Exercise with matrix A) For which $a,b$ is the matrix $A=\begin{bmatrix}
a & 0\\ 
b & b
\end{bmatrix}$ invertible?
B) Calculate $A^{1000}$ where $A$ is the above matrix with $a=1$ and $b=2$.
$$$$
I have done the following:
A) $$\begin{vmatrix}
a & 0\\ 
b & b
\end{vmatrix} \neq 0 \Rightarrow ab \neq 0 \Rightarrow a \neq 0 \text{ and } b \neq 0$$
B) $$A=\begin{bmatrix}
1 & 0\\ 
2 & 2
\end{bmatrix}$$
How can I calculate $A^{1000}$??
Is it maybe  $$A^n=\begin{bmatrix}
1 & 0\\ 
2 \cdot n & 2^n
\end{bmatrix}$$
 A: You may prove by induction that 
$A^n=\begin{bmatrix}
1 & 0\\ 
2^{n+1}-2 & 2^n
\end{bmatrix}$
but that's out of the blue.

Otherwise, notice that $A=I_2+\begin{bmatrix}
0 & 0\\ 
2 & 1
\end{bmatrix}$
Notice that $\forall n, \begin{bmatrix}
0 & 0\\ 
2 & 1
\end{bmatrix}^n=\begin{bmatrix}
0 & 0\\ 
2 & 1
\end{bmatrix}$
And apply Newton binomial theorem (you're allowed to since $I_2$ commutes with any matrix).
$A^{1000}=I_2+ \begin{bmatrix}
0 & 0\\ 
2 & 1
\end{bmatrix}\sum_{k=1}^{1000} \binom{1000}{k}$
A: A) looks good
B) Your approach to find a formula for $A^n$ is good. Try to prove by induction that$$A^n=\begin{bmatrix}
1 & 0\\ 
\sum_{i=1}^{n}2^i & 2^n
\end{bmatrix}$$
A: Your matrix is diagonalizable, because u have 2 different eigenvalues $\lambda_1=1$ and $\lambda_2=2$. Then you find an invertible matrix $S$, such that $SAS^{-1}=D$ with a diagonal matrix D, which contains your eingenvalues:
Now we obtain the following:
$A^{1000}=(S^{-1}DS)^{1000}=S^{-1}DS*S^{-1}DS*....*S^{-1}DS=S^{-1}D^{1000}S$ (Because $S^{-1}S=I$).
But $D^{1000}=\begin{pmatrix} 1 & 0  \\ 0 & 2^{1000}  \\  \end{pmatrix}$. Then it follows that : $A^{1000}=S^{-1}\begin{pmatrix} 1 & 0  \\ 0 & 2^{1000}  \\  \end{pmatrix} S$ 
