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Matrix question

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Original post by Music99
I still get 0. Surely it will always be 0 as the constants on the RHS are all 0...


The constants on the RHS being zero, mean that (0,0,0) will be part of the solution. But it's not all of it.

There is an entire line that satisfies that equation.
Reply 41
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Original post by ghostwalker
The constants on the RHS being zero, mean that (0,0,0) will be part of the solution. But it's not all of it.

There is an entire line that satisfies that equation.


I'm not too sure then, as when I do the Gauss Jordan I just get x=0 y=0 z=0 but don't know how to find the line of solutions.
Original post by Music99
I'm not too sure then, as when I do the Gauss Jordan I just get x=0 y=0 z=0 but don't know how to find the line of solutions.


You need to post some working then.
Reply 43
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Original post by ghostwalker
You need to post some working then.


∣−13−23−1313−23−232313−23∣\begin{vmatrix} \frac{-1}{3} & \frac{-2}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{-2}{3} & \frac{-2}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{-2}{3} \end{vmatrix}

∣1211−2−22313−23∣\begin{vmatrix} 1 & 2 & 1 \\ 1 & -2 & -2 \\ \frac{2}{3} & \frac{1}{3} & \frac{-2}{3} \end{vmatrix}

∣1210−4−32313−23∣\begin{vmatrix} 1 & 2 & 1 \\ 0 & -4 & -3 \\ \frac{2}{3} & \frac{1}{3} & \frac{-2}{3} \end{vmatrix}

∣1210−4−3112−1∣\begin{vmatrix} 1 & 2 & 1 \\ 0 & -4 & -3 \\ 1 & \frac{1}{2} & -1 \end{vmatrix}

∣1210−4−30−32−2∣\begin{vmatrix} 1 & 2 & 1 \\ 0 & -4 & -3 \\ 0 & \frac{-3}{2} & -2 \end{vmatrix}

∣10−1202320−32−2∣\begin{vmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 2 & \frac{3}{2} \\ 0 & \frac{-3}{2} & -2 \end{vmatrix}

∣10−1202320143∣\begin{vmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 2 & \frac{3}{2} \\ 0 & 1 & \frac{4}{3} \end{vmatrix}

∣10−1201160143∣\begin{vmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{1}{6} \\ 0 & 1 & \frac{4}{3} \end{vmatrix}

∣10−1201160076∣\begin{vmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{1}{6} \\ 0 & 0 & \frac{7}{6} \end{vmatrix}

∣10−1201160012∣\begin{vmatrix} 1 & 0 & \frac{-1}{2} \\ 0 & 1 & \frac{1}{6} \\ 0 & 0 & \frac{1}{2} \end{vmatrix}

∣10001160012∣\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & \frac{1}{6} \\ 0 & 0 & \frac{1}{2} \end{vmatrix}

∣100011600−16∣\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & \frac{1}{6} \\ 0 & 0 & \frac{-1}{6} \end{vmatrix}

∣10001000−16∣\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \frac{-1}{6} \end{vmatrix}

∣100010001∣\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}

Hopefully I latex-ed it correctly!
(edited 11 years ago)
Original post by Music99

∣−13−23−1313−23−232313−23∣\begin{vmatrix} \frac{-1}{3} & \frac{-2}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{-2}{3} & \frac{-2}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{-2}{3} \end{vmatrix}



Your initial matrix appears to be incorrect:

A=∣23−23−131323−23231323∣A=\begin{vmatrix} \frac{2}{3} & \frac{-2}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{-2}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \end{vmatrix}

and

A−I=∣−13−23−1313−13−232313−13∣A-I=\begin{vmatrix} \frac{-1}{3} & \frac{-2}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{-1}{3} & \frac{-2}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{-1}{3} \end{vmatrix}

and after multiplying all the rows by 3, we have


∣−1−2−11−1−221−1∣\begin{vmatrix} -1 & -2 & -1 \\ 1 & -1 & -2 \\ 2 & 1 & -1 \end{vmatrix}

Might also explain why the online calculator didn't give anything useful.

Sorry I didn't spot it earlier - I'd just assumed you'd done the initial part correctly.
(edited 11 years ago)
Reply 45
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Original post by ghostwalker
Your initial matrix appears to be incorrect:

A=∣23−23−131323−23231323∣A=\begin{vmatrix} \frac{2}{3} & \frac{-2}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{-2}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \end{vmatrix}

and

A−I=∣−13−23−1313−13−232313−13∣A-I=\begin{vmatrix} \frac{-1}{3} & \frac{-2}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{-1}{3} & \frac{-2}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{-1}{3} \end{vmatrix}

and after multiplying all the rows by 3, we have


∣−1−2−11−1−221−1∣\begin{vmatrix} -1 & -2 & -1 \\ 1 & -1 & -2 \\ 2 & 1 & -1 \end{vmatrix}

Might also explain why the online calculator didn't give anything useful.

