Did I do this right and how do I do part b of this problem about matrices?

Question

Q: "An $n\times n$ matrix $A$ is called invertible if $$AB=I_n=BA$$ for some $n\times n$ matrix $B$. In this case, $B$ is called inverse of $A$.
Suppose an $n\times n$ matrix $A$ is invertible.

(a) Show that the inverse of $A$ is unique.
(b) For any $k\in\Bbb{Z}^+$show that $A^k$ is also invertible and $(A^k)^{-1}=(A^{-1})^k$.

My solution for (a):
Let $B$ and $C$ be inverses of $A$. $$AB=I_n=BA$$ $$AC=I_n=CA$$ $BI_n=B(AC)=(BA)C=I_nC$
$B=C$

Did I do part a correctly and can you help me get started on part b?

Part (a) correct. For part (b) you will use part (a) namely the inverse is unique and every invertible matrix commutes with its inverse. — jijijojo, Dec 28 '19 at 01:34

score 1 · Accepted Answer · answered Dec 28 '19 at 01:37

1

For part (b), compute the products $$A^k(A^{-1})^k,\quad (A^{-1})^kA^k$$ and see what they become.

answered Dec 28 '19 at 01:37

peterwhy

22,256

Did I do this right and how do I do part b of this problem about matrices?

1 Answers1