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Could you help me please and give some tips on how should I prove it (show it).

Prove that Sol(A,b) of quadratic system of linear equations with n×n-Matrix:

$A =\begin{pmatrix} 1 & -1 & 0 & 0 & .. & 0& 0\\ 1 & 1 & -1 & 0& .. & 0 & 0\\ 0 & 1 & 1 & -1 & ..& 0& 0\\ .. & .. & .. & .. & .. & .. &..\\ 0 & 0 & 0 & 0 & .. & 1 & -1\\ 0 & 0& 0& 0& .. & 1 & 1 \end{pmatrix}$

for $n\geq 1$ and all b from $\mathbb{R}^{n}$ has only ONE solution.

Thank you in advance.

  • 2
    Well the row vectors of $A$ are clearly linearly independent. Hence $A$ is invertible and so $Ax = b$ has only one solution.2011-10-20
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    What is a *quadratic system* of linear equations? Does that just mean linear equations with same number of unknowns as equations?2011-10-21

1 Answers 1

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For the system $\mathbf A\mathbf x=\mathbf b$ to have a unique solution, where $\mathbf A$ is the tridiagonal matrix you have, one way of proceeding is to show that $\mathbf A$ is nonsingular. Letting $a_n=\det\mathbf A_n$, where $n$ is the size of $\mathbf A$, you can establish the recursion relation

$$a_n=a_{n-1}+a_{n-2}$$

with initial conditions $a_1=|1|=1$ and $a_2=\begin{vmatrix}1&-1\\1&1\end{vmatrix}=2$, through cofactor expansion on the last column and last row of the matrix. You should be able to recognize a certain famous sequence...

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    thank you very much for your answer. May I ask you: is there any other way, how I can prove it, without determinant?2011-10-25
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    @Lissa: PZZ's comment to your question hints to another method.2011-10-25