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I just wonder if there is any closed form solution for the inverse of matrices with following form, or if it's possible to decompose them.

$ \left[\begin{array}{cccccccccc} {\color{red}1} & {\color{red}x} & {\color{red}x} & {\color{red}x} & {\color{green}y} & {\color{green}y} & {\color{green}y} & {\color{green}y} & {\color{green}y} & {\color{green}y}\\ {\color{red}x} & {\color{red}1} & {\color{red}x} & {\color{red}x} & {\color{green}y} & {\color{green}y} & {\color{green}y} & {\color{green}y} & {\color{green}y} & {\color{green}y}\\ {\color{red}x} & {\color{red}x} & {\color{red}1} & {\color{red}x} & {\color{green}y} & {\color{green}y} & {\color{green}y} & {\color{green}y} & {\color{green}y} & {\color{green}y}\\ {\color{red}x} & {\color{red}x} & {\color{red}x} & {\color{red}1} & {\color{green}y} & {\color{green}y} & {\color{green}{\color{green}y}} & {\color{green}y} & {\color{green}y} & {\color{green}y}\\ {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}1} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z}\\ {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}1} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z}\\ {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}1} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z}\\ {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}1} & {\color{blue}z} & {\color{blue}z}\\ {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}1} & {\color{blue}z}\\ {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}z} & {\color{blue}1} \end{array}\right] $

The diagonal of the matrix is all 1, and matrix is consisted of three parts where the off diagonal entries of the matrix in each part are equal.

Any comment or reference is appreciated.

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  • $\begingroup$ Gauss and Jordan may help you. $\endgroup$ – Lord Soth Apr 13 '13 at 21:27
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Call your big matrix $M$. Let $P$ be a $4\times4$ real orthogonal matrix whose first column is equal to $u=\frac12(1,1,1,1)^T$ and $Q$ be a $6\times6$ real orthogonal matrix whose first column is equal to $v=\frac1{\sqrt{6}}(1,1,1,1,1,1)^T$. Then \begin{align*} &\begin{bmatrix}P^T\\ &Q^T\end{bmatrix} M\begin{bmatrix}P\\ &Q\end{bmatrix}\\ =& \begin{bmatrix} 1+3x&0&0&0&y\sqrt{24}&0&0&0&0&0\\ 0&1-x&0&0&0&0&0&0&0&0\\ 0&0&1-x&0&0&0&0&0&0&0\\ 0&0&0&1-x&0&0&0&0&0&0\\ z\sqrt{24}&0&0&0&1+5z&0&0&0&0&0\\ 0&0&0&0&0&1-z&0&0&0&0\\ 0&0&0&0&0&0&1-z&0&0&0\\ 0&0&0&0&0&0&0&1-z&0&0\\ 0&0&0&0&0&0&0&0&1-z&0\\ 0&0&0&0&0&0&0&0&0&1-z \end{bmatrix} \end{align*} and hence \begin{align*} &\begin{bmatrix}P^T\\ &Q^T\end{bmatrix} M^{-1}\begin{bmatrix}P\\ &Q\end{bmatrix}\\ =& \begin{bmatrix} \tfrac{1+5z}{D}&0&0&0&\tfrac{-y\sqrt{24}}{D}&0&0&0&0&0\\ 0&\tfrac1{1-x}&0&0&0&0&0&0&0&0\\ 0&0&\tfrac1{1-x}&0&0&0&0&0&0&0\\ 0&0&0&\tfrac1{1-x}&0&0&0&0&0&0\\ \tfrac{-z\sqrt{24}}{D}&0&0&0&\tfrac{1+3x}{D}&0&0&0&0&0\\ 0&0&0&0&0&\tfrac1{1-z}&0&0&0&0\\ 0&0&0&0&0&0&\tfrac1{1-z}&0&0&0\\ 0&0&0&0&0&0&0&\tfrac1{1-z}&0&0\\ 0&0&0&0&0&0&0&0&\tfrac1{1-z}&0\\ 0&0&0&0&0&0&0&0&0&\tfrac1{1-z} \end{bmatrix} \end{align*} where $D=(1+3x)(1+5z)-24yz$. This has the same matrix structure as the first displayed equation, except that the entries are different. That is, the diagonal blocks are of the form $aI+bE_{11}$ ($E_{11}$ denotes a matrix of appropriate size, with a $1$ at the top left hand corner and zeroes elsewhere) and its antidiagonal blocks are of the form $cE_{11}$. Therefore $M^{-1}$ should have the same structure as $M$, i.e. its diagonal blocks are of the form $aI+bJ$ ($J$ denotes a matrix of all ones) and its antidiagonal blocks are of the form $cJ$. In fact, by multiplying $P\oplus Q$ on the left and $P^T\oplus Q^T$ on the right and by using the property that $Pe_1=u$ and $Qe_1=v$, we get $$ M^{-1}=\begin{bmatrix} \frac{I_4}{1-x} - \frac{(1+5z)x - 6yz}{(1-x)D}J_{4,4} &-\frac{(1+5z)y - 6yz}{(1-z)D}J_{4,6}\\ -\frac{(1+3x)z - 4zx}{(1-x)D} J_{6,4} &\frac{I_6}{1-z} - \frac{(1+3x)z - 4zy}{(1-z)D} J_{6,6} \end{bmatrix} $$ where $J_{m,n}$ denotes the $m\times n$ matrix with all entries equal to $1$.

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  • $\begingroup$ Thanks for your answer! So for general case, where first $a$ rows are $x$ and $y$ and next $b$ rows are $z$'s (so matrix has $a+b$ rows), then the first column of matrix $P$ should be $u=0.5(1,1,1...,1)$ and first column of matrix $Q$ should be $v=\frac{1}{\sqrt{b}}(1,1,1,1,1,1)$ $\endgroup$ – user54626 Apr 14 '13 at 4:13
  • $\begingroup$ @user54626 Yes (except that the factor in $v$ is $1/\sqrt{6}$, not $1/\sqrt{\color{red}{b}}$, but I take that as a typo of yours). $\endgroup$ – user1551 Apr 14 '13 at 4:20

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