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Ok so I need to find the inverse of matrix $A=\begin{pmatrix} 1& 0 &0 &... & 0&0\\ 2& 1 & 0 & ... &0&0 \\ 0& 2 & 1 & ... & 0&0\\ ...& ... & ... & ... & ...&...\\ 0& 0 & 0 & ... & 1&0\\ 0&0&0&...&2&1 \end{pmatrix}$. I tried to do it on my own and here is the solution. Can you check if it is right?

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    $\begingroup$ You can verify your solution by multiplying it with the original matrix $A$ to see if you get identity matrix. $\endgroup$ – Guangliang Feb 13 '17 at 15:45
  • $\begingroup$ I think it is good. I multiplied them and I got indentity matrix. $\endgroup$ – Ghost Feb 13 '17 at 16:02
  • $\begingroup$ If you got the identity matrix then its correct, be more confident in yourself! $\endgroup$ – petru Feb 13 '17 at 16:13
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Your solution seems correct to me. The standard way of solving this problem is to write the extended matrix $$[A|I]$$ and then use Gauss elimination to get a matrix which is in echelon form. Since your matrix is already in this form (if transposed) you can simply use forward-elimination to get to the form $$[I|A^{-1}].$$ Since every row $j$ of the solution matrix is multiplied by 2 and then subtracted from the subsequent row $j+1$, the coefficients of the inverse are given by $$a^{-1}_{ij}=\begin{cases} 0 & j>i\\ (-2)^{i-j} & j\leq i \end{cases}.$$

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  • $\begingroup$ Ok thanks for the answer. $\endgroup$ – Ghost Feb 13 '17 at 16:14

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