Question:
We have a matrix defines as follow: $\underline{\underline{A_n}} = \begin{pmatrix}
1 & -1 & 0 & ... &0 \\
-1 & 2 & -1 & ...&0\\
0 & -1 & 2 & ... &-1\\
0 & 0 & &... -1& 2 &
\end{pmatrix}$ That means that all the elements (when $i=j$)on the diagonal are equal to $2$ and one element up to it and on his left is equal to $-1$. Except the element in the first row and the first column that is equal to $1$.
Prove that $ \forall n \geq 2 , det(\underline{\underline{A_n}})=1$ after find the $LU$ decomposition of $\underline{\underline{A_n}}$
My answer:
1-
By recursion.
For $n=2$: so $det\begin{pmatrix}
1 &-1 \\
-1 & 2
\end{pmatrix}=1$
For $n$: We have $det(\underline{\underline{A_n}})=1$
For $n+1$: The matrix $\underline{\underline{A_{n+1}}}$ looks like this $\begin{pmatrix}
A_n & & 0 \\
&& -1\\
0 ...& -1 & 2
\end{pmatrix}$ thus WLOG with $n$ even we have $det(A_{n+1})=2det(A_n)-(-1)det(A')=2det(A_n)+det(A')$ with $A'$ a matrix that looks like that: $\begin{pmatrix}
A_{n-1} & 0 \\
&& \\
0 ... -1 & -1
\end{pmatrix}$.
So $det(A')=(-1)det(A_{n-1})+det(A'')$, with $A''$ a matrix that looks like this $A''= \begin{pmatrix}
1 & -1 & 0 & ... &0 \\
-1 & 2 & -1 & ...&0\\
0 & -1 & 2 & ... &0\\
0 & 0 & &... -1& -1 &
\end{pmatrix} = \begin{pmatrix}
A_{n-2} & 0 \\
&& \\
0 ... -1 & -1
\end{pmatrix}$
Thus $det(A'')=det(A_{n-2})-1det(A''')$ and $A'''$ itself can be expressed in the same way as before butwith $A_{n-3}$ at the place of $A_{n-2}$.
By continuing like this we get that $det(A_{n+1})=2det(A_n)-det(A_n)+det(A_{n-1})-det(A_{n-2})+...-det(A_2)=2-1=1$
Rem: if $n$ is even then $n-1$ and $n+1$ are odds.
I would like to know if my demonstration for the first part of the question is correct and mostly to get help in order to answer to the segond part.
Thank you for your help.