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August 29th, 2016, 03:46 PM   #1
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echelon method expressing with infinite solutions

Hi,

Can someone explain to me how you express the answer for using the echelon method when there are an infinite number of solutions?

For example, if you have a matrix of:

1 2 l 3
0 0 l 0

How do I express the answer? For my homework, it has to be ( , y) but I have no idea what goes in the first part and there is not a single example in my book of how to express it.

Last edited by blimper; August 29th, 2016 at 03:54 PM.
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August 29th, 2016, 04:06 PM   #2
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$(1,2)\cdot (x,y) = 3$

$x + 2y = 3$

$y = \dfrac{3-x}{2}$

so all vectors

$c\begin{pmatrix}x \\ \dfrac {3-x}{2}\end{pmatrix},~c \in \mathbb{R}$

are solutions
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August 29th, 2016, 04:31 PM   #3
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I guess what i was thinking is if I had something like:

1 -5/6 l -1/6
0 0 l 0

It would become:

x-5/6y= -1/6

leading to x = 5y-1/6

Where my answer would be
5y-1
( -----, y )
6

(Is there a tutorial for how to express things on here like you did?)
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August 29th, 2016, 04:48 PM   #4
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Quote:
Originally Posted by blimper View Post
I guess what i was thinking is if I had something like:

1 -5/6 l -1/6
0 0 l 0

It would become:

x-5/6y= -1/6

leading to x = 5y-1/6

Where my answer would be
5y-1
( -----, y )
6

(Is there a tutorial for how to express things on here like you did?)
that's equally valid though generally you want to use the first vector element as the independent variable. I would do this example as

$x - \dfrac 5 6 y = -\dfrac 1 6$

$6x -5y = 1$

$y = \dfrac {6x-1}{5}$

so all vectors

$c \begin{pmatrix} x \\ \frac{6x-1}{5}\end{pmatrix},~c \in \mathbb{R}$
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August 30th, 2016, 06:53 PM   #5
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1 2 1 3
0 0 1 0
is the matrix of the system of equations
x+2y+z=3
0x+0y+z=0
in augmented form.
In rref, by elementary row operations which don't change the solution, the matrix is
1 2 0 3
0 0 1 0
from which
z=0,
x+2y=3
y is arbitrary and x=3-2y
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