theorem I posted.
This is copied from tonio's post.

$
A(BC)D=I\Longrightarrow BCD=A^{-1}\Longrightarrow BC=A^{-1}D^{-1}\Longrightarrow B=A^{-1}D^{-1}C^{-1}
$

ABC is invertible and from the 1st rule in your list it's inverse is (ABC)^-1. You can call this D if you wish. It's just shorthand and doesn't change any property of it.

By 1 again, ABCD=I

Now, matrix multiplication is associative. That's a basic property that you should have proved well before this proof.

So ABCD=I implies A(BCD)=I by associative property. By definition of inverse matrices, A and BCD are inverses. So BCD=A^-1 by rule 1. D=(ABC)^-1 is invertible by rule 1 so starting with BCD=A^-1 we get that BCDD^-1=A^-1D^-1 which means BC=A^-1D^-1. Same step now using the fact that C^-1 exists and CC^-1=I. This leaves you with B=A^-1D^-1C^-1 thus B exists.

All I did was explain tonio's post in more detail and reference your rules. The line in Latex sufficiently shows this proof.

9. Thanks.

Let A, B, and C denote n x n matrices. Show that if A and AB are invertible, B is invertible. Same rules apply - we can only use the Theorem I posted.

10. Originally Posted by BrownianMan
Thanks.