Is Matrix A Invertible Given (AB)C Equals the Identity Matrix?

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The discussion centers on the invertibility of matrix A given the equation (AB)C = I, where I is the identity matrix. It is established that A has a right inverse BC, but to prove A's invertibility, both a right and left inverse must be demonstrated. The discussion emphasizes the importance of determinants, stating that A is invertible if and only if det(A) ≠ 0. Additionally, the properties of matrix multiplication, specifically associativity and the implications of square matrices, are highlighted as crucial for understanding the problem.

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Yankel
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Hello all

I am told that (AB)C=I and I am asked if that means A is invertible.
I also knows that A+A^t is defined (irrelevant as far as I understand) and that A has 3 rows, also irrelevant.

what I did is:

(AB)C=A(BC)=I

and then BC is the inverse of A. But according to the definition that ain't sufficient, is it ? Do I need also to show that (BC)A=I ? If so, how ?
 
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Yankel said:
Hello all

I am told that (AB)C=I and I am asked if that means A is invertible.
I also knows that A+A^t is defined (irrelevant as far as I understand) and that A has 3 rows, also irrelevant.

what I did is:

(AB)C=A(BC)=I

and then BC is the inverse of A. But according to the definition that ain't sufficient, is it ? Do I need also to show that (BC)A=I ? If so, how ?

Because for the product of matrices the associative property holds, it will be ...

$\displaystyle (A \cdot B) \cdot C = I \implies A \cdot (B \cdot C) = I\ (1)$

The commutative property are however not true in general, so that ...

$\displaystyle A \cdot (B \cdot C) \ne (B \cdot C) \cdot A\ (2)$

Kind regards

$\chi$ $\sigma$
 
so it could be invertible and it could be not invertible, I have no way to proof one way or another by using algebra only.
 
You know that A has a right inverse BC, and you want to show that BC is also a left inverse for A. This is true, but not that easy to prove. If you know about determinants, the best way to show that A is invertible is to use the fact that this is equivalent to the condition $\det(A)\ne0.$
 
Right, so in order to show that A is invertible, I MUST show that is has both a right inverse AND a left inverse ?
 
Yankel said:
Right, so in order to show that A is invertible, I MUST show that is has both a right inverse AND a left inverse ?

It suffices to make deductions about the determinants of $A$, $(BC)$, and $I$, using that for any finite square matrices $P$ and $Q$, we have:
$$\det(PQ)=\det P \cdot \det Q$$

Note that it is relevant that $A+A^t$ is defined and that $A$ has 3 rows, since it implies that A is a finite, square matrix.
Furthermore, the fact that $A(BC)=I$, implies that $(BC)$ is also a square matrix of the same dimensions as $A$.
You need this, because otherwise the determinants would not be defined.
 

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