Let $A$ be a square matrix of order $n$ where the elements are defined as $a(i,i)=2 ,a(i+1,i)=a(i,i+1)=-1$ for all $i=1 \ldots n-1$.Also a(i,j)=0 otherwise.
Find the inverse of the matrix.
Is there any elegant way to find the inverse completely?
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Please make sure I edited your question properly. – Amzoti Aug 29 '13 at 04:18
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Do you intend the other elements to be all $0$? – Robert Israel Aug 29 '13 at 05:06
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Also since there's no index $n+1$, you want $a(i+1,i) = a(i,i+1) = -1$ for $i = 1 \ldots n-1$, not $1 \ldots n$. – Robert Israel Aug 29 '13 at 05:08
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I have just edited it.. – sayak Aug 29 '13 at 05:48
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For your particular $A$, I think the easiest way to find its inverse is by simple inspection. Computer experiments suggest that the $(i,j)$-th entry of $A^{-1}$ should be $\frac{1}{n+1}(n+1-\max(i,j))\min(i,j)$, and you can check whether this is indeed correct.
More generally, as the other answer points out, your $A$ is a special case of a symmetric tridiagonal matrix and explicit formula for its inverse is known. Actually, your $A$ is nicer -- it is a symmetric tridiagonal Toeplitz matrix. Inverses of matrices of this class can be expressed in terms of Chebyshev polynomials of the second kind. See, e.g. corollary 4.2 of
C.M. da Fonseca and J. Petronilho, Explicit inverses of some tridiagonal matrices, Linear Algebra and its Applications, 325(2001): 7-21.
user1551
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See:
G.Y.Hu and R.F.O’Connell, Analytical inversion of symmetric tridiagonal matrices. J. Phys. A: Math. Gen. {\bf 29}(1996), pp.1511–1513.
Boris Novikov
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