Matrices and their inverses Having a bit of an issue answering this question. The answer seems simple but I am quite unsure. I think the answer is true, as it is a true/false question. The question is:
If a square matrix B is invertible, then its inverse is also invertible - True or False? This is all the information I have. I do not know how to go about this, i've just been trying to read laws on invertible matrices but I am really confused. Going to extra classes tomorrow for it, as I said I joined the class late and now I suddenly have an assignment to hand in, so I was just looking for an answer! 
Please help me out! Thanks
 A: A matrix $B$ is invertible if there exists a matrix $X$ so that $BX=XB=I$; That matrix $X$ is then denoted $B^{-1}$.
Now can you find a matrix $X$ so that $B^{-1}X=XB^{-1}=I$?
A: I will assume the definition that an $n \times n$ matrix $A$ is invertible if there exists an $n \times n$ matrix $B$ such that $AB = BA = I$. (As mentioned by Ryo, it is in fact sufficient that $AB = I$ to get $BA = I$ as a corallary, but you won't need this if you assume the definition with $AB = BA = I$, so just forget about it for now.)
The reason we call this matrix 'the' inverse of $A$ is because there can only be one matrix $B$ satisfying these conditions. Indeed: suppose $B_1$, $B_2$ satisfy $AB_1 = AB_2 = B_1A = B_2A = I$. Then we have
$$
B_1 = IB_1 = B_2AB_1 = B_2I = B_2.
$$
Therefore it makes sense to call a matrix $B$ 'the' inverse of $A$ as soon as it satisfies $AB = BA = I$. It is completely defined by just these two equalities!
Okay, so suppose $B$ is the inverse of $A$. To show that $B$ is invertible, we need to find a matrix $C$ such that $BC = CB = I$. But we already know such a matrix: $A$! That's why $B$ (the inverse of $A$) is invertible and in fact that the inverse of $B$ (i.e. the inverse of the inverse of $A$) is equal to $A$.
A: Alternatively (to the approach above), consider the equivalent expressions (By definition of a matrix inverse):
$B^{-1} * (B^{-1})^{-1} = I$
Because both sides are equivalent,
$B * (B^{-1} * (B^{-1})^{-1}) = B * I$
I'll give you the right half, which simplifies by the definition of the identity matrix.
$ ??? = B $ 
For the left half, what property of invertible matrices & matrix multiplication allows you to declare that
$ ??? = B $
A: True.
Definition: Matrix $B$ is said to be invertible iff there exists a matrix $X$ such that $BX = XB = I$, and in this case, $X$ is called an inverse of $B$.
Now, let $B$ be a square matrix.  Assume that
there exists a square matrix $X$
such that $BX = I$. According to the the theorem proven here
Matrices: left inverse is also right inverse?
$XB = I$.  Therefore, $B$ is invertible and $X$ is an inverse of it.
Now, from Definition, $X$ is invertible and $B$ is an inverse of it.
Therefore, an inverse is invertible, too.
(I haven't proved the uniqueness of inverse, so I kept saying "an inverse of", but uniqueness can be proved easily, I think.)
