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I have the following positive definite matrix given by

\begin{equation*} \begin{pmatrix} a&b&c&0&0&0&0&0&0&0 \\ b&a&b&c&0&0&0&0&0&0 \\ c&b&a&b&c&0&0&0&0&0 \\ 0&c&b&a&b&c&0&0&0&0 \\ 0&0&c&b&a&b&c&0&0&0 \\ 0&0&0&c&b&a&b&c&0&0 \\ 0&0&0&0&c&b&a&b&c&0 \\ 0&0&0&0&0&c&b&a&b&c \\ 0&0&0&0&0&0&c&b&a&b \\ 0&0&0&0&0&0&0&c&b&a \end{pmatrix}_{10\times 10} \end{equation*} $a>0; \quad b\,\,and \,\, c \in \mathbb{R},$

You can think of the above matrix as a variance-covariance matrix, so the diagonal elements are variances and the off-diagonal elements are covariances

what I am interested in is finding the inverse for such a matrix with dimensions of $n \times n$. I tried to figure out the pattern by computing the inverse of smaller dimensions matrices with the same structure, yet I could not recognize a pattern.

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there is a pattern in these, I picked integer values

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 3 } & 1 & 0 & 0 \\ \frac{ 1 }{ 3 } & \frac{ 1 }{ 4 } & 1 & 0 \\ 0 & \frac{ 3 }{ 8 } & \frac{ 3 }{ 10 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 3 & 0 & 0 & 0 \\ 0 & \frac{ 8 }{ 3 } & 0 & 0 \\ 0 & 0 & \frac{ 5 }{ 2 } & 0 \\ 0 & 0 & 0 & \frac{ 12 }{ 5 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } & 0 \\ 0 & 1 & \frac{ 1 }{ 4 } & \frac{ 3 }{ 8 } \\ 0 & 0 & 1 & \frac{ 3 }{ 10 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 3 & 1 & 1 & 0 \\ 1 & 3 & 1 & 1 \\ 1 & 1 & 3 & 1 \\ 0 & 1 & 1 & 3 \\ \end{array} \right) $$

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrrr} 1 & 0 & 0 & 0 & 0 \\ \frac{ 1 }{ 3 } & 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 3 } & \frac{ 1 }{ 4 } & 1 & 0 & 0 \\ 0 & \frac{ 3 }{ 8 } & \frac{ 3 }{ 10 } & 1 & 0 \\ 0 & 0 & \frac{ 2 }{ 5 } & \frac{ 7 }{ 24 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrrr} 3 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 8 }{ 3 } & 0 & 0 & 0 \\ 0 & 0 & \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & 0 & \frac{ 12 }{ 5 } & 0 \\ 0 & 0 & 0 & 0 & \frac{ 115 }{ 48 } \\ \end{array} \right) \left( \begin{array}{rrrrr} 1 & \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } & 0 & 0 \\ 0 & 1 & \frac{ 1 }{ 4 } & \frac{ 3 }{ 8 } & 0 \\ 0 & 0 & 1 & \frac{ 3 }{ 10 } & \frac{ 2 }{ 5 } \\ 0 & 0 & 0 & 1 & \frac{ 7 }{ 24 } \\ 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrr} 3 & 1 & 1 & 0 & 0 \\ 1 & 3 & 1 & 1 & 0 \\ 1 & 1 & 3 & 1 & 1 \\ 0 & 1 & 1 & 3 & 1 \\ 0 & 0 & 1 & 1 & 3 \\ \end{array} \right) $$

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrrrr} 1 & 0 & 0 & 0 & 0 & 0 \\ \frac{ 1 }{ 3 } & 1 & 0 & 0 & 0 & 0 \\ \frac{ 1 }{ 3 } & \frac{ 1 }{ 4 } & 1 & 0 & 0 & 0 \\ 0 & \frac{ 3 }{ 8 } & \frac{ 3 }{ 10 } & 1 & 0 & 0 \\ 0 & 0 & \frac{ 2 }{ 5 } & \frac{ 7 }{ 24 } & 1 & 0 \\ 0 & 0 & 0 & \frac{ 5 }{ 12 } & \frac{ 34 }{ 115 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrrrr} 3 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{ 8 }{ 3 } & 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{ 5 }{ 2 } & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{ 12 }{ 5 } & 0 & 0 \\ 0 & 0 & 0 & 0 & \frac{ 115 }{ 48 } & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{ 273 }{ 115 } \\ \end{array} \right) \left( \begin{array}{rrrrrr} 1 & \frac{ 1 }{ 3 } & \frac{ 1 }{ 3 } & 0 & 0 & 0 \\ 0 & 1 & \frac{ 1 }{ 4 } & \frac{ 3 }{ 8 } & 0 & 0 \\ 0 & 0 & 1 & \frac{ 3 }{ 10 } & \frac{ 2 }{ 5 } & 0 \\ 0 & 0 & 0 & 1 & \frac{ 7 }{ 24 } & \frac{ 5 }{ 12 } \\ 0 & 0 & 0 & 0 & 1 & \frac{ 34 }{ 115 } \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrrrr} 3 & 1 & 1 & 0 & 0 & 0 \\ 1 & 3 & 1 & 1 & 0 & 0 \\ 1 & 1 & 3 & 1 & 1 & 0 \\ 0 & 1 & 1 & 3 & 1 & 1 \\ 0 & 0 & 1 & 1 & 3 & 1 \\ 0 & 0 & 0 & 1 & 1 & 3 \\ \end{array} \right) $$

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