How to find Inverse-squared of a matrix

Question

Given the matrix $$ A = \begin{bmatrix} 1 & 2 & 0 \\ 2 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$

Find $A^{-2}$

I have solved it using Cayley Hamilton theorem, but it is a long procedure. Please tell me is there another approach to solve it in a shorter way.

Answer 1

After manually calculating the square: $$ A^2=\begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ Which is a diagonal matrix. Diagonal matrices have the very nice property that the inverse of $D=\operatorname{diag}(\lambda_1, \dots, \lambda_n)$ is just $D^{-1}=\operatorname{diag}\left(\lambda_1^{-1}, \dots, \lambda_n^{-1}\right)$ (provided $\lambda_i\ne 0$ for $1\le i\le n$), and hence $(A^2)^{-1}=\operatorname{diag}(1/5, 1/5, 1)$.

Answer 2

First do $A^2$ . You will get

Given the matrix [ $A^2$ = \begin{bmatrix}5& 0&0\\0&5&0\\0&0&1\end{bmatrix} ] Now this is a simple matrix, so calculating its inverse is easy

You will get [ $A^{-2}$ = \begin{bmatrix}1/5& 0&0\\0&1/5&0\\0&0&1\end{bmatrix} ]