Advice on proof in linear algebra. I just wrote my first proof in linear algebra so I'd love some advice on the things that go well and what could be improved upon. It's a proof by induction.
Theorem:
Let $A_n$ be a $n\times n$ matrix of the form:
$\begin{pmatrix}
2 & 1 & 0 & 0 && & \cdots & 0\\
1 & 2 & 1 & 0 && & \cdots & 0\\
0 & 1 & 2 & 1 && & \cdots & 0\\
0 & 0 & 1 & 2 && & \cdots & 0\\
\vdots& & & & \ddots & && \vdots\\
0&\cdots&&&&2&1&0\\
0&\cdots&&&&1&2&1\\
0&\cdots&&&&0&1&2\\
\end{pmatrix}$
Then $det(A_n)=3(n-1)$.
Proof:
We'll give a proof by mathematical induction.
Let $n=2$:
$\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}=4-1=3$.
Let $n=3$:
$\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}-\begin{vmatrix}
1&0\\
1&2\\
\end{vmatrix}=8-2=6$
Let $n=4$:
$\begin{vmatrix}
2&1&0&0\\
1&2&1&0\\
0&1&2&1\\
0&0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}-\begin{vmatrix}
1&0&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}-\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}=12-3=9$
Notice how due to the repetitive nature of our matrix, we'll candevise a recursive formula for our determinant: $|A_n|=2|A_{n-1}|-|A_{n-2}|$.
If we assume: $|A_n|=3(n-1)$, $|A_{n-1}|=3(n-2)$, $|A_{n-2}|=3(n-3)$, then:
$|A_n|=6(n-2)-3(n-3)=3(n-1)$
Since we checked $n=2, 3$; we conclude that $|A_n|=3(n-1)$ for all $n \in \mathbb{N} | n>1$
 A: You could remark some of the easy steps you did, making the proof more understandable yet (even though they are very simple). For example, you could write $3=3(2-1)$ (same with the other 2) to show that the formula holds for the smaller values of $n$. 
Moreover, and more important, you should explain how you used Laplace's formula to derive the recurrent relationship (which involves another induction proof). It is not enough to show that the relationship is true for some values of $n$, but, of course, it will help you to generalize the expression.
A: The general formula you gave is false. First notice the determinant you gave for $n=3$ is wrong, the first minor you computed has value $2\cdot 2- 1=3$, so it is in fact $6-2=4$. This shows the importance of checking the initial values correctly when using induction, you could be led to different answers.
A thing that could help you is use the structure of the matrix to find a relationship between $|A_n|$ and determinants of minors to use induction in a right way. Here is a hint
$$|A_n|=2|A_{n-1}|-|A_{n-2}|.$$
If you notice $|A_1|=2$ you should get in the end $|A_n|=n+1$. Check the first two values $n=1$ $n=2$, and use induction as usual.
A: Let $d_n = \det A_n$. Note that $d_1 = 2, d_2 = 3$.
For the induction step, look at the (clunky) picture below:

Note that $d_n = 2 d_{n-1} -d_{n-2}$.
Now show that $n \mapsto 3(n-1)$ satisfies the equation with the same
initial conditions.
