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Suppose we have n particles connected by a string tied to fixed ends.
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In matrix notation the equation of motion can be written as
$\begin{pmatrix}2\omega_o^2&-\omega_o^2&0&0&.&.&.&0\\ -\omega_o^2&2\omega_o^2&-\omega_o^2&0&.&.&.&0\\ 0&-\omega_o^2&2\omega_o^2&-\omega_o^2&0&0&...&0\\.\\.\\.\\ 0&0&0&.&.&-\omega_o^2&2\omega_o^2&-\omega_o^2\\ 0&0&0&.&.&.&-\omega_o^2&2\omega_o^2&\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\\.\\.\\.\\x_{N-1}\\x_N\end{pmatrix}=\begin{pmatrix}\ddot x_1\\\ddot x_2\\\ddot x_3\\.\\.\\.\\\ddot x_{N-1}\\\ddot x_N\end{pmatrix}\tag{1}$
So, to solve it we have to diagonalize the matrix.

If my matrix is this:
$A=\begin{pmatrix}2\omega_o^2&-\omega_o^2&0&0&.&.&.&-\omega_o^2\\ -\omega_o^2&2\omega_o^2&-\omega_o^2&0&.&.&.&0\\ 0&-\omega_o^2&2\omega_o^2&-\omega_o^2&0&0&...&0\\.\\.\\.\\ 0&0&0&.&.&-\omega_o^2&2\omega_o^2&-\omega_o^2\\ -\omega_o^2&0&0&.&.&.&-\omega_o^2&2\omega_o^2&\end{pmatrix}$
This I have studied how to diagonalize. We can see that the matrix $A$ shares the same eigenvectors with the upper shift matrix (as their commutator is 0)which is given as:
$U=\begin{pmatrix}0&1&0&0&.&.&.&0\\ 0&0&1&0&.&.&.&0\\ 0&0&0&1&0&0&...&0\\.\\.\\.\\ 0&0&0&.&.&0&0&1\\ 1&0&0&.&.&.&0&0&\end{pmatrix}$
The eigenvalues of $U$ is given as $\omega^k$ where $\omega=e^{i\frac{2\pi}{N}}$ and $0\leq k< N$
The eigen vectors are given as $\begin{pmatrix}1\\ \omega^k\\\omega^{2k}\\...\\\omega^{(N-1)k}\end{pmatrix}$

But the matrix given in $(1)$ is different from $A$, then is there any way that we can diagonalize the $A$?

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You can try to solve the eigenvalue problem directly, the calculations are not too hard. You just solve the second order induction and match the boundary conditions.

A faster method is to revert to the periodic case that you know already. To do this, you double the size of the system with periodic boundary conditions so that you have $x_0,…,x_{2N}$ variables ($2N+1$ total). In this system, you want to identify the original one in $x_1…x_N$. You therefore restrict your attention to antisymmetric configurations (indices considered modulo $2N+1$): $$ x_i=-x_{-i} $$ To build a mode for the original chain, you’ll need to superpose two modes of conjugate eigenvalue of the bigger chain. You thus recover directly the modes and spectrum.

Hope this helps.

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The matrix eigenvalue problem leads a recurrence equations with constant coefficients $$ x_{n+1}-2 x_n +x_{n-1}= \Omega^2 x_n, \quad x_{N+1}= x_1 $$ for the periodic case, and $$ x_{n+1}-2 x_n +x_{n-1}= \Omega^2 x_n, \quad x_{0}= x_{N+1}=0 $$ for the fixed end case.

As with all such constant-coefficient recurrence relations, you can try $x_n = e^{ikn}$ and so $\Omega^2$.

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