Krdytkyu

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?

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

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

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}
Pbegin{bmatrix} 1^n & 0 \ 0 & 3^n end{bmatrix}P^{-1}
&=Pbegin{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 \ -2cdot3^n & 3^n end{bmatrix}
\[6px]
&=begin{bmatrix} 3-2cdot3^n & 3^n-1 \ 6(1-3^{n}) & 3^{n+1}-2 end{bmatrix}
end{align}