Q:

Finding $n$-th power of a $2 \times 2$ matrix with $2$ identical eigenvalues

If$$  A =
    \begin{pmatrix}
    3 & -4 \\
    1 & -1 \\
    \end{pmatrix}
$$
prove that 
$$A^k = \begin{pmatrix}
    1+2k & -4k \\
    k & 1-2k \\
    \end{pmatrix}$$
Now the first method I used was using assuming an equation $$x^k = f(x)(Ax^2 + Bx +C) + (px + q)........(i)$$
where $ Ax^2 + Bx + C = 0 $ is characteristic equation and $x$ is eigen value.
After solving the characteristic equation I would get two eigen value $x_1$ & $ x_2$
Putting those two eigen values in $(i)$ will give me two equations and I will be able to find $p$ and $q$.
Replacing the value of $p$ and $q$ in $(i)$ and replacing $x$ with $A$ (Cayley Hamilton Theorem) I should be able to get the value of $A^k$
but the problem is that $A$ has 2 identical eigen value ($i.e. 1$) so $p$ and $q$ will have infinite solutions. Thus I cannot apply this method.
I found this other method of finding $n^th$ power of matrices online which uses diagonalization of matrices
but as $A$ has only 1 eigen value therefore it also has only 1 eigen vector. Thus $A$ cannot be diagonalized.
I can just simply multiply $A$ 2 or 3 times and derive the general formula but I am looking for another way.
Thank you for taking your time and reading this far.

A:

Let's prove this by induction. If $k=1$, the result is clear. Now,
if
$$
A^k = \begin{pmatrix}
    1+2k & -4k \\
    k & 1-2k \\
    \end{pmatrix},
$$therefore
$$
A^{k+1}  =A \times A^k =
    \begin{pmatrix}
    3 & -4 \\
    1 & -1 \\
    \end{pmatrix} \times \begin{pmatrix}
    1+2k & -4k \\
    k & 1-2k \\
    \end{pmatrix}
=\begin{pmatrix}
    1+2(k+1) & -4(k+1) \\
    k+1 & 1-2(k+1) \\
    \end{pmatrix}
$$
after simplifying.

