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**Shelled****Member**- Registered: 2014-04-15
- Posts: 44

Let A be a n*n matrix and

Hi, I'm having trouble with the above question. I've made a start but I'm not sure if I'm approaching it correctly.

First of, there's a hint that suggests that I first consider the product AX. But if I do that, the dimensions of A and X are different so I would need to find the transpose of X?

so is it safe to assume

To find if A is invertible can I also do the following?

AX=XB (where B is some solution of A)

and for the formula:

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

First and second line.

Shelled wrote:

(where B is some solution of A)

A is an n x n matrix and B is n x 1 column vector... so how can they be equal?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,332

AX = X diag( 入i )

V := X^(-1)

AX diag( 1/ 入i )V = X diag( 入i )diag( 1/ 入i ) V = I

thus

A^(-1) = X diag( 1/ 入i ) X^(-1)

**X'(y-Xβ)=0**

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