$A^n$ formula not working; checked arithmetic

Question

I'm trying to find the general formula for a matrix raised to a certain power using diagonalization. My original matrix $A$ is: $$ \begin{bmatrix} -3 & 2 \\ -12 & 7 \end{bmatrix}. $$ My eigenvector matrix $P$ is: $$ \begin{bmatrix} 1/2 & 1/3 \\ 1 & 1 \end{bmatrix}. $$ I found the diagonal matrix $D$ to be: $$ \begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix}. $$ And $P^{-1}$ is: $$ \begin{bmatrix} 6 & -2 \\ -6 & 3 \end{bmatrix}. $$ The final formula I got from multiplying $P D^n P^{-1}$ is $$ \begin{bmatrix} 3- 6^n & -1+3^n \\ 6-18^n & -2+9^n \end{bmatrix}. $$ It checks out in the case of $A^1$ but not $A^0$. Where did I go wrong?

Answer 1

Your calculations for $PD^nP^{-1}$ are wrong. Check them.

Remember that, for instance, $2\cdot 3^n \neq 6^n$.

Answer 2

You can simplify the computation by choosing the matrix $$ P=\begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix} $$ so that $$ P^{-1}=\begin{bmatrix} 3 & -1 \\ -2 & 1 \end{bmatrix} $$ Therefore, with more accurate computations, \begin{align} P\begin{bmatrix} 1^n & 0 \\ 0 & 3^n \end{bmatrix}P^{-1} &=P\begin{bmatrix} 1 & 0 \\ 0 & 3^n \end{bmatrix} \begin{bmatrix} 3 & -1 \\ -2 & 1 \end{bmatrix} \\[6px] &=\begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 3 &-1 \\ -2\cdot3^n & 3^n \end{bmatrix} \\[6px] &=\begin{bmatrix} 3-2\cdot3^n & 3^n-1 \\ 6(1-3^{n}) & 3^{n+1}-2 \end{bmatrix} \end{align}