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!
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.
Did you mean "and C^-1 exists"?
That is only true if B^-1 exists, which is what you are trying to prove.ABC and so (ABC)^-1 exists.
Now (ABC) x (C^-1 x B^-1 x A^-1) = I (by associativity of matrix mutiplication)
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.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.
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.
This cannot be correct since you're using in the "proof" what you must prove, namely: that $\displaystyle B^{-1}$ exists! If you begin by saying "$\displaystyle A^{-1}\,\,and\,\,B^{-1}$ exist" then all the following is already wrong, unless that was a typo and you actually meant $\displaystyle C^{-1}$ instead of $\displaystyle B^{-1}$, but then you use $\displaystyle B^{-1}$ in the next line so it is wrong, not to mention that your conclusion is that $\displaystyle B^1=B$ exists, which of course is trivially true.
I think we could argue as follows: since $\displaystyle ABC$ is invertible there exists an invertible matrix $\displaystyle D$ s.t. $\displaystyle ABC\cdot D=I$, but using associativity of matrix multiplication and the fact that $\displaystyle A^{-1}\,,\,\,C^{-1}$ exist we can get:
$\displaystyle A(BC)D=I\Longrightarrow BCD=A^{-1}\Longrightarrow BC=A^{-1}D^{-1}\Longrightarrow B=A^{-1}D^{-1}C^{-1}$, and as B is the product of invertible matrices then it itself is invertible. Q.E.D.
Tonio
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 $\displaystyle B^{-1}$ 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 $\displaystyle B^{-1}$ ALL along your "proof" and this, as already noted, is incorrect.
Anyway, you already have two different approaches to prove what you want.
Tonio
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.
$\displaystyle
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.
This is the exact same idea! It's actually easier. Look at my previous post. You know A,C and ABC are invertible. You showed B was too. Now you have A and AB being invertible and want to show B is. Same process!
Use the definition of the inverse to show that some D exists where D=(AB)^-1. Then start with ABD=I and from there use the other inverses and grouping to solve for B, then use the last rule on your list to show that B is a product of invertible matrices thus is invertible.