Solving a System of Equations with Matrices

elle
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Hi, I've just been given a set of revision questions for matrices and I'm having problems with the following question:

http://tinypic.com/ejbssz.jpg"

My task is to find the general solution of this system (if there is any). Do I start off by reducing it to row echelon form and then reduced row echelon?

My first step would be to divide the first line by 3 to get a 1 on the far left...but that would mean I would have fractions as well which makes everything so 'messy.' Is that the correct first step to this question? :confused:

Help and advice appreciated! Thank you!
 
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I would start by multiplying the 2nd line by -1 and adding it to the 4th line.
 
Ok I've given it a go and I got the answer:

1 0 1 2 | 1
0 1 -1 -2 | 0
0 0 0 0 | 0
0 0 0 0 | 0

So that gives me no solution?...:confused: Can someone check if that's right. Thank you! :biggrin:
 
Looks correct to me. The matrix is singular, so there isn't a solution.
 
Oh thanks! :biggrin: I know how to attempt other similar questions in my notes now.

I've also got two more questions to ask, if that's ok :rolleyes:

1) How do you determine a matrix is an elementary matrix? For example if you were given:

1 0 0
0 1 9
0 0 1

Is this an elementary matrix? :confused:

2) I don't understand the wording of the following question:

http://tinypic.com/ejek4w.jpg"

Does it mean its already in the echelon form so I don't have to do anything apart from just solving it for the variables? Or do I still have to put it into reduced echelon form?

Sorry for being such a pain and thank you for taking time to read my post :smile:
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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