Next step to show that these matrice expressions are equal?

Question

This is a problem from Discrete Mathematics and its Applications

I know invertible means it is possible to take the inverse of this matrix.

This is definition of a power of a square matrix from my book

I know by definition that the product of a matrix and its inverse is the identity matrix(all ones)

Here's my work so far

I am trying to use matrix rules but am not sure as to what matrix rule I should use to show that the two expressions are the bottom are equal(expressions I got from expanding out terms)

Answer 1

We must show that $\mathbf{A}^n(\mathbf{A}^{-1})^n=\mathbf{I}$, where $\mathbf{I}$ is the $n\times n$ identity matrix. Since matrix multiplication is associative, we can write this product as $$ \mathbf{A}^n\left((\mathbf{A}^{-1})^n\right) = \mathbf{A}(\mathbf{A}\ldots(\mathbf{A}(\mathbf{A}\mathbf{A}^{-1})\mathbf{A}^{-1})\ldots\mathbf{A}^{-1})\mathbf{A}^{-1}. $$ By dropping each $\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}$ from the center as it is obtained, this product reduces to $\mathbf{I}$. Similarly, $\left((\mathbf{A}^{-1})^n\right)\mathbf{A}^n=\mathbf{I}$. Therefore, by definition $(\mathbf{A}^n)^{-1}=(\mathbf{A}^{-1})^n$.

Note that a more formal proof would require mathematical induction, but this is the basic sketch.