Augmented matrix

Posted: Mon, 18 Sep 2006 21:37:07 GMT Post subject: rref augmented matrix (already reduced)


i am given the already reduced matrix of : 1 2 0 3
0 0 1 5
0 0 0 0

i know that the linear system is x1= -3 and x2= 5 and x3= 0 but i need to find how to write the infinite solution of the system using variable t.

All I can come up with is: 1x sub 1 - 2x subs = -3
1x sub 3 = 5
x sub 3 = x sub 3

It doesnt make sense to me when I think I need to let X sub 3 = t

How can i get an infinite solution using the variable t for this system?

[size=3]Hello, dekorn24!

Quote:

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i am given the already reduced matrix:
. . 1 2 0 | 3
. . 0 0 1 | 5
. . 0 0 0 | 0

I need to find how to write the infinite solution of the system using variable t.

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We have: .x + y .= .3
. . . . . . . . . . .z .= .5

Get everthing "in terms of y"
. . . . . . . . . x .= .3 - y
. . . . . . . . . y .= .y
. . . . . . . . . z .= .5

On the right, replace y with t.

And we have: .x .= .3 - t
. . . . . . . . . . .y .= .t
. . . . . . . . . . .z .= .5