Inverse of a Matrix: Find Solution for A

[Buffu]

Homework Statement

Find the inverse of
$A = \begin{bmatrix} 1 & \dfrac12 & & \cdots && \dfrac1n \\\dfrac12 & \dfrac13 && \cdots && \dfrac1{n+1} \\ \vdots & \vdots && && \vdots \\ \dfrac1n & \dfrac1{n+1} && \cdots && \dfrac1{2n-1}\end{bmatrix}$

Homework Equations

The Attempt at a Solution

I obvserved that $A_{ij} = \dfrac{1}{i+j-1}$.

Also I know $I = AA^{-1}$

So jth column of $I$ is $A$ times jth column of $A^{-1}$

So for $j = 1$

$A \times \begin{bmatrix}A^{-1}_{11} \\ \vdots \\ A^{-1}_{n1}\end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ \vdots\\0 \end{bmatrix}$.

Now I don't know what to do. Any clue.

 

[jedishrfu]

This might help:

 

[Buffu]

This might help:

I know how to take inverse just I don't know how to do that in case of matrix like this.

 

[jedishrfu]

I don't see where you're confused. The procedure is the same. Is it that the answer is just wrong?

 

[Buffu]

I don't see where you're confused. The procedure is the same. Is it that the answer is just wrong?

Ok, I tried something,

I did $A_i \to A_i - \dfrac1i A_1$, where $A_1, A_i$ are the rows.

I got,
$\begin{bmatrix} 1 & \dfrac12 & & \cdots && \dfrac1n \\ 0 & \dfrac13 - \dfrac12 && \cdots && \dfrac1{n+1} - \dfrac1n \\ \vdots & \vdots && && \vdots \\ 0 & \dfrac1{n+1} - \dfrac1n && \cdots && \dfrac1{2n-1} - \dfrac1n^2\end{bmatrix} = \begin{bmatrix} 1 & \dfrac12 & & \cdots && \dfrac1n \\ 0 & \dfrac1{12} && \cdots && \dfrac1{n+1} - \dfrac1n \\ \vdots & \vdots && && \vdots\\ 0 & \dfrac{i -1}{2i(i + 1)} && && \vdots\\ \vdots & \vdots &&&& \vdots \\ \\ 0 & \dfrac{n-1}{2n(n+1)} && \cdots && \dfrac1{2n-1} - \dfrac1n^2\end{bmatrix}$

See it is very messy and I don't know what to do now.

 

[Ray Vickson]

Homework Statement

Find the inverse of
$A = \begin{bmatrix} 1 & \dfrac12 & & \cdots && \dfrac1n \\\dfrac12 & \dfrac13 && \cdots && \dfrac1{n+1} \\ \vdots & \vdots && && \vdots \\ \dfrac1n & \dfrac1{n+1} && \cdots && \dfrac1{2n-1}\end{bmatrix}$

Homework Equations

The Attempt at a Solution

I obvserved that $A_{ij} = \dfrac{1}{i+j-1}$.

Google "inverse of special matrix".

 

[Buffu]

Google "inverse of special matrix".

Can you provide the link to the site, I searched the first page of Google but nothing matches.

 

[Ray Vickson]

Can you provide the link to the site, I searched the first page of Google but nothing matches.

Have you looked at all the other articles? I found several, just by searching as I suggested to you.

 

[Buffu]

Have you looked at all the other articles? I found several, just by searching as I suggested to you.

Yes I have looked at each and every article on the first page.

 

[StoneTemplePython]

What's the purpose of this exercise? This is clearly a very special instance of a Hankel matrix which has its own name...

 

[Ray Vickson]

Yes I have looked at each and every article on the first page.

The Wikipedia article has all you need. Look at the entry for "Cauchy Matrix".

 

[WWGD]

Or, just work a few small cases, like $2 \times 2 , 3\times 3$ and come up with an educated guess.