Sorry I didn't spot it earlier - I'd just assumed you'd done the initial part correctly.


Okay so I subbed that into an online tool and it gives: http://matrix.reshish.com/gaussSolution.php

Which Is what I also got... does that seem right?
Original post by Music99
Okay so I subbed that into an online tool and it gives: http://matrix.reshish.com/gaussSolution.php

Which Is what I also got... does that seem right?


That link redirected to the home page, so don't know.
Reply 47
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Original post by ghostwalker
That link redirected to the home page, so don't know.


It comes up with the end matrix looking like

1 0 -1
0 1 1
0 0 0
Original post by Music99
It comes up with the end matrix looking like

1 0 -1
0 1 1
0 0 0


Looks reasonable.
Reply 49
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Original post by ghostwalker
Looks reasonable.


But then from that I don't know how to get the axis of rotation :s-smilie:.
Original post by Music99
But then from that I don't know how to get the axis of rotation :s-smilie:.


You presumably know how to convert linear equations into matrix form. All you need do here is the reverse of that process, and simplify your answer.

I'm finding it hard to believe that you're not familiar with at least some of this.
Reply 51
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Original post by ghostwalker
You presumably know how to convert linear equations into matrix form. All you need do here is the reverse of that process, and simplify your answer.

I'm finding it hard to believe that you're not familiar with at least some of this.


I'm familiar with bits of it but applying it is confusing. So then we have

x-z=0
y+z=0

Which if is correct still gives x+y=0 so x=-y and -y=z so z=0
(edited 11 years ago)
Original post by Music99
I'm familiar with bits of it but applying it is confusing. So then we have

x-z=0
y+z=0

?


So, putting the equations together, we have x=-y=z, and there's your line.
Reply 53
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Original post by ghostwalker
So, putting the equations together, we have x=-y=z, and there's your line.


Ohhh!! Right. THANK YOU SO MUCH!!!! Sorry for being so thick!
Original post by Music99
Ohhh!! Right. THANK YOU SO MUCH!!!!


You're welcome.


Sorry for being so thick!


I think it's the fact that you don't seem to have the understanding of what you're doing and consequently I'm inputting rather more than I comfortable with, and it makes me wonder how useful it is for you really.

Just to finish off this bit, to put your equation for the axis into vector form.

x=-y=z

Let x = some parameter, t say.

then x=t, y = -t, and z=t, from the equation of the line.

So, any point on the line is of the form (t,-t,t) and factoring out the t we have:

The equaiton of the line is r= t(1,-1,1)
Reply 55
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Original post by ghostwalker
Your initial matrix appears to be incorrect:

A=∣23−23−131323−23231323∣A=\begin{vmatrix} \frac{2}{3} & \frac{-2}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{-2}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \end{vmatrix}

and

A−I=∣−13−23−1313−13−232313−13∣A-I=\begin{vmatrix} \frac{-1}{3} & \frac{-2}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{-1}{3} & \frac{-2}{3} \\ \frac{2}{3} & \frac{1}{3} & \frac{-1}{3} \end{vmatrix}

and after multiplying all the rows by 3, we have


∣−1−2−11−1−221−1∣\begin{vmatrix} -1 & -2 & -1 \\ 1 & -1 & -2 \\ 2 & 1 & -1 \end{vmatrix}

Might also explain why the online calculator didn't give anything useful.

Sorry I didn't spot it earlier - I'd just assumed you'd done the initial part correctly.


Hi sorry to drag this thread up again. I was just wondering why was it that we had to multiple each row by 3? Also when I did the Gauss Jordan I thought you were meant to end up with the Identity matrix, but in this case we didn't and the answer was still correct.

I have made another thread about the plane of reflection I would really appreciate it if you could take a look. :colondollar:
Original post by Music99
Hi sorry to drag this thread up again. I was just wondering why was it that we had to multiple each row by 3?


We don't have to, but it's easier to work with.


Also when I did the Gauss Jordan I thought you were meant to end up with the Identity matrix, but in this case we didn't and the answer was still correct.


This is only possible if the matrix is non-singular (i.e. a unique solution), which is not the case here.


I have made another thread about the plane of reflection I would really appreciate it if you could take a look. :colondollar:


Will do.

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