Question

Let A be an invertible matrix. Show that $\mathbf{(}{\mathit{A}}^{\mathbf{n}}{\mathbf{)}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{(}{\mathit{A}}^{\mathbf{-}\mathbf{1}}{\mathbf{)}}^{\mathbf{n}}$ whenever n is a positive integer.

Short answer

By the statement of induction, it is proved that$\mathbf{(}{\mathit{A}}^{\mathbf{n}}{\mathbf{)}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{(}{\mathit{A}}^{\mathbf{-}\mathbf{1}}{\mathbf{)}}^{\mathbf{n}}$

Step-by-step solution

Step 1:

Given: A is an invertible matrix and is a positive integer.

To prove: $\mathbf{(}{\mathit{A}}^{\mathbf{n}}{\mathbf{)}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathbf{(}{\mathit{A}}^{\mathbf{-}\mathbf{1}}{\mathbf{)}}^{\mathbf{n}}$

By the proof of induction:

Proof the case for n = 1.

$(A^n{)}^{-1}=(A^1{)}^{-1}=A^{-1}=(A^{-1}{)}^1=(A^{-1}{)}^n$

Step 2:

Assume that the statement is true for.$n=k:(A^k{)}^{-1}=(A^{-1}{)}^k$

Now, proof of the case: n = k + 1

$\begin{array}{r}{\left({\mathbf{A}}^{k+1}\right)}^{-1}={\left({\mathbf{A}}^k\mathbf{A}\right)}^{-1}\\ ={\mathbf{A}}^{-1}{\left({\mathbf{A}}^k\right)}^{-1}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\\ ={\left({\mathbf{A}}^{-1}\right)}^1{\left({\mathbf{A}}^{-1}\right)}^k\\ ={\left({\mathbf{A}}^{-1}\right)}^{k+1}\end{array}$

Thus, it is proven by the statement of induction.