Diagonalise the following matrix: $$A = \begin{pmatrix}-1&-2&2\\ \:4&3&-4\\ \:0&-2&1\end{pmatrix}$$
I know I have to represent the matrix $A$ as $A = P^{-1}SP$, where $P$ is a matrix with only the eigenvalues of $A$ in the diagonal, and $P$ is the corresponding eigenvector matrix. I don't have any trouble with find the eigenvectors and eigenvalues; $$S=\begin{pmatrix}3&0&0\\ \:0&-1&0\\ \:0&0&1\end{pmatrix}, P = \begin{pmatrix}1&0&1\\ \:-1&1&0\\ \:1&1&1\end{pmatrix}$$
However, I am having a tough time trying to find the inverse of $P$ in a quick way. The issue is that I already spent a long time trying to find $S$ and $P$ and if I try to find $P^{-1}$, using either this method, or this method, takes quite some time. Therefore, I'm hoping that there is a quick way to determine $P^{-1}$.
Essentially - is there another/quicker method to determine $P^{-1}$?
The fastest, reliable method for determining $P^{-1}$ is to use Gaussian elimination (the second method). I strongly recommend against using the first method.
In some cases, including this one, it may be easier to use this easier "trick" method (but please note it is not very reliable in the general case!). If the inverse is $P^{-1} = \begin{bmatrix} \vec{v}_1 & \vec{v}_2 & \vec{v}_3 \end{bmatrix}$, then the $\vec{v}_i$ satsify $P \vec{v}_i = \hat{e}_i$, so spotting a linear combination of the columns of $P$ that gives every basis vector will find the inverse. For example, we can observe without too much trouble that $$\hat{e}_1 = -\vec{u}_1 - \vec{u}_2 + 2 \vec{u}_3, ~~\hat{e}_2 = - \vec{u}_1 + \vec{u}_3, ~~\hat{e}_3 = \vec{u}_1 + \vec{u}_2 - \vec{u}_3 $$ where $\vec{u}_i$ is the $i$th column of $P$. The linear combination yielding the $i$th basis vector should be placed in the $i$th column of $P^{-1}$, i.e. $$P^{-1} = \begin{bmatrix} -1 & -1 & 1 \\ -1 & 0 & 1 \\ 2 & 1 & -1\end{bmatrix}$$ Note that here it was easier to spot these linear combinations as the matrix contained some zeroes.